Free math worksheets

- Compare how much the output changes with how much the input changes. - Keep the two differences in the same order. - Your result should describe output change per one unit of input change.

1. The change in the function value is \(8-2=6\). 2. The change in the input is \(4-1=3\). 3. The average rate of change is \(\frac{6}{3}=2\).

\(2\)

- Evaluate the function at both endpoints of the interval. - Form one ratio using the change in outputs and the corresponding change in inputs. - After calculating, compare the result with the coefficient of \(x\).

1. Evaluate the endpoints: \(f(-2)=13\) and \(f(4)=-5\). 2. The average rate of change is \(\frac{-5-13}{4-(-2)}=\frac{-18}{6}=-3\). 3. The slope of \(f(x)=-3x+7\) is also \(-3\). For a linear function, the average rate of change over any interval equals its slope.

The average rate of change is \(-3\), which is the same as the slope of the line.

- Read the two graph values at the stated input endpoints. - Compare vertical change with the corresponding horizontal change. - Use the sign of your result when describing whether the graph rises or falls on average.

1. From the graph, \(g(-1)=1\) and \(g(3)=9\). 2. The average rate of change is \(\frac{9-1}{3-(-1)}=\frac{8}{4}=2\). 3. Over this interval, \(g(x)\) increases by an average of \(2\) units for each \(1\)-unit increase in \(x\).

The average rate of change is \(2\). On average, \(g(x)\) increases by \(2\) units for each \(1\)-unit increase in \(x\).

- Use only the two table entries at the endpoints of the requested interval. - Keep track of which quantity is the output and which is the input when forming the rate. - The units of the rate come from output units divided by input units.

1. From the table, \(V(2)=140\) and \(V(8)=290\). 2. The average rate of change is \(\frac{290-140}{8-2}=\frac{150}{6}=25\,\text{gallons per minute}\). 3. From minute \(2\) through minute \(8\), the amount of water in the tank increases by an average of \(25\) gallons per minute.

\(25\,\text{gallons per minute}\). From minute \(2\) to minute \(8\), the tank gains an average of \(25\) gallons of water each minute.

- Average rate of change uses only the two endpoint values of the interval. - Evaluate the function at each endpoint before forming a quotient. - Divide the change in output by the change in input.

1. Evaluate the endpoints: \(f(1)=1-3=-2\) and \(f(4)=16-12=4\). 2. Use the average-rate formula: \(\frac{f(4)-f(1)}{4-1}=\frac{4-(-2)}{3}=2\).

\(2\)

- Write the definition of average rate of change using the two stated input values. - Treat the unknown endpoint output as the missing quantity in that ratio. - Check whether the negative rate is consistent with the direction of the output change.

1. The input changes by \(8-2=6\). 2. An average rate of change of \(-1.5\) means \(\frac{f(8)-f(2)}{6}=-1.5\). 3. Substitute \(f(2)=10\): \(\frac{f(8)-10}{6}=-1.5\). 4. Therefore, \(f(8)-10=-9\), so \(f(8)=1\).

\(f(8)=1\)

- Check whether the output difference and input difference describe movement in the same direction along the interval. - Try writing both changes from the left endpoint to the right endpoint. - A second valid setup can be made by reversing both differences, not just one.

1. The student reversed the order of the output difference but did not reverse the order of the input difference. Corresponding changes must use the same direction. 2. Using increasing input order gives \(\frac{f(6)-f(2)}{6-2}=\frac{13-5}{4}=2\). 3. Equivalently, reversing both differences gives the same value: \(\frac{5-13}{2-6}=2\).

The numerator and denominator changes were taken in inconsistent directions. The correct average rate of change is \(2\).

- Treat each interval separately and use its own pair of endpoint values. - Compare the two rates after calculating them. - Recall what must be constant about the rate of change of a linear function.

1. On \([0,2]\), the average rate of change is \(\frac{7-1}{2-0}=3\). 2. On \([2,5]\), the average rate of change is \(\frac{10-7}{5-2}=1\). 3. A single linear function has the same rate of change on every interval. Because \(3\neq1\), the three points cannot all lie on one linear function.

a) \(3\) b) \(1\) c) No. The two intervals have different average rates of change, so the three points are not collinear.

- Read the y-values at the two marked x-values. - Average rate of change is the slope between the endpoint points, not the slope of the curve at every point. - Compare the two endpoint y-values before calculating.

1. From the graph, \(q(-2)=0\) and \(q(2)=0\). 2. The average rate of change is \(\frac{0-0}{2-(-2)}=0\). 3. Therefore, the secant line through the two endpoint points is horizontal, even though the graph between them is curved.

Average rate of change: \(0\). The secant line through the endpoint points is horizontal.

- Treat each interval separately and evaluate the function at its two endpoints. - Keep the endpoint order consistent in both the numerator and denominator. - Use the sign of each result to describe the overall direction of change on that interval.

1. On \([-2,0]\), \(p(-2)=5\) and \(p(0)=1\), so the average rate of change is \(\frac{1-5}{0-(-2)}=-2\). 2. On \([0,3]\), \(p(0)=1\) and \(p(3)=10\), so the average rate of change is \(\frac{10-1}{3-0}=3\). 3. The first rate is negative, so the function decreases overall on the first interval. The second rate is positive, so it increases overall on the second interval.

a) \(-2\) b) \(3\) c) The first interval has a negative average rate and the second has a positive average rate, showing an overall decrease followed by an overall increase.

- For each function, use only its values at the two endpoints of the full interval. - “Halfway between” two rates refers to their numerical midpoint. - Once the rate for \(m\) is known, connect it to the total output change over an input change of \(5\).

1. From the graph, \(h(0)=1\) and \(h(5)=8\). Thus, the average rate of change of \(h\) is \(\frac{8-1}{5-0}=\frac{7}{5}=1.4\). 2. From the table, \(k(0)=4\) and \(k(5)=-1\). Thus, the average rate of change of \(k\) is \(\frac{-1-4}{5-0}=-1\). 3. The rate halfway between \(1.4\) and \(-1\) is \(\frac{1.4+(-1)}{2}=0.2\). 4. Over an input change of \(5\), \(m\) changes by \(0.2\cdot5=1\). Since \(m(0)=3\), \(m(5)=4\).

a) \(h\): \(1.4\); \(k\): \(-1\) b) \(m(5)=4\)

- Read the endpoint altitudes from the marked points before calculating any rates. - A negative average rate means the ending altitude is below the starting altitude. - For an average rate of zero, ask what must be true about the two endpoint altitudes.

1. The graph gives \(h(2)=5\) and \(h(4)=7\). The average rate is \(\frac{7-5}{4-2}=1\,\text{m/min}\). 2. The graph gives \(h(8)=2\). From \(t=4\) to \(t=8\), the average rate is \(\frac{2-7}{8-4}=-\frac{5}{4}\,\text{m/min}\). The negative sign means the drone loses altitude overall during that interval. 3. An average rate of \(0\) from \(t=2\) to \(t=T\) requires \(h(T)=h(2)\). The graph returns to altitude \(5\,\text{m}\) at \(T=6\), so the average rate on \([2,6]\) is \(0\).

a) \(1\,\text{m/min}\) b) \(-\frac{5}{4}\,\text{m/min}\); the drone loses altitude overall. c) \(T=6\,\text{min}\). The graph shows \(h(6)=h(2)=5\), so the secant line from \(t=2\) to \(t=6\) is horizontal and has average rate \(0\).