Free math worksheets

- Compare each equation with \(y = mx + b\). - Pay attention to the sign of each coefficient. - What is the slope of a constant function? - If there is no constant term, what is the y-intercept? - You may need to reorder terms before identifying \(m\) and \(b\).

1. For \(f(x) = -4x + 7\), \(m = -4\) and \(b = 7\), so the y-intercept is \((0, 7)\). 2. For \(g(x) = \frac{2}{5}x\), \(m = \frac{2}{5}\) and \(b = 0\), so the y-intercept is \((0, 0)\). 3. Rewrite \(h(x) = 12 - 3x\) as \(h(x) = -3x + 12\). Then \(m = -3\) and \(b = 12\), so the y-intercept is \((0, 12)\). 4. For \(k(x) = -5\), the slope is \(0\) and \(b = -5\), so the y-intercept is \((0, -5)\).

a) \(m = -4\), \(b = 7\), \((0, 7)\) b) \(m = \frac{2}{5}\), \(b = 0\), \((0, 0)\) c) \(m = -3\), \(b = 12\), \((0, 12)\) d) \(m = 0\), \(b = -5\), \((0, -5)\)

- In a linear rule, locate the output at input \(0\) and the amount the output changes when the input increases by \(1\). - Keep the sign of the rate tied to whether the candle is growing or shrinking. - The physical domain ends when the modeled height reaches \(0\).

1. The initial value is \(8\,\text{in.}\), so the candle is \(8\,\text{in.}\) tall at \(t=0\). The rate of change is \(-1\,\frac{\text{in.}}{\text{h}}\), meaning the height decreases by \(1\,\text{in.}\) each hour. 2. \(h(3)=8-3=5\), and \(h(5.5)=8-5.5=2.5\). 3. Solve \(3=8-t\). This gives \(t=5\), so the candle is \(3\,\text{in.}\) tall after \(5\) hours. 4. \(h(10)=8-10=-2\). A negative candle height is not physically meaningful. The model is reasonable from lighting until the candle reaches height \(0\), so \(0\le t\le8\).

a) Initial value: \(8\,\text{in.}\); rate of change: \(-1\,\frac{\text{in.}}{\text{h}}\), meaning the candle loses \(1\,\text{in.}\) of height each hour. b) \(h(3)=5\,\text{in.}\) and \(h(5.5)=2.5\,\text{in.}\) c) \(5\) hours d) \(h(10)=-2\), which is not physically meaningful; a reasonable domain is \(0\le t\le8\).

- What must be true about the slopes of parallel lines? - What are the coordinates of the origin? - Moving a y-intercept upward changes which parameter in \(y = mx + b\)?

1. From \(g(x) = 1.5x - 3\), the slope is \(m = 1.5\) and the y-intercept is \((0, -3)\). 2. A parallel line has the same slope. Since \(h\) passes through the origin, its y-intercept is \(0\). Thus \(h(x) = 1.5x\). 3. Raising the y-intercept by \(5\) changes it from \(-3\) to \(2\). Thus \(k(x) = 1.5x + 2\).

a) \(m = 1.5\); y-intercept \((0, -3)\) b) \(h(x) = 1.5x\) c) \(k(x) = 1.5x + 2\)

- Which parameter determines where the line crosses the y-axis? - What is the slope of a horizontal line? - What y-intercept must a line through the origin have?

1. Shifting the graph up \(5\) units changes the y-intercept from \(-3\) to \(2\) while the slope stays \(2\). The new equation is \(y = 2x + 2\). 2. A line parallel to the x-axis has slope \(0\). Keeping the y-intercept \(-3\) gives \(y = -3\). 3. A line through the origin has y-intercept \(0\). Keeping the slope \(2\) gives \(y = 2x\).

a) \(b\) increases by \(5\); \(y = 2x + 2\) b) \(m = 0\); \(y = -3\) c) \(b = 0\); \(y = 2x\)

- What do equal slopes tell you about two lines? - Which term in \(y = mx + b\) gives the y-intercept? - Converting equivalent fractions and decimals can make the comparisons easier.

1. The slopes are \(\frac{2}{5} = 0.4\), \(0.4\), \(2.5\), \(\frac{5}{2} = 2.5\), and \(-0.4\). 2. Lines with equal slopes are parallel, so \(f_1\) is parallel to \(f_2\), and \(f_3\) is parallel to \(f_4\). 3. Functions \(f_1\), \(f_3\), and \(f_5\) all have y-intercept \(3\), so their graphs pass through \((0, 3)\).

a) \(f_1 \parallel f_2\) and \(f_3 \parallel f_4\) b) \(f_1\), \(f_3\), and \(f_5\); common y-intercept \((0, 3)\)

- What does the coefficient of \(x\) tell you about the slope? - How can you compare a horizontal change and a vertical change to the slope? - Does the measurement unit matter when both axes use the same scale? - What happens to the signs of \(\Delta x\) and \(\Delta y\) when you move left and down?

1. The slope is \(m = 1.2\), so a correct move must satisfy \(\frac{\Delta y}{\Delta x} = 1.2\). 2. For move 1, \(\frac{6}{5} = 1.2\), so it is correct. 3. For move 2, \(\frac{12}{10} = 1.2\), so it is correct. 4. For move 3, \(\frac{2.4}{2} = 1.2\), so it is correct. 5. For move 4, \(\frac{5}{6} \approx 0.83\), so it is not correct. 6. For move 5, \(\frac{-6}{-5} = 1.2\), so it is correct.

1. Correct 2. Correct 3. Correct 4. Incorrect 5. Correct

- What feature of a situation shows a constant rate of change? - What does the value at time \(0\) represent? - Recall the difference between a general linear function and a proportional function.

1. The volume increases by the same amount, \(8\) gallons, each hour, so the rate of change is constant. Therefore, the relationship is linear. 2. The initial value is \(20\), and the rate is \(8\), so \(V(t)=8t+20\). 3. If the barrel started empty, the function would be \(V(t)=8t\). It would still be linear and would also be proportional because it passes through the origin.

a) The relationship is linear because the volume increases at a constant rate. b) \(V(t)=8t+20\). c) \(V(t)=8t\); it is still linear and is also proportional.

- Which number in the equation gives the slope? - Use \(m = \frac{\Delta y}{\Delta x}\) for each proposed move. - Remember that moving left makes \(\Delta x\) negative and moving down makes \(\Delta y\) negative.

1. The slope is \(m = -\frac{3}{4} = -0.75\). 2. Method 1 gives \(\frac{-3}{4} = -0.75\), so it is correct. 3. Method 2 gives \(\frac{-0.75}{1} = -0.75\), so it is mathematically correct, although a fractional grid square may be harder to plot exactly. 4. Method 3 gives \(\frac{6}{-8} = -0.75\), so it is correct.

All three methods are mathematically correct. Methods 1 and 3 are easiest to plot accurately on standard graph paper because they use whole grid squares.

- What must be true about the slopes between pairs of points if all three points lie on one line? - Find the slope between two different pairs of points. - Use change in \(y\) divided by change in \(x\).

1. The slope from \(P\) to \(Q\) is \(\frac{3 - 5}{2 - (-2)} = -\frac{1}{2}\). 2. The slope from \(Q\) to \(R\) is \(\frac{1 - 3}{5 - 2} = -\frac{2}{3}\). 3. Since \(-\frac{1}{2} \neq -\frac{2}{3}\), the slope is not constant, so the three points are not collinear.

No. The slopes are \(-\frac{1}{2}\) and \(-\frac{2}{3}\), so the three points do not lie on the same line.

- Compare the slopes of the four boundary lines. - Equal slopes tell you that two distinct lines are parallel, and horizontal lines have slope \(0\). - Identify the quadrilateral property determined by two pairs of parallel opposite sides.

1. Lines \(g_1\) and \(g_2\) both have slope \(0.5\), so they are parallel. 2. Lines \(g_3\) and \(g_4\) are horizontal, so both have slope \(0\) and are parallel. 3. The enclosed quadrilateral therefore has two pairs of opposite sides that are parallel, so it is a parallelogram.

A parallelogram. The lines form two pairs of parallel sides: \(g_1\parallel g_2\) and \(g_3\parallel g_4\).

- What must be true about the graph of a proportional relationship? - Consider the role of \(b\) in \(y=mx+b\). - What output does a proportional function have when \(x=0\)?

1. A linear function has the form \(y=mx+b\). A proportional function has the form \(y=kx\), which is the special case \(b=0\). 2. Therefore, the statement is false. Only linear functions whose graphs pass through the origin are proportional. 3. For example, \(y=2x+5\) is linear but not proportional because its y-intercept is \(5\).

The statement is false. A linear function \(y=mx+b\) is proportional only when \(b=0\). For example, \(y=2x+5\) is linear but not proportional.

- What coordinate is zero when a graph crosses each coordinate axis? - How do the horizontal and vertical steps in a slope triangle relate to the coefficient of \(x\)? - Which part of a slope-intercept equation determines where the line crosses the y-axis? - If the entire line shifts up or down, which parameter changes?

1. At the y-intercept, \(x = 0\), so the point is \((0, -3)\). At the x-intercept, \(0 = 1.5x - 3\), so \(x = 2\) and the point is \((2, 0)\). 2. A move of 2 units right and 3 units up gives slope \(\frac{3}{2} = 1.5\), so the student's statement is correct. 3. Changing the constant term from \(-3\) to \(1\) increases the y-intercept by \(4\) while leaving the slope unchanged. The entire line shifts up 4 units.

a) x-intercept: \((2, 0)\); y-intercept: \((0, -3)\) b) The statement is correct because \(\frac{3}{2} = 1.5\). c) The line shifts up 4 units.

- Isolate \(y\) first so the coefficient and constant term have their standard linear-function meanings. - Connect a change in input to a change in output through the rate \(m\). - Test the point only after you have identified the function rule.

1. Add \(x\) to both sides: \(\frac{1}{2}y=x+1\). Multiply by \(2\): \(y=2x+2\). 2. The rate of change is \(m=2\), and the initial value is \(b=2\). 3. A change of \(3\) in \(x\) changes \(y\) by \(m\cdot3=2\cdot3=6\). 4. At \(x=5\), the function gives \(y=2\cdot5+2=12\), so \(P(5,12)\) lies on the graph.

a) \(y=2x+2\) b) Rate of change: \(2\); initial value: \(2\) c) \(y\) increases by \(6\). d) Yes.

- What is the x-coordinate of every point on the y-axis? - What is the y-coordinate of every point on the x-axis? - Substitute the appropriate zero and solve.

1. For the y-intercept, set \(x = 0\). Then \(y = 4\), so the y-intercept is \((0, 4)\). 2. For the x-intercept, set \(y = 0\): \(0 = \frac{2}{5}x + 4\). Then \(-4 = \frac{2}{5}x\), so \(x = -10\). The x-intercept is \((-10, 0)\).

x-intercept: \((-10, 0)\); y-intercept: \((0, 4)\)

- Look at each line from left to right. Does it rise, fall, stay horizontal, or stay vertical? - What does a horizontal line tell you about the change in \(y\)? - What happens to the slope formula for a vertical line? - To find an exact slope, choose two grid points on the line and compare rise to run.

1. In (a), the line rises from left to right, so the slope is positive. Using \((0, -1)\) and \((1, 1)\), \(m = \frac{1 - (-1)}{1 - 0} = 2\). 2. In (b), the line falls from left to right, so the slope is negative. Using \((0, 2)\) and \((2, 1)\), \(m = \frac{1 - 2}{2 - 0} = -\frac{1}{2}\). 3. In (c), the line is horizontal, so its rise is \(0\) for any nonzero run. Therefore, \(m = 0\). 4. In (d), the line is vertical, so the run is \(0\). Because division by zero is undefined, the slope is undefined.

a) Positive; \(m = 2\) b) Negative; \(m = -\frac{1}{2}\) c) Zero; \(m = 0\) d) Undefined

- Where does each line cross the y-axis? - Use two grid points on each line to calculate rise over run. - What must be true about the graph and equation of a proportional relationship? - What do parallel nonvertical lines have in common?

1. For each line, read the y-intercept and use two grid points to find the slope. For \(f\), the y-intercept is \(1\). Using \((0, 1)\) and \((2, 4)\), \(m = \frac{4 - 1}{2 - 0} = \frac{3}{2}\), so \(f(x) = \frac{3}{2}x + 1\). For \(g\), the y-intercept is \(-2\). Using \((0, -2)\) and \((2, 1)\), \(m = \frac{1 - (-2)}{2 - 0} = \frac{3}{2}\), so \(g(x) = \frac{3}{2}x - 2\). For \(h\), the y-intercept is \(0\). Using \((0, 0)\) and \((2, 3)\), \(m = \frac{3 - 0}{2 - 0} = \frac{3}{2}\), so \(h(x) = \frac{3}{2}x\). 2. The function \(h\) is proportional because its graph passes through the origin and its equation has the form \(y = mx\). 3. All three lines are parallel because they have the same slope, \(\frac{3}{2}\), and different y-intercepts.

a) \(f(x) = \frac{3}{2}x + 1\), \(g(x) = \frac{3}{2}x - 2\), \(h(x) = \frac{3}{2}x\) b) \(h\). Its graph passes through \((0, 0)\), and its equation has no nonzero constant term. c) The three lines are parallel because they all have slope \(\frac{3}{2}\).

- A horizontal segment means the group's distance from the lodge is unchanged. - Speed is distance divided by time. - The return begins when the graph starts decreasing. - Use the distance at the start of the return segment to determine the return speed.

1. Rest stop A is \(4\) miles from the lodge. The horizontal segment from \(t=2\) to \(t=3\) shows a \(1\)-hour rest. 2. On the first segment, the speed is \(\frac{4}{2}=2\,\text{mph}\). 3. The return trip begins at \(t=5\) hours. The graph returns to distance \(0\) at \(t=6\), so the outing lasts \(6\) hours. 4. The group returns \(6\) miles in \(1\) hour, so its average return speed is \(6\,\text{mph}\).

a) \(4\) miles; \(1\) hour b) \(2\,\text{mph}\) c) The return begins after \(5\) hours; the outing lasts \(6\) hours. d) \(6\,\text{mph}\)

- First identify where each graph crosses the y-axis. - Use a slope triangle to find the rise and run for each graph. - Check whether each line rises or falls from left to right. - Compare the slope and y-intercept you found with the given equations.

