Free math worksheets

- A function assigns exactly one output to each input. - Repeated outputs are allowed. - Look for an input that appears with two different outputs.

1. a) Yes. Each input in \(\{1,2,3,4\}\) is paired with exactly one output. 2. b) Yes. Each input has exactly one output. Different inputs may share the same output without violating the definition of a function. 3. c) No. The input \(5\) is paired with both \(2\) and \(4\).

a) Yes; it is a function. b) Yes; it is a function. c) No; \(x=5\) has two different outputs.

- A function must assign exactly one output to each input. - For each assignment, ask whether one input can be paired with two separate outputs. - Treat each individual plate number in part b as its own output.

1. a) This is a function because each date has one recorded daily-high value at the fixed station. 2. b) This is not necessarily a function because one owner may have several registered vehicles and therefore may be assigned several separate license-plate outputs. 3. c) This is a function because each US state has exactly one state capital.

a) Function b) Not a function c) Function

- Check the arrows leaving each input, one input at a time. - A function may send different inputs to the same output, but one input cannot point to two different outputs. - For part c, change only the arrows connected to the input that violates the rule.

1. A function must assign exactly one output to each input. 2. Input \(2\) points to both \(5\) and \(6\), so the correspondence is not a function. 3. Removing either \(2\rightarrow5\) or \(2\rightarrow6\) leaves input \(2\) with exactly one output, while inputs \(1\) and \(3\) already have one output each.

a) No. Input \(2\) has two outputs. b) \(2\) c) Either \(2\rightarrow5\) or \(2\rightarrow6\)

- Substitute each distance for \(x\). - The base fare stays the same for every trip. - Multiply before adding.

1. For \(x=5\), \(T(5)=3.50+1.80\cdot5=3.50+9.00=12.50\). 2. For \(x=12\), \(T(12)=3.50+1.80\cdot12=3.50+21.60=25.10\). 3. For \(x=0.5\), \(T(0.5)=3.50+1.80\cdot0.5=3.50+0.90=4.40\).

a) \(\$12.50\) b) \(\$25.10\) c) \(\$4.40\)

- Substitute each input for \(x\). - Multiply before subtracting. - Compare your results with the completed entries to check the pattern.

1. For \(x=1\), \(T(1)=7\cdot1-5=2\). 2. For \(x=5\), \(T(5)=7\cdot5-5=30\). 3. For \(x=20\), \(T(20)=7\cdot20-5=135\).

The missing values are \(2\) for \(x=1\), \(30\) for \(x=5\), and \(135\) for \(x=20\).

- Substitute the given input into each function. - Evaluate inside parentheses first. - Be careful when multiplying two negative numbers.

1. \(T_1(-0.75)=4\cdot(-0.75+1.5)=4\cdot0.75=3\). 2. \(T_2(-0.75)=2-6\cdot(-0.75)=2+4.5=6.5\). 3. Since \(6.5>3\), \(T_2\) has the greater value.

\(T_1(-0.75)=3\) and \(T_2(-0.75)=6.5\). Therefore, \(T_2\) has the greater value.

- Substitute each input for \(x\). - Multiply before subtracting. - Simplify fractions during the calculation. - Equal quantities have a difference of \(0\).

1. \(T(3)=\frac{2}{3}(3)-\frac{1}{2}=2-\frac{1}{2}=\frac{3}{2}\). 2. \(T\left(\frac{3}{4}\right)=\frac{2}{3}\cdot\frac{3}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0\). 3. \(T(6)=\frac{2}{3}(6)-\frac{1}{2}=4-\frac{1}{2}=\frac{7}{2}\). 4. \(T\left(\frac{1}{4}\right)=\frac{2}{3}\cdot\frac{1}{4}-\frac{1}{2}=\frac{1}{6}-\frac{3}{6}=-\frac{1}{3}\).

a) \(\frac{3}{2}\) b) \(0\) c) \(\frac{7}{2}\) d) \(-\frac{1}{3}\)

- Substitute the input for \(x\). - Use parentheses when substituting a negative number. - Convert \(\frac{1}{2}\) to a decimal if helpful.

1. \(T(8.9)=-15.4+8.9=-6.5\). 2. \(T(-3.7)=-15.4+(-3.7)=-19.1\). 3. \(T(15.4)=-15.4+15.4=0\). 4. Since \(\frac{1}{2}=0.5\), \(T\left(\frac{1}{2}\right)=-15.4+0.5=-14.9\).

a) \(-6.5\) b) \(-19.1\) c) \(0\) d) \(-14.9\)

- Find the entries where the output is \(0\). - The y-intercept has input \(x=0\). - Match the input \(4\) with its output.

1. Zeros occur where \(g(x)=0\). The table shows \(g(-1)=0\) and \(g(2)=0\), so the zeros are \(-1\) and \(2\). 2. The y-intercept occurs when \(x=0\). Since \(g(0)=-4\), the y-intercept is \((0, -4)\). 3. The table shows \(g(4)=5\).

1) \(-1\) and \(2\) 2) \((0, -4)\) 3) \(5\)

- Apply the definition of a function to the time values. - Compare the differences between consecutive temperatures. - For part c, compare the highest and lowest recorded temperatures.

1. Each recorded time is paired with exactly one temperature, so the data define a function. 2. From \(3{:}00\) p.m. to \(6{:}00\) p.m., the change is \(39-46=-7\,^\circ\text{F}\). From \(6{:}00\) p.m. to \(9{:}00\) p.m., the change is \(31-39=-8\,^\circ\text{F}\). The greatest decrease occurred from \(6{:}00\) p.m. to \(9{:}00\) p.m. 3. The highest temperature is \(46\,^\circ\text{F}\) and the lowest is \(28\,^\circ\text{F}\). The spread is \(46-28=18\,^\circ\text{F}\).

a) Each time has exactly one recorded temperature. b) From \(6{:}00\) p.m. to \(9{:}00\) p.m.; the temperature decreased by \(8\,^\circ\text{F}\). c) \(18\,^\circ\text{F}\)

- In a list of ordered pairs, repeated x-values with different y-values violate the function rule. - Change one of the two points whose x-coordinate is \(2\). - To test the reverse assignment, switch the roles of \(x\) and \(y\).

