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Posttest: Exponential and Logarithmic Functions

Match each transformation of function [tex]f[/tex] to a feature of the transformed function.

- [tex]y[/tex]-intercept at (0,2)
- Asymptote of [tex]y=2[/tex]
- [tex]y[/tex]-intercept at (0,4)
- Function decreases as [tex]x[/tex] increases

[tex]\[
\begin{array}{l}
j(x)=f(x+2) \rightarrow \square \\
g(x)=2f(x) \rightarrow \square \\
h(x)=f(x)+2 \rightarrow \square \\
m(x)=-f(x) \rightarrow \square
\end{array}
\][/tex]

Sagot :

GinnyAnswer
Let's relate each transformation of the function [tex]\( f \)[/tex] to a feature of the transformed function.

1. [tex]\( y \)[/tex]-intercept at [tex]\( (0, 2) \)[/tex]
- The function [tex]\( h(x) = f(x) + 2 \)[/tex] is a vertical shift of [tex]\( f(x) \)[/tex] upwards by 2 units. For any input [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] will be increased by 2. Consequently, the [tex]\( y \)[/tex]-intercept of the original function [tex]\( f(0) \)[/tex] is now [tex]\( f(0) + 2 \)[/tex]. Hence, if the new [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 2) \)[/tex], it perfectly matches with [tex]\( h(x) = f(x) + 2 \)[/tex].

2. Asymptote of [tex]\( y = 2 \)[/tex]
- The function [tex]\( h(x) = f(x) + 2 \)[/tex] also impacts horizontal asymptotes. If the original function [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex], then adding 2 to [tex]\( f(x) \)[/tex] shifts the asymptote to [tex]\( y = 2 \)[/tex]. Therefore, the asymptote of [tex]\( y = 2 \)[/tex] corresponds to [tex]\( h(x) = f(x) + 2 \)[/tex].

3. [tex]\( y \)[/tex]-intercept at [tex]\( (0, 4) \)[/tex]
- The function [tex]\( g(x) = 2f(x) \)[/tex] illustrates a vertical stretch of the function [tex]\( f(x) \)[/tex] by a factor of 2. Therefore, if the original function [tex]\( f(x) \)[/tex] had a [tex]\( y \)[/tex]-intercept at [tex]\( f(0) \)[/tex], the new [tex]\( y \)[/tex]-intercept would be [tex]\( 2 \cdot f(0) \)[/tex]. Thus, if the new [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 4) \)[/tex], then [tex]\( f(0) \)[/tex] must have been [tex]\( 4/2 = 2 \)[/tex]. So, [tex]\( g(x) = 2f(x) \)[/tex] corresponds to a [tex]\( y \)[/tex]-intercept at [tex]\( (0, 4) \)[/tex].

4. Function decreases as [tex]\( x \)[/tex] increases
- The function [tex]\( m(x) = -f(x) \)[/tex] represents a reflection across the [tex]\( x \)[/tex]-axis. If [tex]\( f(x) \)[/tex] was increasing, [tex]\( -f(x) \)[/tex] will be decreasing because flipping the function vertically inverts its growth properties. Thus, [tex]\( m(x) = -f(x) \)[/tex] corresponds to a function that decreases as [tex]\( x \)[/tex] increases.

Summarizing the transformations and their corresponding features:
[tex]\[ \begin{array}{l} h(x)=f(x)+2 \rightarrow y\text{-intercept at }(0, 2) \\ h(x)=f(x)+2 \rightarrow \text{asymptote of } y=2 \\ g(x)=2 f(x) \rightarrow y\text{-intercept at } (0, 4) \\ m(x)=-f(x) \rightarrow \text{function decreases as } x \text{ increases} \end{array} \][/tex]
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