- 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 \(\frac{6}{3}=2\).

\(2\)

- Compare the two changes in water amount. - How much time passes in each interval?

1. From noon to 1 p.m., the change is \(15-18=-3\) gallons. 2. From 1 p.m. to 2 p.m., the change is \(12-15=-3\) gallons. 3. Both intervals last one hour. The amount changes by \(-3\) gallons per hour.

Yes. The water amount decreases at \(3\) gallons per hour (rate \(-3\) gallons per hour).

- The total amount of growth happened over more than one week. - What calculation gives growth per week rather than growth over the whole period?

1. The total growth of \(12\,\text{cm}\) happened over \(3\) weeks, so the average rate was \(12\div3=4\,\text{cm}\) per week. 2. The student confused the total change over the full interval with the change per week.

\(4\,\text{cm}\) per week. The \(12\,\text{cm}\) is the total growth over \(3\) weeks, not the weekly rate.

- 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.

- The time intervals do not have the same length. - For each interval, divide the change in distance by the time that passed.

1. From hour \(0\) to hour \(2\), the distance changes by \(10-4=6\) miles over \(2\) hours, or \(3\) miles per hour. 2. From hour \(2\) to hour \(5\), the distance changes by \(19-10=9\) miles over \(3\) hours, also \(3\) miles per hour. 3. The average rates are equal even though the intervals have different lengths.

Yes. Both intervals have an average rate of \(3\) miles per hour.

- Use the graph's first and last points to find the full-interval average. - Then compare the rise and run on each straight section. - Does an average rate have to match the change on every part?

1. Read the endpoints \((0,1)\) and \((4,5)\). Their secant slope is \(\frac{5-1}{4-0}=1\). 2. From \(x=0\) to \(x=2\), the rise is \(1\) over a run of \(2\), so the rate is \(0.5\). 3. From \(x=2\) to \(x=4\), the rise is \(3\) over a run of \(2\), so the rate is \(1.5\). The two rates differ from each other and from the full-interval average.

Average rate: \(1\) unit of \(y\) per unit of \(x\). No; the rates on the two halves are \(0.5\) and \(1.5\).

- Compare the elevations at the two mile markers. - What units should the final rate have? Check that your calculation produces them.

1. The elevation change is \(370-250=120\) feet over \(4-1=3\) miles. 2. The average rate is \(120\div3=40\) feet per mile. 3. The rate is positive because the trail's elevation increases overall.

\(40\) feet per mile. The positive sign means the elevation rises along this part of the trail.

- Write the change in output divided by the change in input. - One endpoint output is unknown. Keep it as a symbol while solving.

1. Let the missing value be \(f(6)\). The average-rate formula gives \(\frac{f(6)-19}{6-2}=-4\). 2. Multiply by \(4\): \(f(6)-19=-16\), so \(f(6)=3\). 3. Checking, \(\frac{3-19}{4}=-4\).

\(f(6)=3\).

- 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. - Reversing both differences preserves the quotient; reversing only one does not.

1. Emir 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.

- 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 \(0\).