1. The input \(x=2\) occurs in \(P_2(2, 6)\) and \(P_4(2, 1)\) with two different outputs. Therefore, \(M\) does not represent a function. 2. One possible change is to replace \(P_4(2, 1)\) with \(P_4(4, 1)\). The new x-values are \(1\), \(2\), \(3\), and \(4\), so each input has exactly one output. 3. In the new set, the output \(y=3\) occurs at both \(x=1\) and \(x=3\). Therefore, the reverse assignment is not a function.

a) \(x=2\) is paired with both \(y=6\) and \(y=1\). b) One answer is \(P_4(4, 1)\). c) No. The input \(y=3\) in the reverse assignment would have outputs \(x=1\) and \(x=3\).

- Treat each x-value as an input and calculate the corresponding function output. - For an ordered pair to belong to the function, its second coordinate must equal the function output for its first coordinate. - For part c), focus on how many outputs the rule assigns to one fixed input.

1. \(f(-2)=-1.5\cdot(-2)+3=6\), so \(A(-2,6)\) is an input-output pair of \(f\). 2. \(f(4)=-1.5\cdot4+3=-3\), so \(B(4,-3)\) is an input-output pair of \(f\). 3. Solve \(f(k)=0\): \(0=-1.5k+3\), so \(k=2\). 4. The rule gives exactly one value of \(f(x)\) for every allowed input \(x\). That one-output-per-input property makes \(f\) a function.

a) \(f(-2)=6\) and \(f(4)=-3\); both \(A\) and \(B\) are input-output pairs of \(f\). b) \(k=2\) c) Each input is assigned exactly one output by the rule.

- Write the coordinate that stays constant on each axis. - A function may assign only one y-value to each x-value. - Apply the vertical line test.

1. The x-axis has equation \(y=0\). It is the graph of the linear function \(f(x)=0\), because every real input has exactly one output, \(0\). 2. The y-axis has equation \(x=0\). It is not the graph of \(y=f(x)\) because the input \(x=0\) is paired with infinitely many y-values. Equivalently, it fails the vertical line test.

a) \(y=0\); yes, it is the graph of a linear function. b) \(x=0\); no, it is not the graph of \(y=f(x)\).

- Calculate the output of the rule for each listed input before judging any ordered pair. - An ordered pair belongs to the function only when its y-value matches the function output. - Seeing a number written in a proposed pair does not make it an output of the function.

1. \(g(2)=1.2\cdot2+0.5=2.9\). 2. \(g(-1)=1.2(-1)+0.5=-0.7\). 3. \(g(5)=1.2\cdot5+0.5=6.5\). 4. Therefore, \((2,2.9)\) and \((5,6.5)\) are input-output pairs of \(g\), while \((-1,-1)\) is not because \(g(-1)=-0.7\). 5. The number \(-1\) in \((-1,-1)\) is a proposed second coordinate, not another output produced by the rule. For input \(-1\), the rule gives only \(-0.7\).

a) \(g(2)=2.9\), \(g(-1)=-0.7\), and \(g(5)=6.5\) b) \((2,2.9)\) and \((5,6.5)\) c) The rule assigns the input \(-1\) only the output \(-0.7\); the proposed pair \((-1,-1)\) does not belong to \(g\).

- Substitute each input for \(x\). - Use parentheses around negative inputs. - Work with either fractions or decimals consistently.

1. \(T(3)=-4\cdot3+\frac{1}{2}=-12+0.5=-11.5\). 2. \(T(-1.5)=-4\cdot(-1.5)+0.5=6.5\). 3. \(T\left(\frac{3}{8}\right)=-4\cdot\frac{3}{8}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1\).

a) \(-11.5\) b) \(6.5\) c) \(-1\)

- Evaluate each function separately at the requested input. - Apply exponents before multiplication and subtraction. - For part c), compare the two outputs at the same input and subtract the smaller from the larger.

1. For \(y=3\), \(T_1(3)=5\cdot3-15=0\) and \(T_2(3)=3^2-3\cdot3=0\). 2. For \(y=6\), \(T_1(6)=5\cdot6-15=15\) and \(T_2(6)=6^2-3\cdot6=18\). 3. At \(y=6\), \(T_2\) is greater by \(18-15=3\).

a) \(T_1(3)=0\) and \(T_2(3)=0\) b) \(T_1(6)=15\) and \(T_2(6)=18\) c) \(T_2\) is greater by \(3\).

- In \(K(s)\), the input is mileage and the output is cost. - For a known input, substitute the mileage into the function rule. - For a known output, set the function equal to that cost and solve for the input.

1. \(K(80)=42+0.15\cdot80=54\). This means an \(80\)-mile day costs \(\$54.00\). 2. \(K(240)=42+0.15\cdot240=78\), so a \(240\)-mile day costs \(\$78.00\). 3. Solve \(60=42+0.15s\). Then \(18=0.15s\), so \(s=120\). A one-day cost of \(\$60.00\) corresponds to driving \(120\) miles.

a) \(K(80)=54\), meaning \(80\) miles costs \(\$54.00\). b) \(K(240)=78\), so the cost is \(\$78.00\). c) \(s=120\), meaning a \(\$60.00\) day corresponds to \(120\) miles.

- A relation defines \(y\) as a function of \(x\) when each allowed x-value corresponds to exactly one y-value. - Try solving each equation for \(y\). - A single x-value that produces two different y-values is enough to show that a relation is not a function.

1. Equation (1) gives \(y^3=27-x\), so \(y=\sqrt[3]{27-x}\). Every real x-value produces exactly one real y-value, so it defines a function. 2. For equation (2), choose \(x=0\). Then \(y^2=16\), so \(y=4\) or \(y=-4\). One x-value has two y-values, so the equation does not define \(y\) as a function of \(x\). 3. Equation (3) gives \(y=|x|+2\). Every real x-value produces exactly one y-value, so it defines a function.

