Question

evaluate independent practice
learning goal lesson reflection (circle one)
i can graph exponential functions that are expressed symbolically. this means i can show intercepts and end - behavior
□ complete the previous problems, check your solutions, then complete the lesson checkpoint below.
□ complete the lesson reflection above by circling your current understanding of the learning goal(s).
graph each exponential function. identify a, b, the y - intercept, and the end behavior of the graph.

1. f(x)=5(2)^x

x -2 -1 0 1 2
f(x)

1. f(x)=-2(0.8)^x

x -2 -1 0 1 2
f(x)

Explanation:

Step1: Identify the form of exponential function

The general form of an exponential function is $y = a(b)^x$. For $f(x)=5(2)^x$, $a = 5$ and $b = 2$. For $f(x)=-2(0.8)^x$, $a=-2$ and $b = 0.8$.

Step2: Find the y - intercept

The y - intercept is found by setting $x = 0$. For $y = 5(2)^x$, when $x = 0$, $y=5(2)^0=5\times1 = 5$. For $y=-2(0.8)^x$, when $x = 0$, $y=-2(0.8)^0=-2\times1=-2$.

Step3: Calculate function values for the tables

For $f(x)=5(2)^x$:

• When $x=-2$, $f(-2)=5(2)^{-2}=5\times\frac{1}{4}=\frac{5}{4}=1.25$.
• When $x=-1$, $f(-1)=5(2)^{-1}=5\times\frac{1}{2}=2.5$.
• When $x = 1$, $f(1)=5(2)^1 = 10$.
• When $x = 2$, $f(2)=5(2)^2=5\times4 = 20$.

For $f(x)=-2(0.8)^x$:

• When $x=-2$, $f(-2)=-2(0.8)^{-2}=-2\times\frac{1}{0.64}=-\frac{2}{0.64}=-\frac{200}{64}=-\frac{25}{8}=-3.125$.
• When $x=-1$, $f(-1)=-2(0.8)^{-1}=-2\times\frac{1}{0.8}=-\frac{2}{0.8}=-\frac{20}{8}= - 2.5$.
• When $x = 1$, $f(1)=-2(0.8)^1=-1.6$.
• When $x = 2$, $f(2)=-2(0.8)^2=-2\times0.64=-1.28$.

Step4: Determine end - behavior

For $y = 5(2)^x$: As $x
ightarrow-\infty$, $y = 5(2)^x
ightarrow0$ (since $b = 2>1$ and the exponent is negative). As $x
ightarrow\infty$, $y = 5(2)^x
ightarrow\infty$.
For $y=-2(0.8)^x$: As $x
ightarrow-\infty$, since $b = 0.8<1$ and the exponent is negative, $y=-2(0.8)^x
ightarrow-\infty$. As $x
ightarrow\infty$, $y=-2(0.8)^x
ightarrow0$.

For the function $f(x)=5(2)^x$:

$x$	$-2$	$-1$	$0$	$1$	$2$

For the function $f(x)=-2(0.8)^x$:

$x$	$-2$	$-1$	$0$	$1$	$2$

Answer:

For $f(x)=5(2)^x$: $a = 5$, $b = 2$, y - intercept is $(0,5)$, as $x
ightarrow-\infty,y
ightarrow0$ and as $x
ightarrow\infty,y
ightarrow\infty$.
For $f(x)=-2(0.8)^x$: $a=-2$, $b = 0.8$, y - intercept is $(0, - 2)$, as $x
ightarrow-\infty,y
ightarrow-\infty$ and as $x
ightarrow\infty,y
ightarrow0$.