Lesson 5

The goal of this warm-up is for students to use the properties of exponents to reason about how the value of an exponential function increases over different input intervals. In particular, in order to fully answer the questions, students need to generalize from specific examples to any input increase of the given amount. Students will build on this type of thinking with a simple exponential function when they reason about fractional intervals for more complex functions later in the lesson.

Invite students to share their reasoning for the factors that align with each of the 3 intervals. If students determined the factors using only specific examples, such as calculating at the growth factor between \(x=0\) and \(x=2\) as 16 since \(g(2)=16\boldcdot g(0)\), ask them to consider how they know that the growth factor is 16 for any interval of size 2 units. Help students translate their verbal reasoning into equations, such as \(4^{x+2}=4^2 \boldcdot 4^x\) or \(\dfrac{4^{x+2}}{4^x}=\dfrac{4^2 \boldcdot 4^x}{4^x}=4^2\).

In a previous lesson, students found the growth factor of an exponential function over a fractional interval. They then rewrote a given exponential function to represent how the quantity would change over a different unit of input. For example, given a quantity expressed as a function of time in weeks, they expressed it as a function of time in days.

That work continues here, but the focus shifts from finding the growth factor for a particular interval to noticing how a function changes over any equal intervals of input. For example, how does the function value change every time the input increases by 0.5—say, from 1 to 1.5, or from 0.25 to 0.75?

To find the factor by which the function \(f\) changes when the input increases by 0.5, students may:

• Use a calculator to divide the numerical values of \(f(0.5)\) and \(f(0)\) (or other equal intervals).
• Find the factor by which \(f\) changes when the input increases by 1 (say from \(f(0)\) to \(f(1)\)), and then take the square root for that factor.

Monitor for students who use each strategy to share during the whole-class discussion.

Making spreadsheet technology available gives students an opportunity to choose appropriate tools strategically (MP5).

A linear model for \(f\), based on the values for 0 and 1 year, predicts a value of $14,800 after 0.5 year, which is very close to the value found with an exponential model. If students find the value for 0.5 year and think that \(f\) may be linear, ask them to calculate what value of the machine the linear model would predict after 2 and 3 years. Then ask them if this agrees with the values provided in the table. 

Invite previously identified students to share how they found the decay factor for the half-year intervals in the order listed in the Activity Narrative. Students may notice that the numerical values found from dividing two function values or from estimating square roots may not be identical, raising the question of whether the decay factor is really equal over equal intervals. One advantage to finding an expression for the factor of change, namely \(0.85^{(0.5)}\), is that it allows us to see that the decay factor is the same whenever the input is increased by 0.5.

If not mentioned in students’ explanation, clarify that if we reason about the decay factor by writing expressions, we can see that they are indeed equal. For example, we know that the decay factor over one year is 0.85, so the change from, say, \(f(0)\) to \(f(0.5)\) could be written as: \(\dfrac{16,\!000 \boldcdot (0.85)^{(0.5)}}{16,\!000}\), which equals \(0.85^{(0.5)}\). The change from \(f(1.5)\) to \(f(2)\) could be written as: \(\dfrac{16,\!000 \boldcdot (0.85)^{2}}{16,\!000 \boldcdot (0.85)^{(1.5)}}\), which also equals \(0.85^{(0.5)}\). Using exact expressions here is important as the quotients of the approximate values in the table are close to \(0.85^{(0.5)}\) but differ slightly because of rounding: for example \(\frac{14,751.27}{16,000}\) and \(\frac{13,600}{14,571.27}\) are very close to one another and to \((0.85)^{(0.5)}\) but all three quantities are slightly different.

Make sure students see that the process of finding the factor of decay every half a year can be generalized, that is, that the factor is the same no matter which half-year you examine: \(\dfrac {f(t + 0.5)} {f(t)} = \dfrac {16,\!000 \cdot (0.85)^{(t+0.5)}}{16,\!000 \cdot (0.85)^t}=\dfrac {16,\!000 \cdot (0.85)^t \cdot (0.85)^{(0.5)}}{16,\!000 \cdot (0.85)^t}=(0.85)^{(0.5)}\approx0.92\).

In the previous activity, students use the fact that an exponential function changes by the same factor over equal intervals in order to fill out a table of values, and they explain why this is always true using an equation for the function. In this activity, students continue to use the equal factors over equal intervals property of exponential functions in a less scaffolded situation. They are given the coordinates of points on a graph and need to use this information to figure out the value of the given function for a different input.

Students may be uncomfortable with the fact that the vertical axis is not labeled. Remind them that the problem gives them the information they need about how much medicine is left after \(\frac{1}{4}\) hour and \(\frac{3}{4}\) hour. The graph is mainly intended to help visualize the situation.

Invite some students to share how they calculated the amount of medicine left after \(1\frac{3}{4}\) hours. Make sure students see that:

• The interval between \(\frac{3}{4}\) and \(1\frac{3}{4}\) can be seen as two half-hour intervals. So if we know the growth factor of another half-hour interval, say \(\frac{1}{4}\) to \(\frac{3}{4}\), we can find the factor from \(\frac{3}{4}\) to \(1\frac{3}{4}\) by squaring that growth factor.
• The graph can be useful to check how reasonable the estimate of 0.9 mg is for the \(y\)-coordinate of \(C\). This is \(\frac{1}{4}\) of the \(y\)-coordinate value for \(B\) and this makes sense looking at the graph.

Consider asking students how they could find the decay factor for \(\frac{1}{4}\) hour. Since there are two quarter hours in \(\frac{1}{2}\) hour, the decay factor for \(\frac{1}{2}\) hour is the square of the decay factor for \(\frac{1}{4}\) hour.