(1) Function (2) Not a function (3) Function

- A relation defines \(y\) as a function of \(x\) only when each allowed x-value corresponds to exactly one y-value. - Try isolating \(y\). - Equations involving \(y^2\) or \(|y|\) may produce two y-values. - Check values that make the coefficient of \(y\) equal to \(0\).

1. For a), solving for \(y\) gives \(y=x^2+5\). Each x-value gives exactly one y-value, so this is a function. 2. For b), choose \(x=3\). Then \(y^2=4\), so \(y=2\) or \(y=-2\). One x-value has two y-values, so this is not a function. 3. For c), the equation reduces to \(-2x=6\), so \(x=-3\). When \(x=-3\), every real value of \(y\) satisfies the equation. Therefore, this is not a function. 4. For d), choose \(x=0\). Then \(|y|=3\), so \(y=3\) or \(y=-3\). Therefore, this is not a function. 5. For e), choose \(x=2\). Then \(0y=0\), which is true for every real \(y\). Therefore, this is not a function.

a) Function: \(y=x^2+5\) b) Not a function c) Not a function d) Not a function e) Not a function

- Use the definition: each input may have only one output. - A single input with two possible outputs is enough to disprove that a relation is a function. - For an equation already solved for \(y\), ask whether the right-hand side produces one value or more than one value for each input.

1. For a), choose \(x=4\). Both \(y=2\) and \(y=-2\) satisfy the relation, so one input has two outputs. It is not a function. 2. In b), the explicit formula gives exactly one value of \(y\) for every real \(x\), so it is a function. 3. In c), the expression \(2^x\) gives exactly one positive value for every real \(x\), so it is a function. 4. In d), the polynomial expression gives exactly one real output for every real input, so it is a function.

a) Not a function; for example, \(x=4\) gives \(y=2\) and \(y=-2\). b) Function; each real input gives one output. c) Function; each real input gives one output. d) Function; each real input gives one output.

- To evaluate a graphically defined function, start with the input on the horizontal axis and read its one output. - For part b), look for every place the graph reaches the same horizontal level. - A function may repeat an output for different inputs; the restriction is on one input having more than one output. - Reverse the roles of input and output when considering part d).

1. Reading the graph gives \(T(2)=-12\) and \(T(14)=-4\). 2. The horizontal level \(-8\) meets the graph at \(t=8\) and \(t=20\). 3. Every time input on the displayed domain corresponds to exactly one temperature output, so \(T\) is a function of time. 4. The reverse relation is not a function because the input temperature \(-8\) would have two outputs, \(8\) and \(20\).

a) \(T(2)=-12\) and \(T(14)=-4\) b) \(t=8\) and \(t=20\) c) Each time input has exactly one temperature output. d) No. In the reverse relation, the input \(-8\) would have two outputs, \(8\) and \(20\).

- Recall that a function assigns exactly one output to each input. - Read the graph at \(t = 0, 2, 4, 6, 8, 10,\) and \(12\). - To find when the depth is \(4\,\text{m}\), trace horizontally from \(h = 4\) to the graph. - For the reverse relation, ask whether one depth can occur at two different times.

1. Read the plotted values: <table> <tr><th>Time \(t\) (h)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(8\)</td><td>\(10\)</td><td>\(12\)</td></tr> <tr><th>Depth \(h\) (m)</th><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td><td>\(5\)</td><td>\(4\)</td><td>\(2\)</td><td>\(1\)</td></tr> </table> 2. The relation from time to depth is a function because each time is paired with exactly one depth. 3. The graph has depth \(4\,\text{m}\) at \(t = 4\,\text{h}\) and \(t = 8\,\text{h}\). 4. The reverse relation is not a function because the input depth \(4\,\text{m}\) would be paired with two different times.

a) <table> <tr><th>Time \(t\) (h)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(8\)</td><td>\(10\)</td><td>\(12\)</td></tr> <tr><th>Depth \(h\) (m)</th><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td><td>\(5\)</td><td>\(4\)</td><td>\(2\)</td><td>\(1\)</td></tr> </table> Yes. Time to depth is a function because each time has exactly one depth. b) The depth is \(4\,\text{m}\) at \(t = 4\,\text{h}\) and \(t = 8\,\text{h}\). The reverse relation is not a function because one depth can correspond to more than one time.

- Identify which coordinate stays constant on each line. - A vertical line has a constant x-coordinate; a horizontal line has a constant y-coordinate. - Apply the vertical line test.

1. Line \(g\) is vertical at \(x=3\), so one equation is \(x+0y=3\). 2. Line \(g\) is not the graph of a function \(y=f(x)\) because the input \(x=3\) corresponds to infinitely many y-values. It fails the vertical line test. 3. Line \(h\) is horizontal at \(y=-2\), so one equation is \(0x+y=-2\). 4. Line \(h\) is the graph of the constant function \(f(x)=-2\), because each input has exactly one output.

a) \(g:\ x+0y=3\); \(h:\ 0x+y=-2\) b) \(g\) is not a function graph; \(h\) is a function graph.

- Compare an upper semicircle with a full circle. - Apply the vertical line test.

1. The graph is the upper semicircle \(y=\sqrt{4-x^2}\) for \(-2\le x\le2\). 2. Every input in \([-2, 2]\) corresponds to exactly one output on the upper semicircle. Equivalently, every vertical line meets the graph at most once. 3. Therefore, the graph represents a function.

Yes. The graph represents a function because each \(x\in[-2, 2]\) has exactly one y-value.

- Look for a vertical line that intersects the graph more than once. - Check the graph at \(x=2\).

1. A graph represents a function only if every input \(x\) has at most one output \(y\). 2. The curve is a sideways V. For example, the vertical line \(x=2\) meets the curve at \((2, 3)\) and \((2, 1)\). 3. Therefore, the curve fails the vertical line test and is not the graph of a function.