1. Graph a has y-intercept \(1\). It rises \(2\) units for each \(1\) unit to the right, so its slope is \(2\). Therefore, a matches \(y = 2x + 1\), equation 3. 2. Graph b has y-intercept \(3\). It falls \(1\) unit for every \(2\) units to the right, so its slope is \(-\frac{1}{2}\). Therefore, b matches \(y = -\frac{1}{2}x + 3\), equation 4. 3. Graph c has y-intercept \(-1\). It falls \(2\) units for each \(1\) unit to the right, so its slope is \(-2\). Therefore, c matches \(y = -2x - 1\), equation 1. 4. Graph d has y-intercept \(-2\). It rises \(1\) unit for every \(2\) units to the right, so its slope is \(\frac{1}{2}\). Therefore, d matches \(y = \frac{1}{2}x - 2\), equation 2.

a: 3) \(y = 2x + 1\); b: 4) \(y = -\frac{1}{2}x + 3\); c: 1) \(y = -2x - 1\); d: 2) \(y = \frac{1}{2}x - 2\)

- Where does each line cross the y-axis? - Choose two grid points on each line to form a slope triangle. - Find slope as vertical change divided by horizontal change, including the correct sign. - Substitute the slope and y-intercept into \(y = mx + b\).

1. The graph of \(f\) crosses the y-axis at \((0, -1)\), so \(b_1 = -1\). Using \((0, -1)\) and \((2, 0)\), the slope is \(m_1 = \frac{0 - (-1)}{2 - 0} = \frac{1}{2}\). Therefore, \(f(x) = \frac{1}{2}x - 1\). 2. The graph of \(g\) crosses the y-axis at \((0, 3)\), so \(b_2 = 3\). Using \((0, 3)\) and \((2, 0)\), the slope is \(m_2 = \frac{0 - 3}{2 - 0} = -\frac{3}{2}\). Therefore, \(g(x) = -\frac{3}{2}x + 3\).

a) \(b_1 = -1\), \(m_1 = \frac{1}{2}\), and \(f(x) = \frac{1}{2}x - 1\) b) \(b_2 = 3\), \(m_2 = -\frac{3}{2}\), and \(g(x) = -\frac{3}{2}x + 3\)

- Read the coordinates of the two marked points on each line. - Use change in y divided by change in x to calculate the slope. - Where does each line cross the y-axis? - Substitute the slope and y-intercept into \(y = mx + b\).

1. For \(f\), the marked points are \((0, -1)\) and \((3, 1)\). The slope is \(m = \frac{1 - (-1)}{3 - 0} = \frac{2}{3}\). Since \((0, -1)\) is on the y-axis, \(b = -1\). Therefore, \(f(x) = \frac{2}{3}x - 1\). 2. For \(g\), the marked points are \((-1, 3)\) and \((1, -1)\). The slope is \(m = \frac{-1 - 3}{1 - (-1)} = -2\). Using \((1, -1)\), \(-1 = -2 \cdot 1 + b\), so \(b = 1\). Therefore, \(g(x) = -2x + 1\).

\(f\): \(m = \frac{2}{3}\), \(b = -1\), \(f(x) = \frac{2}{3}x - 1\) \(g\): \(m = -2\), \(b = 1\), \(g(x) = -2x + 1\)

- First read the y-intercept of each line. - Choose two grid points on each line to form a slope triangle. - Compute slope as \(\frac{\Delta y}{\Delta x}\). - A line that falls from left to right has a negative slope.

1. Line \(f\) crosses the y-axis at \((0, 1)\). Using \((0, 1)\) and \((2, 4)\), the slope is \(\frac{4 - 1}{2 - 0} = \frac{3}{2}\). Therefore, \(f(x) = \frac{3}{2}x + 1\). 2. Line \(g\) crosses the y-axis at \((0, -2)\). Using \((0, -2)\) and \((2, -3)\), the slope is \(\frac{-3 - (-2)}{2 - 0} = -\frac{1}{2}\). Therefore, \(g(x) = -\frac{1}{2}x - 2\). 3. Line \(h\) crosses the y-axis at \((0, 3)\) and the x-axis at \((3, 0)\). Its slope is \(\frac{0 - 3}{3 - 0} = -1\). Therefore, \(h(x) = -x + 3\).

\(f(x) = \frac{3}{2}x + 1\), \(g(x) = -\frac{1}{2}x - 2\), and \(h(x) = -x + 3\).

- Move vertically from each time value to the graph, then horizontally to the distance axis. - A horizontal segment means the total distance is not changing. - Compare segment steepness to compare speeds.

1. The graph shows \(4\) miles after \(2\) hours and \(7\) miles after \(5\) hours. 2. Reading the graph at each whole hour gives: <table> <tr><th>Time (h)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><th>Distance (mi)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(6\)</td><td>\(7\)</td></tr> </table> 3. During the first \(3\) hours, the hiker travels at a constant rate of \(2\,\text{mph}\). From hour \(3\) to hour \(4\), the total distance does not change, so the hiker likely rests. During the final hour, the hiker travels at \(1\,\text{mph}\), which is slower than before.

a) After \(2\) hours: \(4\) miles; after \(5\) hours: \(7\) miles b) <table> <tr><th>Time (h)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><th>Distance (mi)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(6\)</td><td>\(7\)</td></tr> </table> c) The hiker travels steadily for \(3\) hours, likely rests for \(1\) hour, and then continues at a slower rate.

- How can two points be used to calculate a slope? - Once you know the slope, substitute one point into \(y = mx + b\). - Pay attention to the scale when reading the point coordinates. - Check that your equation gives the correct y-value for both points.

1. The points are \(P_1(1, 5)\) and \(P_2(4, -1)\). 2. The slope is \(m = \frac{-1 - 5}{4 - 1} = -2\). 3. Substitute \(P_1(1, 5)\) into \(y = -2x + b\): \(5 = -2 \cdot 1 + b\), so \(b = 7\). 4. Therefore, \(g(x) = -2x + 7\).

a) \(m = -2\) b) \(b = 7\) c) \(g(x) = -2x + 7\)

- Look at where each line crosses the y-axis. - Use grid points to compare vertical change with horizontal change. - Which part of \(y = mx + b\) determines the y-intercept?

1. For \(g_1\), the points \((0, 2)\) and \((2, 4)\) give slope \(1\), so \(g_1(x) = x + 2\). 2. For \(g_2\), the points \((0, 2)\) and \((1, 0)\) give slope \(-2\), so \(g_2(x) = -2x + 2\). 3. For \(g_3\), the points \((0, 2)\) and \((4, 4)\) give slope \(\frac{1}{2}\), so \(g_3(x) = \frac{1}{2}x + 2\). 4. For \(g_4\), the points \((0, 2)\) and \((4, 0)\) give slope \(-\frac{1}{2}\), so \(g_4(x) = -\frac{1}{2}x + 2\). 5. All four equations have the same y-intercept, \(b = 2\), so all four lines pass through \((0, 2)\).

\(g_1(x) = x + 2\), \(g_2(x) = -2x + 2\), \(g_3(x) = \frac{1}{2}x + 2\), and \(g_4(x) = -\frac{1}{2}x + 2\). All four lines have y-intercept \(2\).

- A linear equation can be written as \(y = mx + b\). - The value \(b\) tells where the line crosses the y-axis. - Use the grid to determine each line's rise and run. - What must be the same for two nonvertical lines to be parallel?

1. For \(f\), the y-intercept is \(3\). The line falls \(1\) unit for every \(2\) units to the right, so \(m = -\frac{1}{2}\). Thus \(f(x) = -\frac{1}{2}x + 3\). 2. For \(g\), the y-intercept is \(1\) and the slope is also \(-\frac{1}{2}\). Thus \(g(x) = -\frac{1}{2}x + 1\). 3. For \(h\), the y-intercept is \(-2\) and the slope is again \(-\frac{1}{2}\). Thus \(h(x) = -\frac{1}{2}x - 2\). 4. Nonvertical lines are parallel when they have the same slope. All three equations have slope \(-\frac{1}{2}\).

\(f(x) = -\frac{1}{2}x + 3\), \(g(x) = -\frac{1}{2}x + 1\), and \(h(x) = -\frac{1}{2}x - 2\). The lines are parallel because all three have slope \(-\frac{1}{2}\).

- The initial height is the graph's y-intercept. - Use two points to find the rate of change. - The candle is burned down when its height is \(0\).

1. At \(t=0\), the graph has a height of \(9\) inches, so the candle's initial height is \(9\,\text{in.}\). 2. Using \((0, 9)\) and \((4, 3)\), the slope is \(m=\frac{3-9}{4-0}=-1.5\). The height decreases by \(1.5\,\text{in./h}\), so the burn rate is \(1.5\,\text{in./h}\). 3. The candle is completely burned down when \(h=0\). Solve \(0=-1.5t+9\) to get \(t=6\).

a) \(9\,\text{in.}\) b) \(1.5\,\text{in./h}\) c) \(6\) hours

- A horizontal line has slope \(0\) and an equation of the form \(y = c\). - For the other lines, notice whether each line rises or falls from left to right.

1. Line \(f\) is horizontal at \(y = 2\), so its slope is \(0\) and its equation is \(f(x) = 2\). 2. Line \(g\) has y-intercept \(1\). Moving \(1\) unit right corresponds to moving \(1\) unit down, so the slope is \(-1\). Thus \(g(x) = -x + 1\). 3. Line \(h\) has y-intercept \(-3\). Moving \(1\) unit right corresponds to moving \(1\) unit up, so the slope is \(1\). Thus \(h(x) = x - 3\).

\(f(x) = 2\), \(g(x) = -x + 1\), and \(h(x) = x - 3\).

- Look for where the charge values stop decreasing and begin increasing. - For the minimum, compare the listed charge levels rather than the time values. - Average rate of change is the change in output divided by the corresponding change in input.

1. The charge level increases from hour \(4\) through hour \(8\), so the battery was charging during that interval. 2. The smallest table value is \(10\%\), which occurs at \(t=4\,\text{h}\). 3. From hour \(4\) to hour \(7\), the charge level increases from \(10\%\) to \(90\%\). The average rate of change is \(\frac{90-10}{7-4}=\frac{80}{3}\approx26.7\) percentage points per hour.

a) From hour \(4\) through hour \(8\) b) At \(t=4\,\text{h}\), when the charge level was \(10\%\) c) About \(26.7\) percentage points per hour

- The y-intercept is the value of \(y\) when \(x = 0\). Where does each graph cross the y-axis? - A negative slope means a line falls from left to right. Compare the steepness of the two decreasing lines. - A linear function with no constant term passes through \((0, 0)\).

1. For \(a(x) = -\frac{2}{3}x + 1\), the y-intercept is \(1\) and the slope is negative with magnitude less than \(1\). Graph p has those features, so p represents \(a\). 2. For \(b(x) = \frac{3}{2}x - 2\), the y-intercept is \(-2\) and the slope is positive. Graph q has those features, so q represents \(b\). 3. For \(c(x) = -2x\), the y-intercept is \(0\), so the line passes through the origin, and its negative slope is steeper than the slope of \(a\). Graph r has those features, so r represents \(c\).

p: \(a(x) = -\frac{2}{3}x + 1\); q: \(b(x) = \frac{3}{2}x - 2\); r: \(c(x) = -2x\)

- Find the point where the line crosses the y-axis. - Choose two grid points on the line. - Compare the vertical change with the horizontal change between those points.

1. The line crosses the y-axis at \((0, 1)\), so \(b = 1\). 2. Using \((0, 1)\) and \((2, 2)\), the slope is \(m = \frac{2 - 1}{2 - 0} = \frac{1}{2}\). 3. Therefore, the equation is \(g(x) = \frac{1}{2}x + 1\).

a) \(b = 1\) b) \(m = \frac{1}{2}\) c) \(g(x) = \frac{1}{2}x + 1\)

- Start by finding where each line crosses the y-axis. - Choose grid points with easy-to-read coordinates for the slope triangle. - Compare the vertical change with the horizontal change. - Use the direction of the line to determine the sign of the slope.

1. Line \(f\) crosses the y-axis at \((0, 4)\). Using \((0, 4)\) and \((1, 1)\), the slope is \(m = \frac{1 - 4}{1 - 0} = -3\). Therefore, \(f(x) = -3x + 4\). 2. Line \(h\) crosses the y-axis at \((0, -3)\). Using \((0, -3)\) and \((5, -1)\), the slope is \(m = \frac{-1 - (-3)}{5 - 0} = \frac{2}{5}\). Therefore, \(h(x) = \frac{2}{5}x - 3\).

\(f(x) = -3x + 4\) and \(h(x) = \frac{2}{5}x - 3\).

- Identify the y-intercept in each equation and compare it with the graph. - Check whether each line rises or falls from left to right. - Compare the steepness of each graph with the coefficient of \(x\).

1. Graph \(g_1\) crosses the y-axis at \(3\) and falls \(2\) units for every \(1\) unit to the right, so its slope is \(-2\). Thus \(g_1\) matches equation 1, \(y = -2x + 3\). 2. Graph \(g_2\) crosses the y-axis at \(1\) and rises \(1\) unit for every \(1\) unit to the right, so its slope is \(1\). Thus \(g_2\) matches equation 3, \(y = x + 1\). 3. Graph \(g_3\) crosses the y-axis at \(-2\) and rises \(1\) unit for every \(3\) units to the right, so its slope is \(\frac{1}{3}\). Thus \(g_3\) matches equation 2, \(y = \frac{1}{3}x - 2\).

\(g_1\): 1) \(y = -2x + 3\); \(g_2\): 3) \(y = x + 1\); \(g_3\): 2) \(y = \frac{1}{3}x - 2\)

- Find a point where each line crosses the y-axis. - What is special about the equation of a line through the origin? - What is the slope of a horizontal line? - Use two points to calculate slope as change in y divided by change in x.

1. Line \(h_1\) passes through \((0, 0)\) and \((4, 1)\), so its slope is \(\frac{1}{4}\). Thus \(h_1(x) = \frac{1}{4}x\). 2. Line \(h_2\) has y-intercept \(4\) and passes through \((4, 2)\). Its slope is \(\frac{2 - 4}{4 - 0} = -\frac{1}{2}\). Thus \(h_2(x) = -\frac{1}{2}x + 4\). 3. Line \(h_3\) is horizontal at \(y = -3\), so \(h_3(x) = -3\). 4. Line \(h_4\) has y-intercept \(4\) and passes through \((-2, 0)\). Its slope is \(\frac{4 - 0}{0 - (-2)} = 2\). Thus \(h_4(x) = 2x + 4\).

\(h_1(x) = \frac{1}{4}x\), \(h_2(x) = -\frac{1}{2}x + 4\), \(h_3(x) = -3\), and \(h_4(x) = 2x + 4\).

- First check the scale used on each axis. - Choose two points on the line that lie exactly on grid intersections. - The point where the line crosses the vertical axis gives the initial value. - Substitute the requested time into your function.

1. Two convenient points are \((0, 10)\) and \((4, 20)\). 2. The slope is \(m = \frac{20 - 10}{4 - 0} = 2.5\). The y-intercept is \(10\), so \(V(t) = 2.5t + 10\). 3. \(V(6) = 2.5 \cdot 6 + 10 = 25\). The graph also contains the point \((6, 25)\).

a) \(V(t) = 2.5t + 10\) b) \(25\,\text{gal}\)

- Isolate \(y\) so the coefficient of \(x\) and the constant term can be interpreted separately. - Translate the rate into a coordinated horizontal and vertical change rather than only naming the coefficient. - The initial-value point occurs when the input is \(0\); use that point as the starting location for part c).

