Full solution

Q. The graph of

y=g(x)+2

in the

x y

-plane has a

y

-intercept of

1

. If

g

is an exponential function, which of the following could define

g

\newline

Choose

1

answer:

\newline

(A)

g(x)=\left(\frac{1}{4}\right)^{x}-1

\newline

(B)

g(x)=2\left(\frac{2}{3}\right) x+1

\newline

(C)

g(x)=2^{x}+3

\newline

(D)

g(x)=3^{x}-2

Set $x=0$ and solve: To find the y-intercept of the function $y=g(x)+2$ , we set $x=0$ and solve for $y$ . $\newline$ $y = g(0) + 2$ $\newline$ We are given that the y-intercept is $1$ , so: $\newline$ $1 = g(0) + 2$
Find $g(0) = -1$ : Solving for $g(0)$ , we get: $\newline$ $g(0) = 1 - 2$ $\newline$ $g(0) = -1$ $\newline$ So, the function $g(x)$ must satisfy the condition $g(0) = -1$ .
Check each option: Now we will check each option to see which one satisfies $g(0) = -1$ .
(A) $g(x)=\left(\frac{1}{4}\right)^x - 1$
$g(0) = \left(\frac{1}{4}\right)^0 - 1$
$g(0) = 1 - 1$
$g(0) = 0$
This does not satisfy $g(0) = -1$ .
Check each option: Now we will check each option to see which one satisfies $g(0) = -1$ . $\newline$ (A) $g(x)=\left(\frac{1}{4}\right)^x - 1$ $\newline$ $g(0) = \left(\frac{1}{4}\right)^0 - 1$ $\newline$ $g(0) = 1 - 1$ $\newline$ $g(0) = 0$ $\newline$ This does not satisfy $g(0) = -1$ .(B) $g(x)=2\left(\frac{2}{3}\right)^x + 1$ $\newline$ $g(0) = 2\left(\frac{2}{3}\right)^0 + 1$ $\newline$ $g(0) = 2(1) + 1$ $\newline$ $g(0) = 2 + 1$ $\newline$ $g(x)=\left(\frac{1}{4}\right)^x - 1$ $0$ $\newline$ This does not satisfy $g(0) = -1$ .
Check each option: Now we will check each option to see which one satisfies $g(0) = -1$ .
(A) $g(x)=\left(\frac{1}{4}\right)^x - 1$
$g(0) = \left(\frac{1}{4}\right)^0 - 1$
$g(0) = 1 - 1$
$g(0) = 0$
This does not satisfy $g(0) = -1$ .
(B) $g(x)=2\left(\frac{2}{3}\right)^x + 1$
$g(0) = 2\left(\frac{2}{3}\right)^0 + 1$
$g(0) = 2(1) + 1$
$g(0) = 2 + 1$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $0$
This does not satisfy $g(0) = -1$ .
(C) $g(x)=\left(\frac{1}{4}\right)^x - 1$ $2$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $3$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $4$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $5$
This does not satisfy $g(0) = -1$ .
Check each option: Now we will check each option to see which one satisfies $g(0) = -1$ .
(A) $g(x)=\left(\frac{1}{4}\right)^x - 1$
$g(0) = \left(\frac{1}{4}\right)^0 - 1$
$g(0) = 1 - 1$
$g(0) = 0$
This does not satisfy $g(0) = -1$ .
(B) $g(x)=2\left(\frac{2}{3}\right)^x + 1$
$g(0) = 2\left(\frac{2}{3}\right)^0 + 1$
$g(0) = 2(1) + 1$
$g(0) = 2 + 1$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $0$
This does not satisfy $g(0) = -1$ .
(C) $g(x)=\left(\frac{1}{4}\right)^x - 1$ $2$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $3$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $4$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $5$
This does not satisfy $g(0) = -1$ .
(D) $g(x)=\left(\frac{1}{4}\right)^x - 1$ $7$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $8$
$g(x)=\left(\frac{1}{4}\right)^x - 1$ $9$
$g(0) = -1$
This satisfies $g(0) = -1$ .