No. At \(x=2\), the graph has two y-values, \(3\) and \(1\).

- Evaluate all three functions at the same input before comparing their outputs. - In the equation \(g(t)=w(t)+s(t)\), what must the solar output be when \(g(t)=w(t)\)? - For the function question, fix one input time and count how many outputs the rule for \(g\) can produce.

1. At noon, the graph gives \(w(12)=2\,\text{MW}\) and \(s(12)=3.6\,\text{MW}\). Therefore, \(g(12)=2+3.6=5.6\,\text{MW}\). 2. Since \(g(t)=w(t)+s(t)\), equality \(g(t)=w(t)\) occurs exactly when \(s(t)=0\). The solar graph is zero from midnight through \(6{:}00\) a.m. and from \(6{:}00\) p.m. through midnight. 3. For each time input \(t\), the functions \(w\) and \(s\) each give exactly one output. Adding those two outputs gives exactly one value of \(g(t)\), so \(g\) is also a function.

a) \(w(12)=2\,\text{MW}\), \(s(12)=3.6\,\text{MW}\), and \(g(12)=5.6\,\text{MW}\) b) Midnight–\(6{:}00\) a.m. and \(6{:}00\) p.m.–midnight, because those are the times when \(s(t)=0\). c) Each input \(t\) produces one wind output and one solar output, so their sum is one unique output \(g(t)\).

- Evaluate the function rule first; do not decide only by how close a marked point looks to the line. - Compare each point's second coordinate with the output for its first coordinate. - For part c), apply the definition that one input cannot have two different outputs in a function.

1. \(f(2)=1.5\cdot2-2=1\). 2. \(f(-1)=1.5(-1)-2=-3.5\). 3. \(f(4)=1.5\cdot4-2=4\). 4. Therefore, \(A(2,1)\) and \(B(-1,-3.5)\) are input-output pairs of \(f\), while \(C(4,3.5)\) is not. 5. For input \(4\), the rule assigns the single output \(4\). Allowing \(3.5\) as another output for that same input would contradict the one-output-per-input property of a function.

a) \(f(2)=1\), \(f(-1)=-3.5\), and \(f(4)=4\) b) \(A\) and \(B\) c) The only output for input \(4\) is \(4\); \((4,3.5)\) is not a pair of \(f\).

- Move an imaginary vertical line across each graph. - A function graph may be intersected at most once by any vertical line.

1. Graph a) is a function because every vertical line intersects the parabola at most once. 2. Graph b) is not a function because vertical lines through \(-2<x<2\) intersect the circle twice. 3. Graph c) is a function because every vertical line intersects the line at most once. 4. Graph d) is not a function because vertical lines through \(0<x\le4\) intersect the sideways V twice.

Graphs a) and c) are functions. Graphs b) and d) are not functions.

- The domain consists of the values used as inputs. - The range consists of distinct output values; repeated outputs are not repeated when writing a set. - For the reversed relation, imagine swapping the two entries in every row and then apply the one-output-per-input rule again.

1. The domain is the set of input values: \(\{-2,0,3,5\}\). 2. The range is the set of output values that occur: \(\{1,4,9\}\). The repeated output \(4\) is listed only once in the set. 3. The inputs \(-2\) and \(3\) both have output \(4\). 4. After reversing the pairs, input \(4\) would be paired with outputs \(-2\) and \(3\). Because one reversed input would have two outputs, the reversed relation would not be a function.

a) \(\{-2,0,3,5\}\) b) \(\{1,4,9\}\) c) \(-2\) and \(3\) d) No. In the reversed relation, input \(4\) would have two outputs, \(-2\) and \(3\).

- Evaluate both functions for each input. - Organize the outputs in matching rows. - Look for a column with equal outputs.

1. The outputs for \(A(x)\) are \(16,18,20,22,\) and \(24\). 2. The outputs for \(B(x)\) are \(12,16,20,24,\) and \(28\). 3. Both functions have output \(20\) when \(x=5\).

a) <table> <tr><th>\(x\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th><th>\(7\)</th></tr> <tr><th>\(A(x)\)</th><td>\(16\)</td><td>\(18\)</td><td>\(20\)</td><td>\(22\)</td><td>\(24\)</td></tr> <tr><th>\(B(x)\)</th><td>\(12\)</td><td>\(16\)</td><td>\(20\)</td><td>\(24\)</td><td>\(28\)</td></tr> </table> b) \(x=5\); both outputs are \(20\).

- Use as many three-scoop deals as possible, then compare with other combinations. - Check whether any fractional number of scoops is a valid input.

1. For \(7\) scoops, buy two three-scoop cups and one single scoop: \(2(\$4.00)+\$1.50=\$9.50\). Seven single scoops would cost \(7(\$1.50)=\$10.50\). 2. The lowest prices are: <table><tr><td>Number of scoops</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td></tr><tr><td>Lowest price</td><td>\(\$1.50\)</td><td>\(\$3.00\)</td><td>\(\$4.00\)</td><td>\(\$5.50\)</td><td>\(\$7.00\)</td><td>\(\$8.00\)</td><td>\(\$9.50\)</td><td>\(\$11.00\)</td></tr></table> 3. The domain consists of whole numbers because the shop sells whole scoops. Values between consecutive whole numbers do not represent possible purchases, so the graph would contain separate points rather than connected line segments.

a) Two three-scoop cups and one single scoop, costing \(\$9.50\) b) <table><tr><td>Number of scoops</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td></tr><tr><td>Lowest price</td><td>\(\$1.50\)</td><td>\(\$3.00\)</td><td>\(\$4.00\)</td><td>\(\$5.50\)</td><td>\(\$7.00\)</td><td>\(\$8.00\)</td><td>\(\$9.50\)</td><td>\(\$11.00\)</td></tr></table> c) The relation is discrete because only whole numbers of scoops can be purchased.