1. From \(x+2y=4\), subtract \(x\) to get \(2y=-x+4\), then divide by \(2\): \(y=-\frac{1}{2}x+2\). 2. The rate of change is \(-\frac{1}{2}\). Thus, when \(x\) increases by \(2\), \(y\) decreases by \(1\). The initial value is \(2\), so the initial-value point is \((0,2)\). 3. Starting at \((0,2)\), increasing \(x\) by \(2\) and decreasing \(y\) by \(1\) gives \((2,1)\). Repeating the same change gives \((4,0)\).

a) \(y=-\frac{1}{2}x+2\) b) Rate of change: \(-\frac{1}{2}\); for every increase of \(2\) in \(x\), \(y\) decreases by \(1\). Initial value: \(2\), represented by \((0,2)\). c) \((2,1)\) and \((4,0)\)

- The intersection point is a convenient point to use when finding each slope. - Compare each line's y-intercept with the shared point \(S\).

1. Line \(f\) crosses the y-axis at \(-2\). From \((0, -2)\) to \(S(2, 1)\), the rise is \(3\) and the run is \(2\), so \(m = \frac{3}{2}\). Thus \(f(x) = \frac{3}{2}x - 2\). 2. Line \(g\) crosses the y-axis at \(4\). From \((0, 4)\) to \(S(2, 1)\), the rise is \(-3\) and the run is \(2\), so \(m = -\frac{3}{2}\). Thus \(g(x) = -\frac{3}{2}x + 4\).

Line \(f\): \(m = \frac{3}{2}\), \(b = -2\); line \(g\): \(m = -\frac{3}{2}\), \(b = 4\)

- What is the slope of a line parallel to the x-axis? - For line \(b\), compare how the y-value changes as the x-value changes.

1. Line \(a\) is horizontal, so its slope is \(0\). It crosses the y-axis at \(-3\), so \(a(x) = -3\). 2. Line \(b\) crosses the y-axis at \(2\). Using \((0, 2)\) and \((-1, 5)\), the slope is \(m = \frac{5 - 2}{-1 - 0} = -3\). Therefore, \(b(x) = -3x + 2\).

Line \(a\): \(a(x) = -3\); line \(b\): \(b(x) = -3x + 2\)

- Where does the line cross the vertical axis? That value represents the starting fee. - How much does the cost increase for each additional mile? - Use the form \(y = mx + b\).

1. The line crosses the y-axis at \(4\), so the starting fee is \(\$4\). 2. Using \((0, 4)\) and \((3, 10)\), the slope is \(m = \frac{10 - 4}{3 - 0} = 2\). This means the taxi charges \(\$2\) per mile. 3. Therefore, the equation is \(y = 2x + 4\).

\(y = 2x + 4\)

- Compare the y-intercept in the equation with where each graph crosses the y-axis. - Does a positive slope make a graph rise or fall from left to right? - Use a slope triangle: if you move one unit to the right, how far should the line move vertically? - You can also substitute an x-value into the equation and check whether the resulting point lies on a candidate graph.

1. The y-intercept is \(1\), so the correct graph must cross the y-axis at \(1\). This eliminates graph c). 2. The slope is positive, so the line must rise from left to right. This eliminates graph d). 3. A slope of \(0.5 = \frac{1}{2}\) means the line rises \(1\) unit for every \(2\) units to the right. Graph b) has this slope, while graph a) is steeper. 4. Therefore, graph b) represents the function.

b)

- First find where the graph crosses the y-axis. Which parameter is this in \(y = mx + b\)? - Use two points that lie exactly on grid intersections to make a slope triangle. - Check the sign of the slope. Does the line rise or fall from left to right? - Compare the slope you found with the slopes in the answer choices.

1. The line crosses the y-axis at \((0, 3)\), so the y-intercept is \(3\). This eliminates D. 2. The line falls from left to right, so its slope is negative. This eliminates B. 3. Use \((0, 3)\) and \((2, 0)\): \(m = \frac{0 - 3}{2 - 0} = -\frac{3}{2} = -1.5\). 4. The equation is \(g(x) = -1.5x + 3\), which is choice A.

A: \(g(x) = -1.5x + 3\)

- A line that rises from left to right has what sign of slope? - Look at where each line crosses the y-axis. Is that point above or below \(0\)?

1. The graph of \(f\) rises from left to right, so \(m_1 > 0\). It crosses the y-axis below the origin, so \(b_1 < 0\). 2. The graph of \(g\) falls from left to right, so \(m_2 < 0\). It crosses the y-axis above the origin, so \(b_2 > 0\).

For \(f\): \(m_1 > 0\) and \(b_1 < 0\); for \(g\): \(m_2 < 0\) and \(b_2 > 0\).

- First identify where the line crosses the y-axis. - Use two grid points to calculate rise over run. - Check whether the line rises or falls from left to right to determine the sign of the slope.

1. The line crosses the y-axis at \(2\), so its y-intercept is \(2\). All four choices have this y-intercept. 2. Use \((0, 2)\) and \((4, 0)\): \(m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}\). 3. The equation is \(y = -\frac{1}{2}x + 2\), which is choice D.

D) \(y = -\frac{1}{2}x + 2\)

- Compare the slopes implied by consecutive table values. - Use the linear change between the two measurements surrounding \(20\) minutes. - One hour is \(60\) minutes. - A less steep distance-time graph represents a lower speed.

1. Anna's points \((0, 0)\), \((15, 3)\), \((30, 6)\), \((45, 9)\), and \((60, 12)\) lie on one line. Ben's points \((0, 0)\), \((15, 4)\), \((30, 7)\), \((45, 9)\), and \((60, 10)\) form a piecewise-linear graph with segments that become less steep. 2. Anna travels \(3\) miles every \(15\) minutes. After \(20\) minutes, she has traveled \(3+\frac{5}{15}(3)=4\) miles. Between minutes \(15\) and \(30\), Ben travels \(3\) miles, so after \(20\) minutes he has traveled \(4+\frac{5}{15}(3)=5\) miles. Ben is ahead by \(1\) mile. 3. Anna travels \(12\) miles in \(1\) hour, so her average speed is \(12\,\text{mph}\). 4. Anna's graph is a line with constant slope. Ben's line segments become less steep, so his speed decreases. Fatigue, a headwind, or an uphill section could explain the change.

a) Anna's graph is a line; Ben's graph is a piecewise-linear graph whose segments become less steep. b) Ben is ahead by \(1\) mile. c) \(12\,\text{mph}\) d) Ben starts faster but slows down, while Anna maintains a constant speed. A possible cause is fatigue, wind, or an uphill section.

- Divide the change in battery level by the elapsed time. - Check whether the older model gains the same amount in every interval. - Find how many percentage points the older model still needs after \(60\) minutes.

1. From \(0\) to \(20\) minutes, the new model increases by \(45-5=40\) percentage points, so its average rate is \(40\div20=2\) percentage points per minute. From \(40\) to \(60\) minutes, it increases by \(90-72=18\) percentage points, so its average rate is \(18\div20=0.9\) percentage point per minute. 2. The older model gains a constant \(10\) percentage points every \(10\) minutes. The new model charges faster at first, but its charging rate decreases as its battery level rises. 3. The older model gains \(1\) percentage point per minute. At \(60\) minutes, it needs \(35\) more percentage points, so it needs \(35\) more minutes. It reaches \(100\%\) after \(95\) minutes. 4. Charging systems often reduce current near a full charge to limit heat and battery stress.

a) \(0\)–\(20\) min: \(2\) percentage points per minute; \(40\)–\(60\) min: \(0.9\) percentage point per minute b) The older model charges at a constant rate; the new model charges quickly at first and then more slowly. c) After \(95\) minutes d) The charging current may be reduced to limit heat and protect the battery.

- What does substituting \(t=0\) tell you? - Compare the units used in the equation with the unit in part c). - Which number in the equation represents the change in height for each additional minute?

1. At \(t=0\), the model gives \(h=3\cdot0+10=10\). The tank already had \(10\,\text{cm}\) of water, so the claim is false. 2. The coefficient of \(t\) is \(3\), so the water level rises \(3\,\text{cm}\) per minute. 3. Convert \(1.60\,\text{m}\) to \(160\,\text{cm}\). Solve \(160=3t+10\): \(150=3t\), so \(t=50\). 4. The water reaches the stated height after \(50\) minutes.

a) The claim is false. At \(t=0\), the water height is \(10\,\text{cm}\). b) \(3\,\text{cm}\) per minute c) \(50\) minutes

- Divide the total change by the number of years. - A year halfway between two table entries has the midpoint of their values under a linear model. - Compare the recent rate with the rate required by the prediction.

1. From 2000 to 2012, the total increase is \(42-32=10\,\text{kg}\) over \(12\) years. The average yearly increase is \(10\div12\approx0.83\,\text{kg}\) per year. 2. The year 2018 is halfway between 2016 and 2020, so linear interpolation gives \((44+45)\div2=44.5\,\text{kg}\). 3. From 2012 to 2020, the amount increased by \(45-42=3\,\text{kg}\) over \(8\) years, or \(3\div8=0.375\,\text{kg}\) per year. 4. To rise from \(45\,\text{kg}\) in 2020 to more than \(60\,\text{kg}\) in 2030 would require an average increase greater than \(15\div10=1.5\,\text{kg}\) per year. That is much faster than the recent rate, so the prediction does not appear realistic from these data.

a) \(\approx0.83\,\text{kg}\) per year b) \(44.5\,\text{kg}\) c) The prediction does not appear realistic. The recent rate was \(0.375\,\text{kg}\) per year, while exceeding \(60\,\text{kg}\) would require more than \(1.5\,\text{kg}\) per year.

- Substitute each input into the function. - For part c), set the output equal to \(0\) and solve. - Compare the outputs for inputs that differ by \(1\). - Relate the change in output to the coefficient of \(x\).

1. \(T(-1.5)=4\cdot(-1.5)-6=-6-6=-12\). 2. \(T\left(\frac{3}{4}\right)=4\cdot\frac{3}{4}-6=3-6=-3\). 3. Set \(4x-6=0\). Then \(4x=6\), so \(x=1.5\). 4. \(T(2)=4\cdot2-6=2\) and \(T(3)=4\cdot3-6=6\). The output increases by \(6-2=4\) when \(x\) increases by \(1\).

a) \(-12\) b) \(-3\) c) \(x=1.5\) d) \(T(2)=2\), \(T(3)=6\); the output increases by \(4\).

- Compare the vertical and horizontal changes between the two intercepts. - Use the slope and y-intercept in \(y = mx + b\). - What must be true about the slopes of parallel lines? - Set \(k(x) = 0\) to find its x-intercept.

1. From \((-2, 0)\) to \((0, 4)\), the rise is \(4\) and the run is \(2\), so \(m = \frac{4}{2} = 2\). 2. Line \(h\) has slope \(2\) and y-intercept \(4\), so \(h(x) = 2x + 4\). 3. A parallel line has the same slope, so \(k(x) = 2x - 2\). Its x-intercept satisfies \(0 = 2x - 2\), giving \(x = 1\). Therefore, the x-intercept is \((1, 0)\).

a) \(m = 2\) b) \(h(x) = 2x + 4\) c) \(k(x) = 2x - 2\), with x-intercept \((1, 0)\).

- Evaluate the function at \(t=0\). - The coefficient of \(t\) gives the change in volume per minute. - Find when the volume reaches the barrel's capacity. - Changing only the initial value shifts a line vertically.

1. The constant \(5\) is the initial volume, \(5\) gallons. The coefficient \(0.5\) is the filling rate, \(0.5\) gallon per minute. 2. \(V(12)=0.5(12)+5=11\), so the barrel contains \(11\) gallons. 3. The barrel is full when \(0.5t+5=25\). Solving gives \(t=40\), so a reasonable domain is \(0\le t\le40\). 4. The graph is a line segment from \((0, 5)\) to \((40, 25)\). If the barrel were initially empty, the graph would have the same slope, start at \((0, 0)\), and reach capacity at \((50, 25)\).

a) Initial volume: \(5\) gallons; filling rate: \(0.5\) gallon per minute b) \(11\) gallons c) \(0\le t\le40\) d) With an empty barrel, the line would have the same slope, start at \((0, 0)\), and extend to \((50, 25)\).

- Compare each linear model with the structure \(y=mx+b\), and keep track of the units attached to the coefficient of time. - Ask what the output is at \(t=0\) and what that output represents in the trip. - For a requested elevation, set the model's output equal to that elevation and solve for time. - For the second hiker, combine the starting elevation with the signed hourly change before comparing valley-arrival times.

1. In \(h(t)=5000-800t\), the rate of change is \(-800\,\text{ft/h}\). The negative sign means the hiker's elevation decreases by \(800\,\text{ft}\) each hour. The initial value is \(h(0)=5000\,\text{ft}\), the elevation at departure. 2. \(h(2.5)=5000-800(2.5)=3000\), so the elevation is \(3000\,\text{ft}\). 3. For \(2200\,\text{ft}\), solve \(2200=5000-800t\). This gives \(t=3.5\) hours. For the valley, solve \(0=5000-800t\), giving \(t=6.25\) hours. Therefore the reasonable domain is \([0,6.25]\) hours and the range is \([0,5000]\) feet. 4. The second hiker's model is \(g(t)=4000-600t\). Its rate of change is \(-600\,\text{ft/h}\), and its initial value is \(4000\,\text{ft}\). Solving \(g(t)=0\) gives \(t=\frac{20}{3}\approx6.67\) hours. Since \(6.25<6.67\), the first hiker reaches the valley first.

a) Rate of change: \(-800\,\text{ft/h}\), meaning the elevation decreases by \(800\,\text{ft}\) each hour. Initial value: \(5000\,\text{ft}\), the departure elevation. b) \(3000\,\text{ft}\) c) \(3.5\) hours to reach \(2200\,\text{ft}\); \(6.25\) hours to reach the valley; domain \([0,6.25]\), range \([0,5000]\). d) \(g(t)=4000-600t\); rate \(-600\,\text{ft/h}\); initial value \(4000\,\text{ft}\). The first hiker reaches the valley first.

- Where can you read the y-intercept directly from each equation? - How does slope connect a horizontal change to a vertical change? - What does the size of the slope tell you about a line's steepness?

