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The graph of
y=g(x)+2 in the
xy-plane has a
y-intercept of 1 . If
g is an exponential function, which of the following could define
g ?
Choose 1 answer:
(A)
g(x)=((1)/(4))^(x)-1
(B)
g(x)=2((2)/(3))x+1
(C)
g(x)=2^(x)+3
(D)
g(x)=3^(x)-2

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

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Write equations of cosine functions using properties

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 graph of $y=g(x)+2$ , we set $x=0$ and solve for $y$ .
$y = g(0) + 2$
We are given that the y-intercept is $1$ , so:
$1 = g(0) + 2$
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 $g(0) = -1$ .
Option (A) calculation: $g(x)=2\left(\frac{2}{3}\right)^x + 1$ $g(0) = 2\left(\frac{2}{3}\right)^0 + 1$ $g(0) = 2(1) + 1$ $g(0) = 2 + 1$ $g(0) = 3$ This does not satisfy $g(0) = -1$ .
Option (B) calculation: $g(x)=2^x + 3$ $g(0) = 2^0 + 3$ $g(0) = 1 + 3$ $g(0) = 4$ This does not satisfy $g(0) = -1$ .
Option (C) calculation: $g(x)=3^x - 2$ $g(0) = 3^0 - 2$ $g(0) = 1 - 2$ $g(0) = -1$ This satisfies $g(0) = -1$ .

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