- Identify the lightest and heaviest possible combinations. - Find how much the weight changes when one workbook is replaced by one textbook. - Decide whether a fractional number of textbooks is possible.

1. The minimum occurs with \(6\) workbooks: \(6(0.25)=1.5\,\text{lb}\). The maximum occurs with \(6\) textbooks: \(6(1.5)=9\,\text{lb}\). 2. Replacing one workbook with one textbook increases the total weight by \(1.5-0.25=1.25\,\text{lb}\). The completed table is: <table><tr><td>Number of textbooks</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr><tr><td>Total weight (lb)</td><td>\(1.5\)</td><td>\(2.75\)</td><td>\(4\)</td><td>\(5.25\)</td><td>\(6.5\)</td><td>\(7.75\)</td><td>\(9\)</td></tr></table> 3. The number of textbooks must be a whole number. Inputs such as \(1.5\) textbooks have no meaning, so the graph is discrete and the points should not be connected.

a) Minimum: \(1.5\,\text{lb}\); maximum: \(9\,\text{lb}\) b) <table><tr><td>Number of textbooks</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr><tr><td>Total weight (lb)</td><td>\(1.5\)</td><td>\(2.75\)</td><td>\(4\)</td><td>\(5.25\)</td><td>\(6.5\)</td><td>\(7.75\)</td><td>\(9\)</td></tr></table> c) No. Only whole-number inputs are meaningful.

- 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 \(4\).

- Substitute the given input for \(x\). - Evaluate the expression inside parentheses before multiplying. - Simplify fractions before multiplying when possible. - Check the sign of the final result.

1. Substitute \(x=\frac{3}{4}\): \(T\left(\frac{3}{4}\right)=\frac{2}{3}\cdot\frac{3}{4}-4\cdot\left(\frac{3}{4}+\frac{1}{2}\right)\). 2. The first term is \(\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}\). 3. Inside the parentheses, \(\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\). 4. Then \(4\cdot\frac{5}{4}=5\). 5. Therefore, \(T\left(\frac{3}{4}\right)=\frac{1}{2}-5=-\frac{9}{2}=-4.5\).

\(-\frac{9}{2}\), or \(-4.5\)

- Substitute each given input into both functions. - Compare the outputs at the smaller and larger inputs. - Test integer values between \(2\) and \(10\).

1. For \(k=2\), \(T_1(2)=2\cdot2+10=14\) and \(T_2(2)=40-3\cdot2=34\). 2. For \(k=10\), \(T_1(10)=2\cdot10+10=30\) and \(T_2(10)=40-3\cdot10=10\). 3. \(T_1\) increases as \(k\) increases, while \(T_2\) decreases. 4. Testing \(k=6\) gives \(T_1(6)=22\) and \(T_2(6)=22\), so the outputs are equal.

a) For \(k=2\): \(T_1=14\), \(T_2=34\). For \(k=10\): \(T_1=30\), \(T_2=10\). b) \(T_1\) increases and \(T_2\) decreases. c) \(k=6\); both outputs are \(22\).

- Substitute the requested number of minutes into each cost rule. - For part b), compare the two outputs for the same input. - For part c), identify which plan's cost must match the fixed monthly fee, then solve for the number of minutes.

1. For \(m=50\), \(K_A=5+0.12\cdot50=11\) and \(K_B=0.20\cdot50=10\). 2. For \(m=100\), \(K_A=5+0.12\cdot100=17\) and \(K_B=0.20\cdot100=20\). Plan A costs less. 3. Set Plan B's cost equal to \(5\): \(0.20m=5\). Then \(m=5\div0.20=25\).

a) Plan A: \(\$11.00\); Plan B: \(\$10.00\) b) Plan A; it costs \(\$17.00\), compared with \(\$20.00\) for Plan B. c) \(25\) minutes

- Substitute the given input into both functions. - Equal function values must have identical outputs. - Follow the order of operations when evaluating.

1. \(T_1(3)=24-4\cdot3=12\) and \(T_2(3)=2\cdot3+6=12\). 2. \(T_1(5)=24-4\cdot5=4\) and \(T_2(5)=2\cdot5+6=16\). Since \(4\ne 16\), the claim is false. 3. \(T_1(10)=24-4\cdot10=-16\).

a) \(T_1(3)=12\) and \(T_2(3)=12\) b) The claim is false because \(4\ne 16\). c) \(T_1(10)=-16\)

- Think of a graph that never meets the x-axis. - Apply the function rule at \(x=0\). - Interpret what the coordinate \(y=0\) means.

1. Statement a is false. For example, the constant function \(f(x)=2\) never equals \(0\). 2. Statement b is false. A y-axis intersection has input \(x=0\), and a function can assign at most one output to that input. 3. Statement c is true. If \((4, 0)\) lies on the graph, then \(f(4)=0\), which means \(4\) is a zero.

a) False; for example, \(f(x)=2\) has no zero. b) False; a function can have at most one output at \(x=0\). c) True; \((4, 0)\) means \(f(4)=0\).

- Look for an interval in which the battery level increases. - Divide the total decrease by the elapsed time. - For the reverse assignment, check whether one battery percentage appears at more than one time.

1. The battery level rises from \(44\%\) at \(4{:}00\) p.m. to \(88\%\) at \(6{:}00\) p.m., so the phone was probably charging during that interval. 2. From \(7{:}00\) a.m. to \(12{:}00\) p.m. is \(5\) hours. The battery decreases by \(100-60=40\) percentage points, so the average use is \(40\div5=8\) percentage points per hour. 3. The reverse assignment is not a function because the battery level \(60\%\) corresponds to both \(12{:}00\) p.m. and \(2{:}00\) p.m.

a) From \(4{:}00\) p.m. to \(6{:}00\) p.m. b) \(8\) percentage points per hour c) No. The input \(60\%\) would have two outputs: \(12{:}00\) p.m. and \(2{:}00\) p.m.