1. Both functions have y-intercept \(1\). When \(x = 0\), both have output \(1\), so the common point is \((0, 1)\). 2. For \(f\), \(\Delta y = m\Delta x = \frac{2}{3} \cdot 6 = 4\). 3. For \(g\), \(\Delta y = 1.5 \cdot 6 = 9\). 4. Since \(1.5 > \frac{2}{3}\), line \(g\) is steeper. The larger vertical change for the same horizontal change confirms this.

a) \((0, 1)\) b) For \(f\), \(\Delta y = 4\); for \(g\), \(\Delta y = 9\). c) \(g\) is steeper because its slope is larger.

- What must be true about the slopes of parallel lines? - Which parameter gives the y-intercept? - Compare the positive slopes to determine which line rises fastest.

1. Parallel lines have the same slope. Both \(g_1\) and \(g_2\) have slope \(1.5\), so they are parallel. 2. The y-intercept is determined by \(b\). Functions \(g_1\), \(g_3\), and \(g_4\) all have \(b = 2\), so they pass through \((0, 2)\). 3. Among the positive slopes, \(3\) is the greatest. Therefore, \(g_4\) increases most steeply.

a) \(g_1\) and \(g_2\) b) \(g_1\), \(g_3\), and \(g_4\); the common y-intercept is \((0, 2)\). c) \(g_4\)

- How can you find the vertical change when the slope and horizontal change are known? - To compare a fraction and a decimal, rewrite them in the same form. - Which value tells you how steep a line is: the height of one chosen slope triangle or the slope itself?

1. For line \(k\), \(\Delta y = \frac{2}{3} \cdot 3 = 2\). 2. For line \(l\), \(\Delta y = 0.6 \cdot 5 = 3\). 3. The slopes are \(\frac{2}{3}\) and \(0.6\). Since \(\frac{2}{3} \approx 0.667 > 0.6\), line \(k\) is steeper. The larger height of the second slope triangle does not make \(l\) steeper because the two triangles have different horizontal widths.

a) Line \(k\): 2 units; line \(l\): 3 units b) Line \(k\) is steeper because \(\frac{2}{3} > 0.6\).

- What does the constant term in slope-intercept form tell you about the graph? - What is the y-coordinate of an x-intercept? - When are two lines parallel? - What do different slopes imply about two lines?

1. The parameter \(b\) is the y-intercept. Decreasing \(b\) shifts the line downward without changing its slope. 2. An x-intercept of \(6\) means \(y = 0\) when \(x = 6\). Thus \(0 = -1.5 \cdot 6 + b\), so \(b = 9\). 3. Every line in the family has slope \(-1.5\), while \(h\) has slope \(2\). Because the slopes are different, the lines are not parallel, so each line in the family intersects \(h\) exactly once.

a) The line shifts downward as \(b\) decreases. b) \(b = 9\) c) The slopes are different, \(-1.5 \neq 2\), so the lines intersect exactly once.

- What does the constant term tell you about where each graph crosses the y-axis? - How does the sign of the slope affect whether a line increases or decreases? - For two positive slopes, how does the larger slope affect steepness? - What must be the same for two distinct lines to be parallel?

1. Every graph has y-intercept \(-2\), so every graph passes through \((0, -2)\). 2. A line decreases from left to right when its slope is negative, so \(m < 0\). 3. The line with \(m = 2\) is steeper than the line with \(m = 1\) because \(2 > 1\). 4. A parallel line must have slope \(0.5\). With y-intercept \(3\), its equation is \(g(x) = 0.5x + 3\).

a) \((0, -2)\) b) \(m < 0\) c) The graph with \(m = 2\) is steeper. d) \(g(x) = 0.5x + 3\)

- Pay close attention to the sign of the given slope. - What signs do \(\Delta x\) and \(\Delta y\) have for each move? - How can you compare the decimal ratio with the fraction in the given slope?

1. Sarah's move has \(\Delta x = 3\) and \(\Delta y = 2\), so the slope would be \(\frac{2}{3}\). This is positive, so Sarah is not correct. 2. Tom's move has \(\Delta x = -1.5\) and \(\Delta y = 1\), so the slope is \(\frac{1}{-1.5} = -\frac{2}{3}\). Tom is correct.

Tom is correct. Sarah's move gives slope \(\frac{2}{3}\), while Tom's move gives slope \(-\frac{2}{3}\).

- Find the numerical slope represented by each description. - How do moving left or down affect the signs of \(\Delta x\) and \(\Delta y\)? - How are slope, \(\Delta x\), and \(\Delta y\) related?

1. Description (1) gives \(m = \frac{-2}{5} = -0.4\). 2. Description (2) gives \(m = \frac{1}{-2.5} = -0.4\). 3. Description (3) gives \(m = -2.5\), so it is different from the first two. 4. For \(m = -0.4\) and \(\Delta x = 10\), \(\Delta y = -0.4 \cdot 10 = -4\). Thus, \(y\) decreases by 4 units.

Descriptions (1) and (2) represent the same slope, \(m = -0.4\). When \(x\) increases by 10 units, \(\Delta y = -4\), so \(y\) decreases by 4 units.

- Use the relationship \(\Delta y = m\Delta x\). - How can you rewrite \(1.25\) as a fraction to choose whole-number changes? - For a positive slope, what direction must \(y\) change when you move left?

1. For \(f\), \(\Delta y = 1.25 \cdot 4 = 5\). 2. For \(g\), \(\Delta y = 0.8 \cdot 4 = 3.2\). 3. Since \(1.25 = \frac{5}{4}\), choose \(\Delta x = -4\). Then \(\Delta y = 1.25 \cdot (-4) = -5\). One suitable triangle moves 4 units left and 5 units down.

a) For \(f\), \(\Delta y = 5\). For \(g\), \(\Delta y = 3.2\). b) One possible slope triangle is 4 units left and 5 units down.

- What does the constant term in slope-intercept form tell you about a graph? - Think about the vertical change for a horizontal change of 1 as \(m\) gets larger. - Substitute the coordinates of \(P\) into \(f(x) = mx + 3\) to find \(m\).

1. Both functions have y-intercept \(3\), so both graphs pass through \((0, 3)\). 2. As \(m\) increases, the graph of \(f\) rises more steeply and the graph of \(g\), whose slope is \(-m\), falls more steeply. 3. Since \(P(2, 7)\) lies on \(f\), \(7 = 2m + 3\). Thus, \(m = 2\), so \(g(x) = -2x + 3\). For example, \(g(1) = 1\), so \(Q(1, 1)\) is on the graph.

1. Both graphs have y-intercept \(3\), so they share \((0, 3)\). 2. \(f\) becomes steeper increasing, while \(g\) becomes steeper decreasing. 3. \(m = 2\); \(g(x) = -2x + 3\); one possible point is \(Q(1, 1)\).

- Compare the slopes between two pairs of points. - Use change in \(y\) divided by change in \(x\), even when the coordinates are decimals. - What does it mean if the two slopes are different?

1. The slope from \(A\) to \(B\) is \(\frac{9 - 4}{4 - 1.5} = \frac{5}{2.5} = 2\). 2. The slope from \(B\) to \(C\) is \(\frac{13.5 - 9}{6 - 4} = \frac{4.5}{2} = 2.25\). 3. Since \(2 \neq 2.25\), the points are not collinear. Three noncollinear points form a triangle.

The points form a triangle because the slopes \(2\) and \(2.25\) are different.

- Is the candle's height increasing or decreasing as it burns? - Use the two height measurements to find the change per hour. - Think about what the function value at time \(0\) represents.

1. The initial height is the y-intercept, so \(b = 10\). 2. The slope is \(m = \frac{8.5 - 10}{3 - 0} = -0.5\) inch per hour. Thus \(h(t) = -0.5t + 10\). 3. The y-intercept \(10\) is the candle's initial height. The slope \(-0.5\) means its height decreases by \(0.5\) inch each hour. 4. Evaluate at \(t = 5\): \(h(5) = -0.5 \cdot 5 + 10 = 7.5\) inches.

a) \(h(t) = -0.5t + 10\) b) \(b = 10\) is the initial height, and \(m = -0.5\) means the candle loses \(0.5\) inch of height per hour. c) \(7.5\) inches

- How much does the fare change when the distance increases? - What cost would remain when the distance is \(0\) miles? - Substitute the requested distance into your function.

1. From \((5, 14.50)\) and \((12, 28.50)\), the slope is \(m = \frac{28.50 - 14.50}{12 - 5} = 2\) dollars per mile. 2. Substitute \((5, 14.50)\) into \(C(x) = 2x + b\): \(14.50 = 2 \cdot 5 + b\), so \(b = 4.50\). 3. Therefore, \(C(x) = 2x + 4.50\). 4. The slope means the fare increases by \(\$2.00\) per mile, and the y-intercept is the \(\$4.50\) starting fare. 5. Evaluate at \(x = 20\): \(C(20) = 2 \cdot 20 + 4.50 = 44.50\).

a) \(C(x) = 2x + 4.50\) b) The slope is a charge of \(\$2.00\) per mile, and the y-intercept is the \(\$4.50\) starting fare. c) \(\$44.50\)

- Evaluate each function at \(x=0\) to find its initial value. - Identify the coefficient that gives the rate of change. - In a proportional relationship, doubling the input should double the output.

1. Only \(f(x)=1.5x\) is proportional because \(f(0)=0\), so its graph passes through \((0, 0)\). 2. For \(f\), the rate of change is \(1.5\) and the initial value is \(0\). For \(g\), the rate of change is \(1.5\) and the initial value is \(4\). For \(h\), the rate of change is \(0\) and the initial value is \(3\). 3. \(g(2)=1.5\cdot 2+4=7\), and \(g(4)=1.5\cdot 4+4=10\). 4. Since \(10\ne 2\cdot 7\), doubling the input does not double the output for \(g\).

a) Only \(f(x)\), because its initial value is \(0\). b) \(f\): rate \(1.5\), initial value \(0\). \(g\): rate \(1.5\), initial value \(4\). \(h\): rate \(0\), initial value \(3\). c) \(g(2)=7\) and \(g(4)=10\), so the doubling rule does not hold.

- Separate each cost into a fixed part and a usage-dependent part. - What should the cost be at zero usage in a proportional relationship? - How does a fixed starting value affect the y-intercept?

1. Provider A is \(y_A=0.40x\). Provider B is \(y_B=0.30x+10\). 2. Provider A is proportional because it has the form \(y=kx\) and \(y_A(0)=0\). 3. Provider B is not proportional because \(y_B(0)=10\), so the output-to-input ratio is not constant. 4. The base charge shifts Provider B's graph upward; its y-intercept is \(10\) instead of \(0\).

a) Provider A: \(y=0.40x\). Provider B: \(y=0.30x+10\). b) Only Provider A is proportional. c) The base charge shifts Provider B's graph upward so it crosses the y-axis at \(10\), not at the origin.

- What amount of water is in the barrel at time \(0\)? - How does the amount change each minute? - What form represents a quantity decreasing at a constant rate? - What volume represents an empty barrel?

1. The initial volume is \(60\) gallons and the volume decreases by \(2\) gallons per minute, so \(V(t) = 60 - 2t\). 2. The y-intercept \(60\) is the starting volume. The slope \(-2\) means the volume decreases by \(2\) gallons each minute. 3. Evaluate at \(t = 12\): \(V(12) = 60 - 2 \cdot 12 = 36\) gallons. 4. The barrel is empty when \(V(t) = 0\): \(0 = 60 - 2t\), so \(t = 30\) minutes.

a) \(V(t) = 60 - 2t\) b) The y-intercept \(60\) is the starting volume, and the slope \(-2\) means the barrel loses \(2\) gallons of water per minute. c) \(36\) gallons d) \(30\) minutes

- Calculate slopes for different parts of the table. - With a constant inflow, a wider section causes the water level to rise more slowly. - Compare the two rates directly.

1. From \(0\) to \(10\) minutes, the rate is \(\frac{5-0}{10-0}=0.5\) inch per minute. From \(20\) to \(30\) minutes, it is \(\frac{12.5-10}{30-20}=0.25\) inch per minute. The rate is not constant over the entire table, so the relation is not linear. 2. On \([0, 20]\), the rate is \(\frac{10-0}{20-0}=0.5\) inch per minute. On \([20, 50]\), the rate is \(\frac{17.5-10}{50-20}=\frac{7.5}{30}=0.25\) inch per minute. 3. The water rises twice as fast in the lower section. Because the rainfall rate is constant, the smaller rate above \(10\) inches indicates that the barrel becomes wider there.

a) No b) \([0, 20]\): \(0.5\) inch per minute; \([20, 50]\): \(0.25\) inch per minute c) The barrel becomes wider above a height of \(10\) inches.

- Speed is the slope over each time interval. - What does a constant distance over time mean? - Check whether one rate describes every interval.

1. The interval speeds are \((3-0)\div(1-0)=3\), \((6-3)\div(2-1)=3\), \((6-6)\div(2.5-2)=0\), \((8-6)\div(3-2.5)=4\), and \((14-8)\div(4.5-3)=4\) miles per hour. 2. One proportional function cannot describe the entire hike because the rate of change is not constant. 3. The unchanged distance from \(2\) to \(2.5\) hours indicates a \(30\)-minute break. 4. Before the break, the group walked at \(3\) miles per hour; after the break, it walked at \(4\) miles per hour.

a) \(3\), \(3\), \(0\), \(4\), and \(4\) miles per hour. b) No. The rate of change is not constant. c) The group took a \(30\)-minute break. d) The group walked faster after the break: \(4\) miles per hour instead of \(3\).

- Which part of a linear function gives the starting value? - What does the slope represent in this context? - Substitute the given time into each function to compare the water heights. - To find when the heights are equal, set the two function expressions equal.

1. The initial heights are the values at \(t = 0\): Tank A starts at \(20\,\text{cm}\), and Tank B starts at \(10\,\text{cm}\). 2. The rates of change are the slopes: Tank A rises at \(1.5\,\text{cm/min}\), and Tank B rises at \(2\,\text{cm/min}\). 3. After \(15\) minutes, \(h_A(15) = 1.5 \cdot 15 + 20 = 42.5\,\text{cm}\), while \(h_B(15) = 2 \cdot 15 + 10 = 40\,\text{cm}\). Tank A is higher. 4. Set the heights equal: \(1.5t + 20 = 2t + 10\). Then \(10 = 0.5t\), so \(t = 20\) minutes.

a) Tank A: \(20\,\text{cm}\); Tank B: \(10\,\text{cm}\) b) Tank A: \(1.5\,\text{cm/min}\); Tank B: \(2\,\text{cm/min}\) c) Tank A, with \(42.5\,\text{cm}\) compared with \(40\,\text{cm}\) d) \(20\) minutes

- Evaluate the function at \(t=0\). - Convert half an hour to minutes. - Set the function equal to \(100\) and solve. - Two points determine the line segment.

