How does h(x) = 9^x change over the interval from x = 0 to x = 1? Choices: (A)h(x) increases by 800% (B)h(x) increases by 9% (C)h(x) increases by 9 (D)h(x) decreases by 9%

Calculate h(0): Calculate h(0) by substituting x = 0 into h(x) = 9^x. h(0) = 9^0 h(0) = 1Find h(1): Now, find h(1) by substituting x = 1 into h(x) = 9^x. h(1) = 9^1 h(1) = 9Determine increase: Determine if h(x) increases or decreases from x = 0 to x = 1. Since h(0) = 1 and h(1) = 9, h(x) increases.Calculate percentage increase: Calculate the percentage increase from x = 0 to x = 1. % increase = [(h(1) - h(0)) / h(0)] * 100% % increase = [(9 - 1) / 1] * 100% % increase = [8 / 1] * 100% % increase = 8 * 100% % increase = 800%

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How does $h(x) = 9^x$ change over the interval from $x = 0$ to $x = 1$ ? $\newline$ Choices: $\newline$ (A) $h(x)$ increases by $800\%$ $\newline$ (B) $h(x)$ increases by $9\%$ $\newline$ (C) $h(x)$ increases by $9$ $\newline$ (D) $h(x)$ decreases by $9\%$

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Math Problems

Algebra 2

Exponential functions over unit intervals

Full solution

Q. How does

h(x) = 9^x

change over the interval from

x = 0

x = 1

\newline

Choices:

\newline

(A)

h(x)

increases by

800\%

\newline

(B)

h(x)

increases by

9\%

\newline

(C)

h(x)

increases by

9

\newline

(D)

h(x)

decreases by

9\%

Calculate $h(0)$ : Calculate $h(0)$ by substituting $x = 0$ into $h(x) = 9^x$ . $\newline$ $h(0) = 9^0$ $\newline$ $h(0) = 1$
Find $h(1)$ : Now, find $h(1)$ by substituting $x = 1$ into $h(x) = 9^x$ .
$h(1) = 9^1$
$h(1) = 9$
Determine increase: Determine if $h(x)$ increases or decreases from $x = 0$ to $x = 1$ . Since $h(0) = 1$ and $h(1) = 9$ , $h(x)$ increases.
Calculate percentage increase: Calculate the percentage increase from $x = 0$ to $x = 1$ .
% increase = $\left[\frac{h(1) - h(0)}{h(0)}\right] \times 100\%$
% increase = $\left[\frac{9 - 1}{1}\right] \times 100\%$
% increase = $\left[\frac{8}{1}\right] \times 100\%$
% increase = $8 \times 100\%$
% increase = $800\%$