- Substitute the given x-value into the function to find its output. - Moving vertically changes the y-coordinate but not the x-coordinate. - To test a point, evaluate the function at the point's x-coordinate and compare the result with its y-coordinate.

1. \(f(4) = 1.5 \cdot 4 - 2 = 4\), so \(P(4, 4)\). 2. Moving \(5\) units vertically upward keeps the x-coordinate the same and adds \(5\) to the y-coordinate. Thus \(Q(4, 9)\). 3. \(g(4) = 1.5 \cdot 4 + 3 = 9\). This matches the y-coordinate of \(Q\), so \(Q\) lies on the graph of \(g\).

1) \(y_P = 4\) 2) \(Q(4, 9)\) 3) Yes, because \(g(4) = 9\).

- Find a number in \(D\) with more than one distinct prime factor. - The revised rule must produce exactly one output even for \(n=1\). - Ask how many absolute values one rational number has.

1. The input \(6\) has the distinct prime factors \(2\) and \(3\), so it would receive two different outputs. Therefore, the rule does not define a function. 2. One valid revision is: “Assign each \(n\) the number of its distinct prime factors.” This rule assigns exactly one output to every input, including assigning \(0\) to \(1\). 3. Every rational number has exactly one absolute value, so \(x\mapsto|x|\) is a function.

a) No. For example, \(6\) would have outputs \(2\) and \(3\). b) Example: Assign each \(n\) the number of its distinct prime factors, with \(1\mapsto0\). c) Yes. Every rational number has exactly one absolute value.

- Pay close attention to which endpoints are included in each interval. - A function assigns exactly one output to every input in its domain. - In the reverse direction, ask whether a postage value determines one exact weight.

1. Apply the rate interval that contains each weight. The postage values are \(\$0.85\), \(\$0.85\), \(\$1.00\), \(\$1.00\), and \(\$1.60\). 2. Each weight in \(0<m\le16\) belongs to exactly one rate interval, so it has exactly one postage value. Therefore, weight \(\rightarrow\) postage is a function. 3. The reverse assignment is not a function. For example, the postage value \(\$1.00\) corresponds to every weight in \(1<m\le2\), not to one unique weight.

a) <table><tr><td>Weight (oz)</td><td>\(0.5\)</td><td>\(1\)</td><td>\(1.5\)</td><td>\(2\)</td><td>\(5\)</td></tr><tr><td>Postage</td><td>\(\$0.85\)</td><td>\(\$0.85\)</td><td>\(\$1.00\)</td><td>\(\$1.00\)</td><td>\(\$1.60\)</td></tr></table> b) Yes. Each allowed weight has exactly one postage value. c) No. One postage value corresponds to many possible weights.

- Compare the x-coordinates of the four points. - One input cannot have several outputs in a function. - After switching the coordinates, identify the new input set and the common output.

1. The points \((4, -2)\), \((4, 0)\), \((4, 2)\), and \((4, 5)\) all have x-coordinate \(4\), so they lie on the vertical line \(x=4\). 2. The input \(x=4\) is assigned four different outputs, so \(x\rightarrow y\) is not a function. 3. After switching the coordinates, the inputs are \(-2\), \(0\), \(2\), and \(5\), and every output is \(4\). Thus, \(f(u)=4\) for \(u\in\{-2, 0, 2, 5\}\).

a) The points lie on \(x=4\). b) It is not a function because \(x=4\) has multiple outputs. c) \(f(u)=4\) for \(u\in\{-2, 0, 2, 5\}\).

- Use distance divided by time for each average speed. - For the whole hike, use total distance divided by total time rather than averaging the two speeds. - In part c, express the total time as the fixed outbound time plus the variable return time.

1. The speed to the shelter is \(v_1=\frac{6}{1.5}=4\,\text{mph}\). 2. The return time is \(t_2=\frac{9}{4.5}=2\) hours. 3. The total distance is \(15\) miles, and the total time is \(1.5+2=3.5\) hours. 4. The average speed for the entire hike is \(v_g=\frac{15}{3.5}\approx4.29\,\text{mph}\). 5. If the return time is \(t_2\), the total time is \(1.5+t_2\), so \(v_g=\frac{15}{1.5+t_2}\).

a) \(v_1=4\,\text{mph}\) b) \(t_2=2\) hours, and \(v_g\approx4.29\,\text{mph}\) c) \(v_g=\frac{15}{1.5+t_2}\)

- A negative number squared is positive. - Evaluate both functions for each listed input. - Compare the two outputs in each row.

1. For \(y=-4\), the outputs are \(-5\) and \(12\). 2. For \(y=-2\), the outputs are \(-4\) and \(0\). 3. For \(y=0\), the outputs are \(-3\) and \(-4\). 4. For \(y=2\), the outputs are \(-2\) and \(0\). 5. For \(y=4\), the outputs are \(-1\) and \(12\). 6. Only at \(y=0\) is \(T_1(y)>T_2(y)\).

<table> <tr><th>\(y\)</th><th>\(T_1(y)\)</th><th>\(T_2(y)\)</th></tr> <tr><td>\(-4\)</td><td>\(-5\)</td><td>\(12\)</td></tr> <tr><td>\(-2\)</td><td>\(-4\)</td><td>\(0\)</td></tr> <tr><td>\(0\)</td><td>\(-3\)</td><td>\(-4\)</td></tr> <tr><td>\(2\)</td><td>\(-2\)</td><td>\(0\)</td></tr> <tr><td>\(4\)</td><td>\(-1\)</td><td>\(12\)</td></tr> </table> \(T_1(y)>T_2(y)\) only when \(y=0\).

- Substitute each input into the function rule. - Represent a required output with an equation. - Set two function rules equal when their outputs must match. - Use equivalent operations to isolate the input.