1. The value \(15\) is the initial battery level, \(15\%\). The coefficient \(0.8\) is the charging rate, \(0.8\) percentage point per minute. 2. Half an hour is \(30\) minutes. \(L(30)=0.8(30)+15=39\), so the predicted battery level is \(39\%\). 3. Solve \(0.8t+15=100\). Then \(0.8t=85\), so \(t=106.25\) minutes. 4. Over the first \(60\) minutes, the graph is the line segment from \((0, 15)\) to \((60, 63)\).

a) Initial level: \(15\%\); charging rate: \(0.8\) percentage point per minute b) \(39\%\) c) \(106.25\) minutes d) The line segment through \((0, 15)\) and \((60, 63)\)

- How do you find the slope from two time-temperature points? - Which quantity changes as time passes? - What does the function value at time \(0\) represent? - What does a negative slope mean in this context?

1. Use \((2, 170)\) and \((10, 130)\): \(m = \frac{130 - 170}{10 - 2} = -5\) degrees Fahrenheit per minute. 2. Substitute \((2, 170)\) into \(T(t) = -5t + b\): \(170 = -5 \cdot 2 + b\), so \(b = 180\). Thus \(T(t) = -5t + 180\). 3. The slope \(-5\) means the temperature decreases by \(5^\circ\text{F}\) each minute. The y-intercept \(180\) is the temperature at time \(0\).

a) \(T(t) = -5t + 180\) b) The slope means the temperature decreases by \(5^\circ\text{F}\) per minute, and the y-intercept means the initial temperature was \(180^\circ\text{F}\).

- Rewrite the slopes in the same form before comparing them. - Where is the y-intercept visible in slope-intercept form? - What happens to the slope when it is doubled?

1. The slopes are \(\frac{2}{3}\) and \(0.6\). Since \(\frac{2}{3} \approx 0.667 > 0.6\), \(f\) is steeper. 2. Both functions have y-intercept \(4\), so both graphs pass through \((0, 4)\). 3. Twice the slope of \(f\) is \(2 \cdot \frac{2}{3} = \frac{4}{3}\). Keeping the y-intercept \(4\) gives \(h(x) = \frac{4}{3}x + 4\).

a) \(f\) is steeper because \(\frac{2}{3} > 0.6\). b) Both y-intercepts are \((0, 4)\). c) \(h(x) = \frac{4}{3}x + 4\)

- What is always true about the x-coordinate of a point on the y-axis? - In a linear function, how does \(y\) change when \(x\) changes? Which parameter describes that relationship?

1. For the y-intercept, set \(x = 0\): \(8y = 16\), so \(y = 2\). The y-intercept is \((0, 2)\). 2. For the x-intercept, set \(y = 0\): \(4x = 16\), so \(x = 4\). The x-intercept is \((4, 0)\). 3. Solve for \(y\): \(8y = -4x + 16\), so \(y = -0.5x + 2\). Thus, the slope is \(-0.5\). 4. A slope of \(-0.5\) means \(\Delta y = -0.5\Delta x\). If \(\Delta x = 3\), then \(\Delta y = -1.5\), so \(y\) decreases by \(1.5\) units.

a) x-intercept: \((4, 0)\); y-intercept: \((0, 2)\) b) \(m = -0.5\) c) \(y\) decreases by \(1.5\) units.

- What geometric object is the solution set of one linear equation in two variables? - Compare the slopes between two pairs of points. - What must be true about those slopes if all three points lie on one line?

1. The solution set of a nondegenerate linear equation in two variables is a line, so the three points would have to be collinear. 2. The slope from \((-2, 7)\) to \((1, 1)\) is \(\frac{1 - 7}{1 - (-2)} = -2\). 3. The slope from \((1, 1)\) to \((4, -4)\) is \(\frac{-4 - 1}{4 - 1} = -\frac{5}{3}\). 4. Since the slopes are different, the three points are not collinear and cannot all satisfy the same nondegenerate linear equation.

No. The slopes are \(-2\) and \(-\frac{5}{3}\), so the three points are not collinear.

- How much does the candle's height change over the three-hour interval? - What sign should the rate have when the candle gets shorter? - What does time \(t = 0\) represent?

1. Use the points \((2, 7.5)\) and \((5, 6)\): \(m = \frac{6 - 7.5}{5 - 2} = -0.5\) inch per hour. 2. Substitute \((2, 7.5)\) into \(h(t) = -0.5t + b\): \(7.5 = -0.5 \cdot 2 + b\), so \(b = 8.5\). 3. Therefore, \(h(t) = -0.5t + 8.5\). 4. The y-intercept \(8.5\) is the candle's starting height. The slope \(-0.5\) means the candle loses \(0.5\) inch of height each hour.

a) \(h(t) = -0.5t + 8.5\) b) The y-intercept \(8.5\) is the initial height in inches, and the slope \(-0.5\) means the height decreases by \(0.5\) inch per hour.

- Rewrite the equation so the rate and initial value can be read from \(y=mx+b\). - A rate of \(-\frac{1}{4}\) tells how the output changes for each one-unit increase in input. - For part c), use the initial value as the starting output instead of solving the original equation from scratch.

1. Subtract \(0.5x\): \(2y=4-0.5x\). Divide by \(2\): \(y=-0.25x+2\). 2. The rate of change is \(-0.25=-\frac{1}{4}\), and the initial value is \(2\). 3. Increasing \(x\) from \(0\) to \(4\) changes the output by \(-\frac{1}{4}\cdot4=-1\). Starting from \(2\), the output is \(1\), so \((4,1)\) is on the graph. 4. At \(x=-4\), \(y=-0.25(-4)+2=3\), so \((-4,3)\) is an input-output pair.

a) \(y=-0.25x+2\) b) Rate of change: \(-0.25\); initial value: \(2\) c) \(1\) d) Yes.

- Set one coordinate equal to zero to find each intercept. - Isolate \(y\) to rewrite the equation in slope-intercept form.

1. For the x-intercept, set \(y = 0\): \(2x = 10\), so the point is \((5, 0)\). 2. For the y-intercept, set \(x = 0\): \(-5y = 10\), so the point is \((0, -2)\). 3. Solve for \(y\): \(-5y = -2x + 10\), so \(y = 0.4x - 2\). Therefore, \(m = 0.4\).

a) x-intercept: \((5, 0)\); y-intercept: \((0, -2)\) b) \(y = 0.4x - 2\), with slope \(m = 0.4\)

- Identify the volume at the end of each phase. - Use slope to find the filling and removal rates. - Check both the increasing and decreasing parts of the graph for a volume of \(9\) gallons.

1. The graph consists of line segments through \((0, 0)\), \((20, 12)\), \((30, 12)\), and \((45, 6)\). 2. The filling rate is \(\frac{12-0}{20-0}=0.6\) gallon per minute. 3. During filling, \(0.6t=9\), so \(t=15\) minutes. During removal, the rate is \(\frac{6-12}{45-30}=-0.4\) gallon per minute. Starting at minute \(30\), the volume drops from \(12\) to \(9\) gallons, a decrease of \(3\) gallons, in \(3\div0.4=7.5\) minutes. Thus, the second time is \(t=37.5\) minutes.

a) Line segments through \((0, 0)\), \((20, 12)\), \((30, 12)\), and \((45, 6)\) b) \(0.6\) gallon per minute c) After \(15\) minutes and after \(37.5\) minutes

- Convert all times to hours. - Compare the magnitudes of the slopes for the climbing and descending segments. - The group may pass the target elevation once while climbing and once while descending.

1. The graph consists of line segments through \((0, 1500)\), \((2, 3500)\), \((2.5, 3500)\), and \((4, 2500)\). 2. During the climb, the vertical speed is \(\frac{3500-1500}{2}=1000\,\text{ft/h}\). During the descent, its magnitude is \(\frac{3500-2500}{1.5}=\frac{2000}{3}\approx666.7\,\text{ft/h}\). The group's vertical speed was greater during the climb. 3. During the climb, \(1500+1000t=3000\), so \(t=1.5\) hours. During the descent, \(3500-\frac{2000}{3}(t-2.5)=3000\), so \(t-2.5=0.75\) and \(t=3.25\) hours.

a) Line segments through \((0, 1500)\), \((2, 3500)\), \((2.5, 3500)\), and \((4, 2500)\) b) The climb: \(1000\,\text{ft/h}\) versus about \(666.7\,\text{ft/h}\) during the descent c) After \(1.5\) hours and after \(3.25\) hours

- Divide each height change by the corresponding time change. - The greatest average rate identifies the fastest interval. - For the prediction, start with the day-20 height and add \(10\) days of growth.

1. The rates are: \([0, 5]\): \(\frac{22-12}{5}=2\,\text{cm/day}\); \([5, 15]\): \(\frac{62-22}{10}=4\,\text{cm/day}\); \([15, 20]\): \(\frac{92-62}{5}=6\,\text{cm/day}\); \([20, 30]\): \(\frac{112-92}{10}=2\,\text{cm/day}\). 2. The greatest rate is \(6\,\text{cm/day}\) on \([15, 20]\). 3. Continuing at \(6\,\text{cm/day}\) for \(10\) more days gives \(92+10(6)=152\,\text{cm}\). The actual height is \(112\,\text{cm}\), so growth slowed after day \(20\).

a) \(2\), \(4\), \(6\), and \(2\,\text{cm/day}\) b) \([15, 20]\) c) Predicted: \(152\,\text{cm}\); actual: \(112\,\text{cm}\)

- Compare the changes over equal five-day intervals. - Test the proposed rate against at least one table value. - Average rate of change compares the change in depth with the corresponding change in time.

1. The ordered pairs are \((0, 160)\), \((5, 155)\), \((10, 148)\), \((15, 139)\), and \((20, 128)\). The depth decreases as time increases, and the decreases become larger over successive five-day intervals. 2. A decrease of \(1.5\,\text{ft}\) per day predicts a decrease of \(5\cdot1.5=7.5\) feet by day \(5\), giving \(160-7.5=152.5\) feet. The table gives \(155\) feet, so the model does not match exactly. 3. From day \(15\) to day \(20\), the average rate is \(\frac{128-139}{20-15}=\frac{-11}{5}=-2.2\,\text{ft/day}\).

a) \((0, 160)\), \((5, 155)\), \((10, 148)\), \((15, 139)\), and \((20, 128)\); the water depth decreases at a changing rate. b) No. The model predicts \(152.5\,\text{ft}\) on day \(5\), but the table gives \(155\,\text{ft}\). c) \(-2.2\,\text{ft/day}\)

- What x-value gives a y-intercept? - Substitute the given y-value in part b) and solve. - What y-value gives an x-intercept? - How does the sign of the slope affect a line?

1. For \(g_1\), setting \(x = 0\) gives the y-intercept \((0, -3)\). Rewrite \(g_2\) as \(y = -2x + 4\), so its y-intercept is \((0, 4)\). 2. For \(g_1\), solve \(6 = 1.5x - 3\). Then \(9 = 1.5x\), so \(x = 6\). 3. For the x-intercept of \(g_2\), set \(y = 0\): \(2x = 4\), so \(x = 2\). Therefore, the x-intercept is \((2, 0)\). 4. The slope of \(g_1\) is \(1.5 > 0\), so it is increasing. The slope of \(g_2\) is \(-2 < 0\), so it is decreasing.

a) \(g_1: (0, -3)\); \(g_2: (0, 4)\) b) \(x = 6\) c) \((2, 0)\) d) \(g_1\) is increasing; \(g_2\) is decreasing.

- If a point lies on a line, what must be true when its coordinates are substituted into the equation? - What does a negative slope tell you about how the values change? - How can you interpret the slope as a fraction relating vertical change to horizontal change? - How can you check whether a point lies on the line without drawing the graph?

1. For \(A\), \(y_A = -\frac{2}{3} \cdot 6 + 2 = -2\). 2. For \(B\), solve \(0 = -\frac{2}{3}x_B + 2\). This gives \(x_B = 3\). 3. At \(x = -3\), \(y = -\frac{2}{3} \cdot (-3) + 2 = 4\), so \(C\) lies on the line. 4. With slope \(-\frac{2}{3}\), a horizontal increase of \(3\) gives \(\Delta y = -\frac{2}{3} \cdot 3 = -2\). Thus, \(y\) decreases by 2 units.

a) \(y_A = -2\) b) \(x_B = 3\) c) Yes. d) \(y\) decreases by 2 units.

- Use linear interpolation between consecutive measurements. - Compare consecutive depth values to identify rising and falling intervals. - Find where the linear segments cross the minimum safe depth.

1. At \(4{:}30\) a.m., the time is halfway between \(3{:}00\) a.m. and \(6{:}00\) a.m., so the depth is halfway between \(24\) and \(18\): \(21\,\text{ft}\). At \(10{:}30\) a.m., it is halfway between \(12\) and \(18\): \(15\,\text{ft}\). 2. The level rises from \(12{:}00\) a.m. to \(3{:}00\) a.m., from \(9{:}00\) a.m. to \(3{:}00\) p.m., and from \(9{:}00\) p.m. to \(12{:}00\) a.m. It falls from \(3{:}00\) a.m. to \(9{:}00\) a.m. and from \(3{:}00\) p.m. to \(9{:}00\) p.m. 3. From \(6{:}00\) a.m. to \(9{:}00\) a.m., the depth falls from \(18\) to \(12\), a rate of \(-2\,\text{ft/h}\). It reaches \(16\,\text{ft}\) at \(7{:}00\) a.m. On the rising segment after \(9{:}00\) a.m., it reaches \(16\,\text{ft}\) at \(11{:}00\) a.m. The same pattern repeats \(12\) hours later. Because exactly \(16\,\text{ft}\) is safe, the unsafe intervals exclude their endpoints.

1. \(4{:}30\) a.m.: \(21\,\text{ft}\) \(10{:}30\) a.m.: \(15\,\text{ft}\) 2. Rising: \(12{:}00\) a.m.–\(3{:}00\) a.m., \(9{:}00\) a.m.–\(3{:}00\) p.m., and \(9{:}00\) p.m.–\(12{:}00\) a.m. Falling: \(3{:}00\) a.m.–\(9{:}00\) a.m. and \(3{:}00\) p.m.–\(9{:}00\) p.m. 3. Between \(7{:}00\) a.m. and \(11{:}00\) a.m., excluding the endpoints, and between \(7{:}00\) p.m. and \(11{:}00\) p.m., excluding the endpoints.

- Find each change in battery level per hour. - Interpret positive, negative, and zero rates in context. - Divide the remaining battery level by the projected use rate.

1. The level decreases by \(100-60=40\) percentage points in \(2\) hours, or \(20\) percentage points per hour. 2. From \(10{:}30\) a.m. to \(12{:}30\) p.m., the level decreases by \(90-20=70\) percentage points in \(2\) hours, or \(35\) percentage points per hour. This is greater than the \(20\)-point rate from \(8{:}00\) a.m. to \(10{:}00\) a.m. From \(10{:}00\) a.m. to \(10{:}30\) a.m., the battery is charging rather than being used. 3. The car may have been parked and turned off, so its battery level did not change noticeably. 4. At \(35\) percentage points per hour, the remaining \(20\%\) lasts \(\frac{20}{35}\approx0.5714\) hour, or about \(34.3\) minutes. The model predicts the battery will be empty at about \(2{:}04\) p.m.

