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Exponential Functions


Alignments to Content Standards: F-LE.A.1

Task

The figure below shows the graphs of the exponential functions $f(x)=c\cdot 3^x$ and $g(x)=d\cdot 2^x$, for some numbers $c\gt 0$ and $d\gt 0$. They intersect at the point $(p,q)$.

Graph_of_two_exponen_48b277e9fd77c0a9d0711ecae2300d9c



  1. Which is greater, $c$ or $d$? Explain how you know.
  2. Imagine you place the tip of your pencil at $(p,q)$ and trace the graph of $g$ out to the point with $x$-coordinate $p+2$. Imagine I do the same on the graph of $f$. What will be the ratio of the $y$-coordinate of my ending point to the $y$-coordinate of yours?

IM Commentary

This task requires students to use the fact that the value of an exponential function $f(x)=a\cdot b^x$ increases by a multiplicative factor of $b$ when $x$ increases by one. It intentionally omits specific values for $c$ and $d$ in order to encourage students to use this fact instead of computing the point of intersection, $(p,q)$, and then computing function values to answer the question.

This task is preparatory for standard F.LE.1a.

Solutions

Solution: Exponential Functions

  1. The graph of $f(x)=c\cdot3^x$ is steeper than the graph of $g(x)= d \cdot 2^x$ because the value of $f(x)$ triples each time $x$ is increased by one while the value of $g(x)$ doubles each time $x$ is increased by one. Hence the graph of $f$ is the one that intersects the $y$-axis at a lower value. The graph of $f$ meets the $y$-axis at $f(0) = c\cdot3^0 = c$ while the graph of $g$ meets the $y$-axis at $g(0) = d\cdot 2^0 = d$. We conclude that $c < d$.
  2. Along the graph of $g$ each increase of one unit in the $x$ value multiplies the output of $g$ by 2. So an increase of two units in the $x$ value multiplies the output of $g$ by 4. Similarly, an increase of two units in the $x$ value will multiply the value of $f$ by $3^2=9$. So the ratio of my $y$-coordinate to your $y$-coordinate at our ending points is $\frac{9}{4}$.

Solution: Exponential Functions, Alternate Solution

  1. Noting that $f(p) = g(p)$, we can say that $c \cdot 3^p = d \cdot 2^p$ so $$\frac{c}{d} = \frac{2^p}{3^p}=\left(\frac{2}{3}\right)^p.$$ Since $p > 0$ it follows that $\displaystyle\left(\frac{2}{3}\right)^p < 1$. This means that $\displaystyle\frac{c}{d} < 1$ and $c< d$.
  2. We have $\displaystyle \frac{f(p)}{g(p)} = 1$ since $f(p)=g(p)$. That means that $\displaystyle\frac{c \cdot 3^p}{d \cdot 2^p} = 1$. At the ending points, this ratio becomes $$\frac{c \cdot 3^{p+2}}{d \cdot 2^{p+2}} = \frac{c \cdot 3^p}{d \cdot 2^p} \cdot \frac{3^2}{2^2}= 1*\frac{9}{4} = \frac{9}{4}$$.

cfoster says:

almost 6 years

Should this task be aligned to F-LE.2 instead of F-LE.1?

Kristin says:

almost 6 years

I think the reason that this was chosen to illustrate F-LE.1 is that it heavily relies on the connection between the algebraic representation of a function and its graph. F-LE.2 has more of an emphasis of using functions and function notation in a context, so is somewhat less appropriate here, although in truth there are ways in which this task fits elements of that standard as well as you point out. For that reason I would be inclined to align this to the cluster containing these two standards. I'd be interested to hear from other people on this issue.

cfoster says:

almost 6 years

Stating that c and d are greater than 0 will help the student who tries to consider the cases in which c and d will have different or negative signs.

Carolyn Foster

Kristin says:

almost 6 years

Good call! I made that change--thank you.