1. Evaluate \(T(x) = 5x - 7\) at each input: \(T(-3) = 5 \cdot (-3) - 7 = -22\), \(T(0) = 5 \cdot 0 - 7 = -7\), and \(T(2.5) = 5 \cdot 2.5 - 7 = 5.5\). 2. Solve \(5x - 7 = 13\). Add \(7\): \(5x = 20\). Divide by \(5\): \(x = 4\). 3. Set the outputs equal: \(5x - 7 = 2x + 8\). Subtract \(2x\) and add \(7\): \(3x = 15\). Divide by \(3\): \(x = 5\).

1) The missing outputs are \(-22\), \(-7\), and \(5.5\). 2) \(x = 4\) 3) \(x = 5\)

- Substitute each value of \(m\) into both formulas. - The constant term is the cost at \(0\) minutes. - Compare the two entries in each column.

1. Plan A costs \(\$10.00\), \(\$12.50\), \(\$15.00\), and \(\$17.50\) for the listed inputs. 2. Plan B costs \(\$5.00\), \(\$10.00\), \(\$15.00\), and \(\$20.00\). 3. The plans cost the same at \(100\) minutes. At \(150\) minutes, Plan A is less expensive.

a) <table> <tr><td>Minutes \((m)\)</td><td>\(0\)</td><td>\(50\)</td><td>\(100\)</td><td>\(150\)</td></tr> <tr><td>Plan A</td><td>\(\$10.00\)</td><td>\(\$12.50\)</td><td>\(\$15.00\)</td><td>\(\$17.50\)</td></tr> <tr><td>Plan B</td><td>\(\$5.00\)</td><td>\(\$10.00\)</td><td>\(\$15.00\)</td><td>\(\$20.00\)</td></tr> </table> b) At \(150\) minutes, Plan A is less expensive: \(\$17.50<\$20.00\).

- Look for an x-value that makes more than one y-value possible. - In equation (1), consider when the product \(xy\) can equal zero without restricting \(y\) to one value. - Remember that \(\sqrt{a}\) denotes one nonnegative value, not both square roots. - In equation (2), consider whether opposite y-values can produce the same x-value.

1. In equation (1), let \(x=0\). Then \(0y=0\) is true for every real value of \(y\). Because one x-value corresponds to infinitely many y-values, the equation does not define a function. 2. In equation (2), let \(x=0.5\). Then \(0.5=\frac{1}{y^2+1}\), so \(y^2+1=2\) and \(y=\pm1\). Because one x-value corresponds to two y-values, the equation does not define a function. 3. In equation (3), the square-root symbol denotes the unique nonnegative square root. For every \(x\ge1\), the expression \(\sqrt{x-1}+2\) produces exactly one y-value, so it defines a function.

(1) Not a function (2) Not a function (3) Function

- Solve each equation for \(y\) when possible and compare the resulting shape with the graphs. - Use the vertical line test to decide whether each relation is a function. - A proportional linear function has the form \(y=kx\) and passes through the origin.

1. Graph 1 is Equation A. Solving \(2y-3x=0\) for \(y\) gives \(y=\frac{3}{2}x\), a line through the origin with slope \(\frac{3}{2}\). 2. Graph 2 is Equation B. The relation \(y^2=x\) is \(y=\pm\sqrt{x}\), a sideways parabola. 3. Graph 3 is Equation C. Solving \(x^2+y=4\) gives \(y=4-x^2\), a downward-opening parabola with vertex \((0,4)\). 4. Graph 1 is a function because each \(x\)-value has exactly one \(y\)-value. Graph 2 is not a function because every \(x>0\) has two \(y\)-values. Graph 3 is a function because each \(x\)-value has exactly one \(y\)-value. 5. For Graph 1, \(y=\frac{3}{2}x\), so it is proportional. For Graph 3, \(y=4-x^2\), so it is neither linear nor proportional.

a) A \(\rightarrow\) Graph 1; B \(\rightarrow\) Graph 2; C \(\rightarrow\) Graph 3 b) Graph 1: function; Graph 2: not a function; Graph 3: function c) Graph 1: \(y=\frac{3}{2}x\), proportional; Graph 3: \(y=4-x^2\), neither linear nor proportional

- Interpret increasing, horizontal, and vertical segments in the context. - Time cannot reverse direction in a realistic time graph. - Use the vertical line test.

1. Graph A is realistic. Increasing segments represent walking away from home, and horizontal segments represent stopping, such as at a crosswalk. 2. Graph B is not realistic. At \(t=2\) minutes, it contains a vertical segment, which would place the student at many distances at the same instant. It also moves backward in time from \(t=6\) to \(t=5\), which cannot represent a time-ordered trip. 3. Graph A passes the vertical line test, so it is a function graph. 4. Graph B fails the vertical line test. For example, \(t=2\) has many distances, and some inputs between \(5\) and \(6\) minutes have multiple distances.

a) Graph A is realistic. Graph B is impossible because it includes an instantaneous change in distance and a segment that moves backward in time. b) Graph A is a function graph. Graph B is not a function graph.

- The maximum is the graph's highest point. - Read the graph at the requested time values. - For the reverse relation, check whether one height can be paired with more than one time.

1. The vertex of the graph is \((3, 18)\), so the ball reaches a maximum height of \(18\,\text{ft}\) after \(3\) seconds. 2. The graph gives \(h(1)=10\) and \(h(5)=10\), so the ball is \(10\,\text{ft}\) high at both times. 3. The reverse relation is not a function because most heights correspond to two different times: once while the ball is rising and once while it is falling. For example, a height of \(10\,\text{ft}\) corresponds to both \(t=1\) and \(t=5\).

a) \(18\,\text{ft}\) after \(3\) seconds b) \(10\,\text{ft}\) at both \(1\) second and \(5\) seconds c) No. A single height can correspond to two different times.

- Apply the vertical line test to an x-value near the center of the circle. - For part b), look for a way to divide the shoreline so that each resulting part has at most one point for every x-value.