1. \(20\) percentage points per hour 2. \(10{:}30\) a.m. to \(12{:}30\) p.m., at \(35\) percentage points per hour 3. The car may have been parked and turned off. 4. About \(2{:}04\) p.m.

- How much water was used between the two measurements? - Divide that change by the number of days between the measurements. - Use the daily rate to work backward from day \(5\) to day \(0\). - How many daily amounts fit into the starting amount?

1. Between day \(5\) and day \(12\), the amount of water decreases by \(300 - 230 = 70\) gallons over \(12 - 5 = 7\) days. 2. The daily decrease is \(70 \div 7 = 10\) gallons per day. 3. Five days before there were \(300\) gallons remaining, the tank contained \(300 + 5 \cdot 10 = 350\) gallons. 4. At \(10\) gallons per day, \(350 \div 10 = 35\), so the tank will be empty after \(35\) days.

a) \(10\,\text{gal/day}\) b) \(350\,\text{gal}\) c) \(35\) days

- How many minutes pass between the two battery readings? - How many percentage points does the battery lose during that interval? - Find the loss per minute first, then convert it to a per-hour rate. - How long would a \(20\)-percentage-point drop take at that rate? - How long would a \(100\)-percentage-point drop take at the same rate?

1. From \(10{:}00\) a.m. to \(11{:}20\) a.m. is \(80\) minutes. The battery level decreases by \(20\) percentage points, so the rate is \(20 \div 80 = 0.25\) percentage point per minute, or \(0.25 \cdot 60 = 15\) percentage points per hour. 2. Going from \(100\%\) to \(80\%\) is a decrease of \(20\) percentage points, which takes \(20 \div 0.25 = 80\) minutes. Eighty minutes before \(10{:}00\) a.m. is \(8{:}40\) a.m. 3. Going from \(100\%\) to \(0\%\) takes \(100 \div 0.25 = 400\) minutes, or \(6\) hours \(40\) minutes. Starting at \(8{:}40\) a.m., the battery reaches \(0\%\) at \(3{:}20\) p.m.

a) \(15\) percentage points per hour b) \(8{:}40\) a.m. c) \(3{:}20\) p.m.

- What constant rate connects the two odometer readings? - Find the change in distance divided by the change in time. - How do you convert miles per minute to miles per hour? - Once you know the rate, work backward from the first reading to \(t = 0\).

1. Over the \(50 - 20 = 30\) minutes between readings, the odometer increases by \(100 - 70 = 30\) miles. The rate is \(30 \div 30 = 1\) mile per minute. 2. Convert the rate to miles per hour: \(1 \cdot 60 = 60\,\text{mph}\). 3. In the first \(20\) minutes, the car travels \(20\) miles. Therefore, the odometer reading at \(t = 0\) was \(70 - 20 = 50\,\text{mi}\).

The speed is \(60\,\text{mph}\), and the odometer reading at \(t = 0\) was \(50\,\text{mi}\).

- What does a constant burn rate tell you about the type of relationship? - For each candle, use the change in height divided by the change in time to find the burn rate. - Once you know each rate, work backward to find each starting height. - Compare the two starting heights and the two burn rates.

1. For Candle A, the height decreases by \(8 - 5 = 3\) inches over \(5 - 2 = 3\) hours, so it burns at \(1\,\text{in./h}\). Working back \(2\) hours gives an initial height of \(8 + 2 \cdot 1 = 10\) inches. 2. For Candle B, the height decreases by \(9 - 5.4 = 3.6\) inches over \(6 - 3 = 3\) hours, so it burns at \(1.2\,\text{in./h}\). Working back \(3\) hours gives an initial height of \(9 + 3 \cdot 1.2 = 12.6\) inches. 3. Candle B started taller because \(12.6 > 10\), and it burns faster because \(1.2 > 1\).

Candle B was taller at the start: \(12.6\,\text{in.}\) compared with Candle A's \(10\,\text{in.}\). Candle B also burns faster: \(1.2\,\text{in./h}\) compared with \(1\,\text{in./h}\).

- Substitute each given time into the equation. - What time value represents the moment draining begins? - What water height represents an empty tank?

1. Substitute each time into \(h=120-8t\): \(t=0\): \(h=120-8\cdot0=120\) \(t=5\): \(h=120-8\cdot5=80\) \(t=10\): \(h=120-8\cdot10=40\) \(t=15\): \(h=120-8\cdot15=0\) 2. The constant \(120\) is the initial water height, because it is the value of \(h\) when \(t=0\). 3. The tank is empty when \(h=0\). Since \(120-8\cdot15=0\), the tank is empty after \(15\) minutes.

a) <table><thead><tr><th>Time \(t\) (min)</th><th>Height \(h\) (cm)</th></tr></thead><tbody><tr><td>\(0\)</td><td>\(120\)</td></tr><tr><td>\(5\)</td><td>\(80\)</td></tr><tr><td>\(10\)</td><td>\(40\)</td></tr><tr><td>\(15\)</td><td>\(0\)</td></tr></tbody></table> b) \(120\) is the initial water height in centimeters. c) \(15\) minutes, because substituting \(t=15\) gives \(h=0\).

- Break the trip into riding intervals and the rest interval before finding accumulated distance. - During the break, elapsed time increases but distance traveled does not. - For a target distance, first decide which riding interval contains that target. - Whole-trip average speed uses total distance and total elapsed time, including the break.

1. At \(9{:}00\) a.m., the distance is \(0\). At \(10{:}00\) a.m., it is \(12\) miles. During the next \(1.5\) hours, she travels \(1.5\times8=12\) miles, so at \(11{:}30\) a.m. she has traveled \(24\) miles. The distance remains \(24\) miles through the break until \(12{:}00\) p.m. In the final \(0.5\) hour, she travels \(0.5\times16=8\) miles, so the final distance is \(32\) miles. 2. From \(10{:}00\) a.m. to \(11{:}00\) a.m., she travels \(8\) more miles, so the total is \(12+8=20\) miles. 3. After \(10{:}00\) a.m., she needs \(16-12=4\) more miles at \(8\,\text{mph}\). This takes \(4\div8=0.5\) hour, so she reaches \(16\) miles at \(10{:}30\) a.m. 4. The total elapsed time is \(3.5\) hours. The average speed is \(32\div3.5=\frac{64}{7}\approx9.1\,\text{mph}\).

1) <table><tr><td>Time</td><td>\(9{:}00\) a.m.</td><td>\(10{:}00\) a.m.</td><td>\(11{:}30\) a.m.</td><td>\(12{:}00\) p.m.</td><td>\(12{:}30\) p.m.</td></tr><tr><td>Distance (mi)</td><td>\(0\)</td><td>\(12\)</td><td>\(24\)</td><td>\(24\)</td><td>\(32\)</td></tr></table> 2) \(20\) miles 3) \(10{:}30\) a.m. 4) About \(9.1\,\text{mph}\)

- Look for points on the line that lie exactly on grid intersections. - Use change in volume divided by change in time to find the slope. - What volume does the graph show when \(t = 0\)? - How do the slope and initial value fit into \(V(t) = mt + b\)? - What value of \(V\) represents an empty tank?

1. Two convenient points are \((0, 250)\) and \((5, 150)\). The slope is \(m = \frac{150 - 250}{5 - 0} = -20\,\text{gal/min}\). This means the amount of water decreases by \(20\) gallons each minute. 2. The graph crosses the y-axis at \((0, 250)\), so \(b = 250\). Therefore, \(V(t) = -20t + 250\). 3. The tank is empty when \(V(t) = 0\). Solve \(0 = -20t + 250\), giving \(t = 12.5\) minutes.

a) \(m = -20\,\text{gal/min}\); the tank loses \(20\) gallons of water per minute. b) \(V(t) = -20t + 250\) c) \(12.5\) minutes

- Read the starting point and highest point of the graph. - A horizontal segment represents a constant temperature. - Average rate of change is change in temperature divided by change in time. - For d), write a linear equation for each segment that crosses \(65\,^{\circ}\text{C}\).

1. At \(t = 0\), the temperature is \(20\,^{\circ}\text{C}\). The maximum is \(80\,^{\circ}\text{C}\). 2. The graph is horizontal at \(80\,^{\circ}\text{C}\) from \(t = 3\) to \(t = 6\), so the maximum lasts \(6 - 3 = 3\) minutes. 3. During the first \(3\) minutes, the average rate of change is \(\frac{80 - 20}{3 - 0} = 20\,^{\circ}\text{C}/\text{min}\). 4. On the heating segment, \(T = 20 + 20t\). Solving \(20 + 20t = 65\) gives \(t = 2.25\) minutes. 5. On the first cooling segment, \(T = 80 - 15(t - 6)\). Solving \(80 - 15(t - 6) = 65\) gives \(t = 7\) minutes.

a) Initial temperature: \(20\,^{\circ}\text{C}\); maximum temperature: \(80\,^{\circ}\text{C}\) b) \(3\) minutes c) \(20\,^{\circ}\text{C}/\text{min}\) d) \(t = 2.25\) minutes and \(t = 7\) minutes

- Horizontal graph segments represent constant altitude. - Check both an ascending segment and a descending segment for the requested altitude. - Compute each rate as change in altitude divided by change in time. - The steepest downward segment has the greatest descent-rate magnitude.

1. The balloon remains at \(900\,\text{ft}\) for \(t\in[1, 1.5]\) hours and at \(1800\,\text{ft}\) for \(t\in[3, 4]\) hours. 2. From \(t=1.5\) to \(t=3\), the altitude rises from \(900\) to \(1800\) feet at \(600\,\text{ft/h}\), so it reaches \(1200\,\text{ft}\) at \(t=2\) hours. From \(t=4\) to \(t=5\), the altitude falls from \(1800\) to \(600\) feet at \(-1200\,\text{ft/h}\), so it reaches \(1200\,\text{ft}\) at \(t=4.5\) hours. 3. During the first hour, the balloon rises \(900\) feet, so the ascent rate is \(900\,\text{ft/h}\). 4. The steepest descending segment is from \(t=4\) to \(t=5\), where the balloon drops \(1200\) feet in \(1\) hour. Its descent rate is \(1200\,\text{ft/h}\) downward.

a) \(t\in[1, 1.5]\) h at \(900\,\text{ft}\); \(t\in[3, 4]\) h at \(1800\,\text{ft}\) b) \(t=2\) hours and \(t=4.5\) hours c) \(900\,\text{ft/h}\) d) From \(t=4\) to \(t=5\) hours; \(1200\,\text{ft/h}\) downward

- Describe what the graph shows before calculating. - Pay close attention to the scale on each axis. - What are the coordinates of the axis intercepts? - How can two points be used to calculate the slope? - Which value can you read directly as the y-intercept?

1. The line crosses the y-axis at \((0, 10)\) and the x-axis at \((-2, 0)\). 2. The slope is \(m = \frac{10 - 0}{0 - (-2)} = 5\), and the y-intercept is \(10\). Therefore, \(g(x) = 5x + 10\). 3. \(g(6) = 5 \cdot 6 + 10 = 40\). The graph also shows the point \((6, 40)\), confirming the calculation.

a) \(g(x) = 5x + 10\) b) \(g(6) = 40\)

- Identify what each axis represents. - Opening and closing correspond to when the graph first rises above 0 and later returns to 0. - For a specified number of visitors, imagine drawing a horizontal line through that y-value. - Average rate of change is change in visitors divided by change in time.

1. Before \(8{:}00\) a.m., the graph is at \(0\) visitors. It begins increasing at \(8{:}00\) a.m., so the pool opens then. The graph returns to \(0\) visitors at \(8{:}00\) p.m., so the pool closes then. 2. The highest point occurs at about \(3{:}15\) p.m. The maximum is about \(600\) visitors. 3. A horizontal line at \(400\) visitors intersects the graph at about noon and shortly after \(6{:}00\) p.m. 4. At \(10{:}00\) a.m., the graph shows about \(200\) visitors, and at noon it shows about \(400\) visitors. The estimated average increase is \(\frac{400-200}{12-10}=100\) visitors per hour.

a) Opens at \(8{:}00\) a.m.; closes at \(8{:}00\) p.m. b) About \(3{:}15\) p.m.; about \(600\) visitors c) About noon and shortly after \(6{:}00\) p.m. d) About \(100\) visitors per hour

- Read the y-intercept of each line from the graph. - Use two points on each line to determine its slope. - Recall the slope-intercept form \(y = mx + b\). - At an intersection point, the two functions have the same output for the same input. - After finding the x-coordinate, substitute it into either equation to find the y-coordinate.

1. For \(f\), the y-intercept is \(2\). Using the points \((0, 2)\) and \((2, 3)\), the slope is \(\frac{3 - 2}{2 - 0} = \frac{1}{2}\). Therefore, \(f(x) = \frac{1}{2}x + 2\). 2. For \(g\), the y-intercept is \(-1\). Using the points \((0, -1)\) and \((1, -2)\), the slope is \(\frac{-2 - (-1)}{1 - 0} = -1\). Therefore, \(g(x) = -x - 1\). 3. Set the expressions equal: \(\frac{1}{2}x + 2 = -x - 1\). 4. Then \(\frac{3}{2}x = -3\), so \(x = -2\). 5. Substituting into \(g\) gives \(y = -(-2) - 1 = 1\). Thus, \(S = (-2, 1)\).

a) \(f(x) = \frac{1}{2}x + 2\) and \(g(x) = -x - 1\). b) \(S = (-2, 1)\).

- Check the axis scales before estimating each intersection. - Use the marked points to find the slope and y-intercept of each line. - Recall the slope-intercept form \(y = mx + b\). How can two points be used to find \(m\)? - To find an exact intersection, set the two function expressions equal and solve for \(x\). - Substitute the x-coordinate into either equation to find the corresponding y-coordinate.

1. In graph a), the intersection is about \((1.4, 1.6)\). In graph b), the intersection is about \((-0.8, 2.6)\). 2. For graph a), \(f\) passes through \((0, 3)\) and \((3, 0)\), so its slope is \(-1\) and \(f(x) = -x + 3\). Line \(g\) passes through \((1, 1)\) and \((3, 4)\), so its slope is \(\frac{4 - 1}{3 - 1} = 1.5\). Using \((1, 1)\) gives a y-intercept of \(-0.5\), so \(g(x) = 1.5x - 0.5\). 3. For graph a), set the equations equal: \(-x + 3 = 1.5x - 0.5\). Then \(3.5 = 2.5x\), so \(x = 1.4\). Substitution gives \(y = 1.6\). Thus, \(S = (1.4, 1.6)\). 4. For graph b), \(f\) passes through \((0, 1)\) and \((1, -1)\), so \(f(x) = -2x + 1\). Line \(g\) passes through \((0, 3)\) and \((2, 4)\), so its slope is \(0.5\) and \(g(x) = 0.5x + 3\). 5. For graph b), set the equations equal: \(-2x + 1 = 0.5x + 3\). Then \(-2 = 2.5x\), so \(x = -0.8\). Substitution gives \(y = 2.6\). Thus, \(S = (-0.8, 2.6)\).