1. For most x-values between the leftmost and rightmost points of the circle, a vertical line meets the shoreline twice: once on the upper half and once on the lower half. Therefore, one input would have two outputs, so the entire circle is not the graph of \(y=f(x)\). 2. Divide the circle horizontally into an upper semicircle and a lower semicircle. Each semicircle passes the vertical line test and can be represented as a separate function.

a) The circle is not a function graph because many x-values correspond to two y-values. b) Split it into the upper semicircle and the lower semicircle.

- Apply the vertical line test to each graph. - A proportional linear function must pass through the origin. - For Equation (II), think about what y-values can correspond to one allowed x-value and what shape that creates.

1. Graph a) passes the vertical line test, so it is a function. Graph b) fails the vertical line test because many x-values correspond to two y-values. 2. Graph a) is linear but not proportional because it is a line that does not pass through the origin. 3. Graph a) matches Equation (I), \(y=\frac{1}{2}x+1\). Graph b) matches Equation (II), \(x=y^2-2\), which is a sideways parabola.

1) Graph a) is a function; Graph b) is not. 2) Graph a) is linear but not proportional. 3) Graph a) matches (I); Graph b) matches (II).

- The trip ends when the graph returns to distance \(0\). - A horizontal segment means the distance from the center is unchanged. - Use the constant rate on the final segment to find the value at \(45\) minutes. - Apply the definition of a function or the vertical line test.

1. The graph begins at \(t=0\) and returns to distance \(0\) at \(t=60\) minutes, so the trip lasts \(60\) minutes. 2. At \(t=20\), the graph is on a horizontal segment at \(4\) miles. The courier is probably making a delivery or taking a short break. 3. From minute \(40\) to minute \(60\), the distance decreases linearly from \(6\) miles to \(0\). In \(5\) minutes, it decreases by \(\frac{5}{20}\cdot6=1.5\) miles, so after \(45\) minutes the distance is \(6-1.5=4.5\) miles. 4. Every input time is paired with exactly one distance. Equivalently, every vertical line intersects the graph at most once, so the relation is a function.

a) \(60\) minutes b) \(4\) miles; probably making a delivery or taking a break c) \(4.5\) miles d) Each time corresponds to exactly one distance.

- Locate the graph's highest point. - Compare the graph with a horizontal line at \(15\). - Check the early-morning portion of the graph for part c). - For the reverse relation, ask whether one data-rate value can occur at more than one time.

1. The graph reaches its maximum at about \(8{:}00\) p.m., with a data rate of about \(42\,\text{Mbps}\). 2. The graph is above \(15\,\text{Mbps}\) from about noon to \(2{:}30\) p.m. and from about \(4{:}30\) p.m. to \(11{:}00\) p.m. 3. The claim is false. During the overnight and early-morning hours, the graph is below \(5\,\text{Mbps}\). 4. The reverse relation is not a function because the same data rate occurs at several different times during the day. One input data rate would be paired with more than one output time.

a) About \(8{:}00\) p.m.; about \(42\,\text{Mbps}\) b) About noon–\(2{:}30\) p.m. and \(4{:}30\) p.m.–\(11{:}00\) p.m. c) False; the rate falls below \(5\,\text{Mbps}\) overnight. d) No; one data rate can correspond to several times.

- Determine the two possible values of \(f(x)\) whose squares equal \(4\). - Find where the graph has either of those y-values. - Check which listed interval contains one of the corresponding x-values.

1. The equation \((f(x))^2=4\) means \(f(x)=2\) or \(f(x)=-2\). 2. The graph intersects \(y=2\) at approximately \(x=-3.46\) and \(x=3.46\). 3. The graph never reaches \(y=-2\), because its minimum value is \(-1\). 4. Of the listed intervals, only \([-4, -3]\) contains one of the solutions.

1) \([-4, -3]\)

- “Started hour” means that one minute into a new hour triggers the next charge. - Ask whether one fee can occur for two different parking times. - A reverse function requires the original pricing rule to be one-to-one.

1. A \(30\)-minute or \(60\)-minute stay is within the first started hour, so each costs \(\$1.50\). A \(61\)-minute stay starts a second hour, so it costs \(\$1.50+\$1.00=\$2.50\). A \(120\)-minute stay also costs \(\$2.50\). 2. The reverse assignment is not a function because one fee can correspond to many parking times. For example, \(\$1.50\) corresponds to both \(30\) minutes and \(60\) minutes. 3. A strictly increasing rule such as \(P(t)=kt\), where \(k>0\) and \(t\) is the exact parking time, assigns different fees to different times. Its reverse therefore defines a function.

a) \(30\) minutes: \(\$1.50\) \(60\) minutes: \(\$1.50\) \(61\) minutes: \(\$2.50\) \(120\) minutes: \(\$2.50\) b) The reverse assignment is not a function because one fee corresponds to multiple parking times. c) One possible rule is \(P(t)=kt\) with \(k>0\), using exact time and no rounding.

- Use total round-trip distance and the desired average speed to find the total time allowed. - Subtract the fixed outbound time to determine the time left for the return trip. - For part c, check whether the proposed average leaves any positive amount of time for the return trip.

1. The trip to work takes \(0.5\) hour, so \(v_1=\frac{20}{0.5}=40\,\text{mph}\). 2. A \(50\,\text{mph}\) average over the \(40\)-mile round trip requires a total time of \(\frac{40}{50}=0.8\) hour. 3. The return trip must therefore take \(0.8-0.5=0.3\) hour. 4. The required return speed is \(v_2=\frac{20}{0.3}=\frac{200}{3}\,\text{mph}\approx66.67\,\text{mph}\). 5. An \(80\,\text{mph}\) average over \(40\) miles would require a total time of \(\frac{40}{80}=0.5\) hour. 6. The outbound trip already takes all \(0.5\) hour, leaving \(0\) hour for the return trip. That would require an unattainable infinite speed, so the proposed round-trip average is not possible.

a) \(v_1=40\,\text{mph}\) b) \(v_2=\frac{200}{3}\,\text{mph}\approx66.67\,\text{mph}\) c) No. It would require a return time of \(0\) hours.