1) a) About \((1.4, 1.6)\); b) about \((-0.8, 2.6)\). 2) a) \(f(x) = -x + 3\) and \(g(x) = 1.5x - 0.5\); b) \(f(x) = -2x + 1\) and \(g(x) = 0.5x + 3\). 3) a) \(S = (1.4, 1.6)\); b) \(S = (-0.8, 2.6)\).

- Add the elapsed time \(t\) to \(6{:}00\) a.m. to find the clock time. - Zeros occur where the graph crosses the horizontal axis. - Compare the graph's highest value with its value at \(t=10\). - Average rate of change is change in temperature divided by change in time.

1. At \(t=2\), the time is \(8{:}00\) a.m. and the temperature is \(4\,\text{°F}\). At \(t=8\), the time is \(2{:}00\) p.m. and the temperature is \(-2\,\text{°F}\). 2. The graph crosses the horizontal axis at \(t=1\) and \(t=7.5\), corresponding to \(7{:}00\) a.m. and \(1{:}30\) p.m. These are the zeros, or x-intercepts, of the function. 3. The maximum is \(10\,\text{°F}\) at \(t=5\), and the ending temperature is \(-6\,\text{°F}\). The drop is \(10-(-6)=16\,\text{°F}\). 4. From \(t=0\) to \(t=2\), the average rate of change is \(\frac{4-(-4)}{2-0}=4\,\text{°F/h}\). From \(t=2\) to \(t=5\), it is \(\frac{10-4}{5-2}=2\,\text{°F/h}\). The temperature rises faster on the first interval.

a) \(t=2\): \(8{:}00\) a.m. and \(4\,\text{°F}\); \(t=8\): \(2{:}00\) p.m. and \(-2\,\text{°F}\) b) \(t=1\) and \(t=7.5\); zeros or x-intercepts c) \(16\,\text{°F}\) d) From \(t=0\) to \(t=2\); \(4\,\text{°F/h}\) compared with \(2\,\text{°F/h}\)

- Look for two points where the line passes exactly through grid intersections. - The zero is the x-value where the graph crosses the x-axis. - What y-intercept does every proportional relationship have?

1. From the graph, the line crosses the y-axis at \((0, -3)\), so \(b = -3\). Using \((0, -3)\) and \((4, 0)\), the slope is \(m = \frac{0 - (-3)}{4 - 0} = \frac{3}{4} = 0.75\). 2. Therefore, \(f(x) = 0.75x - 3\). Set \(f(x) = 0\): \(0.75x - 3 = 0\), so \(x = 4\). 3. A proportional relationship has y-intercept \(0\). With the same slope, its equation is \(p(x) = 0.75x\).

a) \(m = 0.75\), \(b = -3\) b) \(x = 4\) c) \(p(x) = 0.75x\)

- At a zero, the y-value is \(0\). - To find the slope exactly, use two points where the line passes through grid intersections.

1. The graph crosses the y-axis at \(-4\), so \(b = -4\). Using \((0, -4)\) and \((5, 0)\), the slope is \(m = \frac{0 - (-4)}{5 - 0} = \frac{4}{5} = 0.8\). Thus \(f(x) = 0.8x - 4\). 2. From the graph, the zero is \(x = 5\). 3. Verify algebraically: \(0.8x - 4 = 0\). Then \(0.8x = 4\), so \(x = 5\).

Equation: \(f(x) = 0.8x - 4\). Zero: \(x = 5\).

- The change in slope occurs when the water reaches the upper cylinder. - Compute slope as change in height divided by change in time. - With the same inflow, water rises faster in a narrower cylinder.

1. The lower section is full when the water reaches \(8\,\text{cm}\). The graph reaches this height at \(t=4\,\text{s}\). 2. In the lower section, the water rises \(8\,\text{cm}\) in \(4\,\text{s}\), or \(2\,\text{cm/s}\). In the upper section, it rises from \(8\) to \(18\,\text{cm}\) in \(2\,\text{s}\), or \(5\,\text{cm/s}\). The water rises faster in the upper section, so that cylinder has a smaller cross-sectional area and therefore a smaller diameter. 3. The first slope is \(m_1=\frac{8-0}{4-0}=2\,\text{cm/s}\). The second slope is \(m_2=\frac{18-8}{6-4}=5\,\text{cm/s}\). Each slope is the rate at which the water height increases.

a) \(4\,\text{s}\) b) The upper section; it has the smaller diameter. c) Lower section: \(2\,\text{cm/s}\); upper section: \(5\,\text{cm/s}\)

- The lake is represented by the greatest distance from home. - A horizontal segment represents a stop. - Convert minutes to hours before calculating miles per hour.

1. The greatest distance shown is \(9\) miles, so the lake is \(9\) miles from her home. 2. The horizontal segment begins after \(30\) minutes and ends after \(45\) minutes. The break lasts \(15\) minutes. 3. On the first segment, Lara travels \(6\) miles in \(30\) minutes, or \(0.5\) hour, so her speed is \(\frac{6}{0.5}=12\,\text{mph}\). On the last segment, she travels \(3\) miles in \(15\) minutes, or \(0.25\) hour, so her speed is \(\frac{3}{0.25}=12\,\text{mph}\). She travels equally fast on both riding segments.

a) \(9\) miles b) After \(30\) minutes; the break lasts \(15\) minutes. c) Both riding segments; \(12\,\text{mph}\)

- Read the first and highest points on the graph. - A decreasing graph segment represents downhill travel. - Divide elevation gain by horizontal distance to compare average steepness.

1. The route starts at \(500\,\text{ft}\) and reaches a maximum elevation of \(1500\,\text{ft}\) at the end. 2. The graph decreases from mile \(6\) to mile \(12\), so that is the downhill section. 3. From mile \(3\) to mile \(6\), the elevation gain is \(1200-600=600\,\text{ft}\) over \(3\) miles, or \(200\,\text{ft/mi}\). From mile \(12\) to mile \(18\), the gain is \(1500-900=600\,\text{ft}\) over \(6\) miles, or \(100\,\text{ft/mi}\). The middle climb is twice as steep on average, so the claim is correct.

a) Starting elevation: \(500\,\text{ft}\); maximum elevation: \(1500\,\text{ft}\) b) Mile \(6\) to mile \(12\) c) The claim is correct: \(200\,\text{ft/mi}\) versus \(100\,\text{ft/mi}\).

- Find the point where all four lines meet. - A point on the y-axis has x-coordinate \(0\). - Since all four lines share the same y-intercept, focus on finding each slope. - What is the slope of a horizontal line?

1. All four lines meet on the y-axis at \(-2\), so \(P(0, -2)\). 2. Because every line passes through \(P(0, -2)\), each has y-intercept \(-2\). For \(h_1\), using \((1, 0)\) gives slope \(2\), so \(h_1(x) = 2x - 2\). For \(h_2\), using \((2, -4)\) gives slope \(-1\), so \(h_2(x) = -x - 2\). For \(h_3\), using \((2, -1)\) gives slope \(\frac{1}{2}\), so \(h_3(x) = \frac{1}{2}x - 2\). For \(h_4\), using \((1, -5)\) gives slope \(-3\), so \(h_4(x) = -3x - 2\). 3. A line parallel to the x-axis has slope \(0\). Passing through \(P(0, -2)\), its equation is \(h_5(x) = -2\).

a) \(P(0, -2)\) b) \(h_1(x) = 2x - 2\); \(h_2(x) = -x - 2\); \(h_3(x) = \frac{1}{2}x - 2\); \(h_4(x) = -3x - 2\) c) \(h_5(x) = -2\)

- Since all the lines pass through the same point on the y-axis, what does that tell you about their y-intercepts? - From point \(A\), use another grid point on each line to calculate its slope. - A negative slope with smaller absolute value is closer to \(0\) and represents a less steep decreasing line.

1. Since every line passes through \(A(0, 1.5)\), each has y-intercept \(b = 1.5\). 2. For \(g_1\), using \((1, -0.5)\), the slope is \(m = \frac{-0.5 - 1.5}{1 - 0} = -2\). Thus \(g_1(x) = -2x + 1.5\). 3. For \(g_2\), using \((1, 2)\), the slope is \(m = \frac{2 - 1.5}{1 - 0} = 0.5\). Thus \(g_2(x) = 0.5x + 1.5\). 4. Line \(g_3\) is horizontal, so its slope is \(0\). Thus \(g_3(x) = 1.5\). 5. For \(g_4\), any slope satisfying \(-2 < m < 0\) works. For example, with \(m = -1\), \(g_4(x) = -x + 1.5\).

a) \(g_1(x) = -2x + 1.5\); \(g_2(x) = 0.5x + 1.5\); \(g_3(x) = 1.5\) b) They all have the same y-intercept, \(b = 1.5\). c) For example, \(g_4(x) = -x + 1.5\). Any slope \(m\) with \(-2 < m < 0\) works.

- What must be true about the slopes of parallel nonvertical lines? - Choose a line with two easy-to-read grid points and calculate rise over run. - The y-intercept is the value \(b\) in \(y = mx + b\).

1. Use two points on one line. On line \(l\), for example, \((0, 1)\) and \((4, -2)\) give \(m = \frac{-2 - 1}{4 - 0} = -\frac{3}{4}\). Since the lines are parallel, they all have this slope. 2. Read the y-intercepts from the graph. Line \(k\) has y-intercept \(3\), line \(l\) has y-intercept \(1\), and line \(n\) has y-intercept \(-1\). Therefore, \(k: y = -\frac{3}{4}x + 3\), \(l: y = -\frac{3}{4}x + 1\), and \(n: y = -\frac{3}{4}x - 1\). 3. A vertical shift changes only the y-intercept \(b\). Shifting up increases \(b\); shifting down decreases \(b\). The slope remains unchanged.

a) \(m = -\frac{3}{4}\) b) \(k: y = -\frac{3}{4}x + 3\); \(l: y = -\frac{3}{4}x + 1\); \(n: y = -\frac{3}{4}x - 1\) c) The y-intercept \(b\) changes, while the slope \(m\) stays the same.

- Since the candle gets shorter, what sign should the slope have? - Check how much one grid step represents on each axis. - Use two easy-to-read points to find the change in height per hour. - Substitute the requested time into your function.

1. Two convenient points are \((0, 8)\) and \((8, 4)\). 2. The slope is \(m = \frac{4 - 8}{8 - 0} = -\frac{1}{2}\). The initial height is \(8\), so \(h(t) = -\frac{1}{2}t + 8\). 3. \(h(12) = -\frac{1}{2} \cdot 12 + 8 = 2\). The graph also shows \((12, 2)\).

a) \(h(t) = -\frac{1}{2}t + 8\) b) \(2\,\text{in.}\)

- Use the coordinates of \(P\) and \(Q\) to find the slope. - Once you know the slope, substitute one point into \(y = mx + b\) to find \(b\). - To test \(R\), substitute its x-coordinate into your equation and compare the result with its y-coordinate.

1. The slope is \(m = \frac{-2 - 4}{2 - (-1)} = -2\). Using \(Q(2, -2)\), \(-2 = -2 \cdot 2 + b\), so \(b = 2\). Therefore, \(g(x) = -2x + 2\). 2. Substitute \(x = 4\): \(g(4) = -2 \cdot 4 + 2 = -6\). This matches the y-coordinate of \(R\), so \(R\) lies on the line.

a) \(g(x) = -2x + 2\) b) Yes, \(R(4, -6)\) lies on \(g\).

- What special point must a proportional graph contain? - Look at where the graph crosses the y-axis. - Choose two points and compare the horizontal and vertical changes. - Check whether the same rate of change holds elsewhere on the line.

1. A proportional relationship must graph as a line through \((0, 0)\). This graph crosses the y-axis at \(2\), so Maya is incorrect. 2. Using \((0, 2)\) and \((2, 3)\), the change in \(y\) is \(1\) when the change in \(x\) is \(2\). The slope is \(\frac{1}{2}=0.5\), and it is constant along the line. Therefore, Jordan is correct.

Maya is incorrect because the line does not pass through \((0, 0)\). Jordan is correct because the constant slope is \(0.5\), so every increase of \(2\) in \(x\) produces an increase of \(1\) in \(y\).

- Use two clear points on each line to determine its slope. - When you set the function expressions equal, collect the x-terms on one side. - Check whether your calculated point is consistent with your estimate from the graph.

1. For \(f\), the graph shows a y-intercept of \(3\) and points such as \((0, 3)\) and \((3, 0)\), so the slope is \(-1\). Thus, \(f(x) = -x + 3\). 2. For \(g\), the graph shows a y-intercept of \(1\) and points such as \((0, 1)\) and \((4, 2)\), so the slope is \(\frac{2 - 1}{4 - 0} = 0.25\). Thus, \(g(x) = 0.25x + 1\). 3. Set the functions equal: \(-x + 3 = 0.25x + 1\). 4. Then \(2 = 1.25x\), so \(x = 1.6\). 5. Substitute to get \(y = -1.6 + 3 = 1.4\). To two decimal places, the intersection is \(S = (1.60, 1.40)\).

Estimate: about \(S = (1.5, 1.5)\). To two decimal places: \(S = (1.60, 1.40)\).

- First determine each line's equation in slope-intercept form \(y = mx + b\). - Read the y-intercept where each line crosses the y-axis. - Use two clear points to calculate each slope. - At the intersection, both functions have the same output for the same input.

1. For \(f\), the marked points \((0, 1)\) and \((4, 3)\) give slope \(\frac{3 - 1}{4 - 0} = 0.5\), and the y-intercept is \(1\). Therefore, \(f(x) = 0.5x + 1\). 2. For \(g\), the marked points \((1, 3)\) and \((3, 1)\) give slope \(\frac{1 - 3}{3 - 1} = -1\). Using \((3, 1)\), the y-intercept is \(4\), so \(g(x) = -x + 4\). 3. Set the functions equal: \(0.5x + 1 = -x + 4\). 4. Then \(1.5x = 3\), so \(x = 2\). 5. Substituting gives \(y = -2 + 4 = 2\). The calculated intersection is \((2, 2)\), which matches the graph.

a) The graph gives an intersection of about \((2, 2)\). b) The equations are \(f(x) = 0.5x + 1\) and \(g(x) = -x + 4\), and the calculated intersection is \((2, 2)\), matching the graph.

- What graph feature identifies a proportional relationship? - A proportional line has a y-intercept of zero. - Use rise over run to find each slope.

1. A proportional graph passes through \((0, 0)\), so line \(p\) is proportional. 2. Line \(p\) also passes through \((2, 3)\), so its slope is \(3 \div 2=1.5\). Its equation is \(y=1.5x\). 3. Line \(q\) crosses the y-axis at \(3\) and passes through \((3, 0)\). Its slope is \((0-3) \div (3-0)=-1\), so its equation is \(y=-x+3\).

Line \(p\) is proportional and has equation \(y=1.5x\). Line \(q\) has equation \(y=-x+3\).

- The initial value is the y-intercept. - A steeper upward segment represents a faster increase in volume. - The amount removed is the decrease from the post-rain maximum to the final volume.

1. At \(t=0\), the tank contains \(50\) gallons. 2. From \(t=0\) to \(t=2\), the volume increases by \(50\) gallons in \(2\) hours, a rate of \(25\,\text{gal/h}\). From \(t=2\) to \(t=3\), it increases by \(50\) gallons in \(1\) hour, a rate of \(50\,\text{gal/h}\). The heaviest rain occurs from hour \(2\) to hour \(3\). 3. After the rain, the volume eventually decreases from \(150\) gallons to \(50\) gallons. The amount removed is \(150-50=100\) gallons.

a) \(50\) gallons b) From hour \(2\) to hour \(3\), because the slope is \(50\,\text{gal/h}\) rather than \(25\,\text{gal/h}\). c) \(100\) gallons

- With constant inflow, a steeper graph means the water height rises faster. - Water rises fastest where the vase has the smallest cross-sectional area.

1. The graph passes through \((2, 1)\), so the water is \(1\,\text{cm}\) high after \(2\) seconds. 2. The graph reaches \(9\,\text{cm}\) at \(t=6\) seconds. 3. With constant inflow, the water rises fastest where the vase is narrowest. The steepest segment is from \(t=2\) to \(t=6\), corresponding to water heights from \(1\,\text{cm}\) to \(9\,\text{cm}\).

1. \(1\,\text{cm}\) 2. \(6\) seconds 3. From \(1\,\text{cm}\) to \(9\,\text{cm}\)

- Identify the filling, waiting, and draining phases. - Rate of change is change in volume divided by change in time. - Use the constant drain rate to work backward from \(20\) gallons.

1. The volume increases linearly from \(0\) to \(20\) gallons in \(4\) minutes. After \(2\) minutes, the volume is \(10\) gallons. 2. The graph is horizontal from \(t=4\) to \(t=10\), so the volume remains unchanged for \(6\) minutes. 3. The fill rate is \(\frac{20}{4}=5\,\text{gal/min}\). 4. Draining begins at \(t=10\) with \(20\) gallons and ends at \(t=14\). The drain rate is \(5\,\text{gal/min}\). Reaching \(5\) gallons requires a decrease of \(15\) gallons, which takes \(3\) minutes. Therefore, the time is \(t=13\) minutes.

1) \(10\) gallons 2) \(6\) minutes 3) \(5\,\text{gal/min}\) 4) \(13\) minutes

- Horizontal segments represent constant elevation. - Use the highest point for the maximum. - On a straight segment, calculate the constant elevation change per mile. - Compare rise divided by run for the two climbs.

1. The drone starts at \(1500\,\text{ft}\) and finishes at \(1300\,\text{ft}\). 2. The graph is horizontal from mile \(2\) to mile \(4\), at \(2100\,\text{ft}\). 3. The maximum elevation is \(2900\,\text{ft}\) at mile \(6\). 4. From mile \(9\) to mile \(12\), elevation decreases from \(1900\) to \(1300\) feet, a rate of \(-200\,\text{ft/mi}\). At mile \(10\), the elevation is \(1900-200=1700\,\text{ft}\). 5. From mile \(0\) to mile \(2\), the average climb is \(\frac{2100-1500}{2}=300\,\text{ft/mi}\). From mile \(4\) to mile \(6\), it is \(\frac{2900-2100}{2}=400\,\text{ft/mi}\). The second climb is steeper.

1. Start: \(1500\,\text{ft}\); finish: \(1300\,\text{ft}\) 2. Mile \(2\) to mile \(4\) 3. \(2900\,\text{ft}\) 4. \(1700\,\text{ft}\) 5. Mile \(4\) to mile \(6\); \(400\,\text{ft/mi}\) versus \(300\,\text{ft/mi}\)

- Determine which graph segment contains the requested time. - Horizontal segments represent station stops. - Convert minutes to hours before calculating miles per hour. - Use the constant rate on the second moving segment.

1. At \(t=60\), the train is on the segment from \((50, 40)\) to \((90, 80)\), which has a rate of \(1\,\text{mi/min}\). Ten minutes after \(t=50\), the distance is \(40+10=50\) miles. 2. The train is stopped from minute \(40\) to minute \(50\) and from minute \(90\) to minute \(100\). The total stop time is \(10+10=20\) minutes. 3. On the first segment, the train travels \(40\) miles in \(40\) minutes, or \(\frac{2}{3}\) hour. Its speed is \(40\div\frac{2}{3}=60\,\text{mph}\). 4. Starting at minute \(50\), the train is \(40\) miles from the start and travels at \(1\,\text{mi/min}\). It needs \(20\) more minutes to reach \(60\) miles, so the time is \(70\) minutes.

1) \(50\) miles 2) \(20\) minutes 3) \(60\,\text{mph}\) 4) \(70\) minutes

- A horizontal graph segment represents no change. - Find a rate of change by dividing the change in water level by the change in time. - Read the initial water level at \(t = 0\). - For the final question, identify the segment where the graph passes from below to above \(160\,\text{cm}\).

1. From \(t = 4\) to \(t = 8\), the graph passes through \((4, 100)\) and \((8, 180)\). Its rate of change is \(\frac{180 - 100}{8 - 4} = 20\,\text{cm/h}\). Therefore, \(h(6) = 100 + 2 \cdot 20 = 140\,\text{cm}\). 2. The graph is horizontal from \(t = 2\) to \(t = 4\), so the water level remains constant for \(2\) hours. 3. During the first \(2\) hours, the level rises from \(40\,\text{cm}\) to \(100\,\text{cm}\). The rate is \(\frac{100 - 40}{2 - 0} = 30\,\text{cm/h}\). 4. The level reaches \(160\,\text{cm}\) between \(t = 4\) and \(t = 8\). Solve \(100 + 20(t - 4) = 160\). Then \(20(t - 4) = 60\), so \(t = 7\).

1) The water level is \(140\,\text{cm}\). 2) It remains constant for \(2\) hours. 3) The water level rises at \(30\,\text{cm/h}\). 4) The water level is \(160\,\text{cm}\) at \(t = 7\,\text{h}\).

- Read values from the graph at the requested times. - A horizontal segment means the elevation is unchanged. - For the first four hours, subtract the starting elevation from the ending elevation. - Compare the rates of change on the descending segments.

1. The graph gives \(h(2)=3000\,\text{ft}\). At \(t=6\), the point is halfway between \((5, 4200)\) and \((7, 2400)\), so \(h(6)=3300\,\text{ft}\). 2. The group remains at the maximum elevation from \(t=4\) to \(t=5\) hours. 3. The elevation rises continuously from \(1200\) feet at the start to \(4200\) feet at \(t=4\). The total gain is \(4200-1200=3000\,\text{ft}\). 4. From \(t=5\) to \(t=7\), the descent rate is \(\frac{2400-4200}{7-5}=-900\,\text{ft/h}\). From \(t=7\) to \(t=8\), it is \(\frac{1200-2400}{8-7}=-1200\,\text{ft/h}\). The group descends fastest from hour \(7\) to hour \(8\).

1. After \(2\) h: \(3000\,\text{ft}\); after \(6\) h: \(3300\,\text{ft}\) 2. From \(4\) to \(5\) hours 3. \(3000\,\text{ft}\) 4. From hour \(7\) to hour \(8\)

- The initial value is the graph's output when time is \(0\). - For each straight segment, compare the change in speed with the elapsed time. - Positive, zero, and negative rates correspond to three different behaviors of the speed function.

1. At \(t=0\), the graph has speed \(0\,\text{mph}\). The initial value is therefore \(0\,\text{mph}\), meaning the cyclist starts from rest. 2. From \(0\) to \(4\) minutes, the rate is \(\frac{8-0}{4-0}=2\,\frac{\text{mph}}{\text{min}}\). 3. From \(4\) to \(12\) minutes, the speed stays at \(8\,\text{mph}\), so the rate is \(0\,\frac{\text{mph}}{\text{min}}\). 4. From \(12\) to \(20\) minutes, the rate is \(\frac{16-8}{20-12}=1\,\frac{\text{mph}}{\text{min}}\). 5. From \(20\) to \(22\) minutes, the rate is \(0\,\frac{\text{mph}}{\text{min}}\). 6. From \(22\) to \(30\) minutes, the rate is \(\frac{0-16}{30-22}=-2\,\frac{\text{mph}}{\text{min}}\). 7. Positive rates on \(0\)–\(4\) and \(12\)–\(20\) mean the cyclist is speeding up. Zero rates on \(4\)–\(12\) and \(20\)–\(22\) mean constant speed. The negative rate on \(22\)–\(30\) means the cyclist is slowing down.

a) \(0\,\text{mph}\); the cyclist starts from rest. b) \(0\)–\(4\): \(2\,\frac{\text{mph}}{\text{min}}\); \(4\)–\(12\): \(0\); \(12\)–\(20\): \(1\,\frac{\text{mph}}{\text{min}}\); \(20\)–\(22\): \(0\); \(22\)–\(30\): \(-2\,\frac{\text{mph}}{\text{min}}\) c) Speeding up: \(0\)–\(4\) and \(12\)–\(20\); constant speed: \(4\)–\(12\) and \(20\)–\(22\); slowing down: \(22\)–\(30\).

- Read the graph value directly above the requested distance. - The x-intercept represents the distance when no fuel remains. - Divide the total fuel used by the total distance, then scale the rate to \(100\) miles. - Use \(y=mx+b\), where \(b\) is the starting amount of fuel.

1. At \(x=200\), the graph has value \(y=9\), so \(9\,\text{gal}\) remain. 2. The graph reaches the x-axis at \(x=500\), so the tank is empty after \(500\) miles. 3. The car uses \(15\,\text{gal}\) over \(500\) miles. The rate is \(15\div500=0.03\,\text{gal/mi}\). Over \(100\) miles, it uses \(0.03\cdot100=3\,\text{gal}\). 4. The initial value is \(15\), and the slope is \(\frac{0-15}{500-0}=-0.03\). Therefore, \(y=-0.03x+15\) for \(0\le x\le500\).

1) \(9\,\text{gal}\) 2) \(500\) miles 3) \(3\,\text{gal}\) per \(100\) miles 4) \(y=-0.03x+15\), for \(0\le x\le500\)

- Initial values occur when \(x=0\). - The phones have the same charge where the two graphs intersect. - Find each rate of change from two points on its line. - Use the form \(y=mx+b\), where \(b\) is the initial charge.

1. The y-intercepts are \(10\%\) for Phone A and \(40\%\) for Phone B, so Phone B starts with the greater charge. 2. The lines intersect at \((30, 55)\). After \(30\) minutes, both phones have a charge of \(55\%\). 3. Phone A's rate of change is \(\frac{100-10}{60-0}=1.5\) percentage points per minute. After \(50\) minutes, its charge is \(10+1.5\cdot50=85\%\). 4. For Phone A, the initial value is \(10\) and the rate is \(1.5\), so \(a(x)=1.5x+10\) for \(0\le x\le60\). For Phone B, the initial value is \(40\) and the rate is \(\frac{90-40}{100}=0.5\), so \(b(x)=0.5x+40\) for \(0\le x\le100\).

1) Phone B; Phone A starts at \(10\%\), and Phone B starts at \(40\%\). 2) After \(30\) minutes; both are at \(55\%\). 3) \(85\%\) 4) \(a(x)=1.5x+10\), for \(0\le x\le60\); \(b(x)=0.5x+40\), for \(0\le x\le100\)

- Read each initial value where \(t=0\). - Find each rate using change in volume divided by change in time. - Equal amounts occur where the two graph curves intersect.

1. At \(t=0\), Tank A contains \(5\,\text{L}\), and Tank B contains \(35\,\text{L}\). 2. Tank A rises from \(5\,\text{L}\) to \(15\,\text{L}\) in \(4\) minutes. Its rate is \(\frac{15-5}{4-0}=2.5\,\text{L/min}\). 3. Tank B falls from \(35\,\text{L}\) to \(15\,\text{L}\) in \(4\) minutes. Its rate of change is \(\frac{15-35}{4-0}=-5\,\text{L/min}\), so it is drained at \(5\,\text{L/min}\). 4. The graphs intersect at \((4, 15)\). After \(4\) minutes, both tanks contain \(15\,\text{L}\).

a) Tank A: \(5\,\text{L}\); Tank B: \(35\,\text{L}\) b) Tank A: \(2.5\,\text{L/min}\); Tank B: \(-5\,\text{L/min}\), or draining at \(5\,\text{L/min}\) c) After \(4\) minutes; \(15\,\text{L}\)

- Put the equation into slope-intercept form before looking at the choices. - In \(y=mx+b\), identify separately what \(m\) and \(b\) tell you about the graph. - A correct choice must match both the direction/steepness and the y-axis crossing.

1. Solve for \(y\): \(x-2y=4\) gives \(-2y=4-x\), so \(y=\frac{1}{2}x-2\). 2. The rate of change is \(m=\frac{1}{2}\), and the initial value is \(b=-2\). 3. Graph a) rises from left to right with slope \(\frac{1}{2}\) and crosses the y-axis at \(-2\). The other graphs have either the wrong slope sign or the wrong initial value.

a) \(y=\frac{1}{2}x-2\); \(m=\frac{1}{2}\), \(b=-2\) b) a)

- For \(g_1\) and \(g_2\), first read the y-intercept. - Use another grid point to calculate the slope. - The intersection is where the two lines cross.

1. Line \(g_1\) has y-intercept \(-2\) and passes through \((4, 0)\). Its slope is \(m = \frac{0 - (-2)}{4 - 0} = \frac{1}{2}\), so \(g_1: y = \frac{1}{2}x - 2\). 2. Line \(g_2\) has y-intercept \(4\) and passes through \((4, 0)\). Its slope is \(m = \frac{0 - 4}{4 - 0} = -1\), so \(g_2: y = -x + 4\). 3. Line \(g_3\) is horizontal at \(y = 1\), so \(g_3: y = 1\). 4. The graph shows \(g_1\) and \(g_2\) crossing at \((4, 0)\).

\(g_1: y = \frac{1}{2}x - 2\); \(g_2: y = -x + 4\); \(g_3: y = 1\) Intersection of \(g_1\) and \(g_2\): \((4, 0)\)