IDNLearn.com: Your go-to resource for finding precise and accurate answers. Ask any question and receive accurate, in-depth responses from our dedicated team of experts.
Sagot :
To match each transformation of function [tex]\( f \)[/tex] to a notable feature of the transformed function, we'll consider how each transformation affects the function and match it with the appropriate feature. Here are the detailed steps:
1. Understanding [tex]\( h(x) = f(x) + 2 \)[/tex]:
- This transformation shifts the function [tex]\( f \)[/tex] vertically up by 2 units.
- If the original [tex]\( f(x) \)[/tex] had a y-intercept at (0,2), then [tex]\( h(x) = f(x) + 2 \)[/tex] will have a y-intercept 2 units higher.
- Therefore, the y-intercept will be at (0,4).
Matching:
- [tex]\( h(x) = f(x) + 2 \)[/tex] matches with [tex]\( y \)[/tex]-intercept at (0,4).
2. Understanding [tex]\( j(x) = f(x + 2) \)[/tex]:
- This transformation shifts the function [tex]\( f \)[/tex] horizontally to the left by 2 units.
- The behavior of the function as [tex]\( x \)[/tex] changes could be shifted similarly. If the function [tex]\( f \)[/tex] decreases as [tex]\( x \)[/tex] increases, the shift to the left won't change the decreasing nature of the function.
Matching:
- [tex]\( j(x) = f(x + 2 \)[/tex] matches with [tex]\( \)[/tex]function decreases as [tex]\( x \)[/tex] increases.
3. Understanding [tex]\( m(x) = -f(x) \)[/tex]:
- This transformation reflects the function [tex]\( f \)[/tex] over the x-axis.
- If [tex]\( f(x) \)[/tex] has an asymptote, reflecting it over the x-axis won't affect the x-position of the asymptote, but its y-value may change based on the nature of [tex]\( f \)[/tex].
Matching:
- [tex]\( m(x) = -f(x) \)[/tex] matches with [tex]\( \)[/tex]asymptote of [tex]\( y = 2 \)[/tex].
4. Understanding [tex]\( g(x) = 2 f(x) \)[/tex]:
- This transformation stretches the function [tex]\( f \)[/tex] vertically by a factor of 2.
- If the original [tex]\( f(x) \)[/tex] had a y-intercept at (0,1), then [tex]\( g(x) = 2 f(x) \)[/tex] will double the y-intercept value to (0,2).
Matching:
- [tex]\( g(x) = 2 f(x) \)[/tex] matches with [tex]\( y \)[/tex]-intercept at (0,2).
Putting it all together, the correct matches are:
[tex]\[ \begin{align*} \boxed{h(x) = f(x) + 2} \ &\longrightarrow \ y \text{-intercept at } (0,4), \\ \boxed{j(x) = f(x + 2)} \ &\longrightarrow \ \text{function decreases as } x \text{ increases}, \\ \boxed{m(x) = -f(x)} \ &\longrightarrow \ \text{asymptote of } y = 2, \\ \boxed{g(x) = 2 f(x)} \ &\longrightarrow \ y \text{-intercept at } (0,2). \end{align*} \][/tex]
1. Understanding [tex]\( h(x) = f(x) + 2 \)[/tex]:
- This transformation shifts the function [tex]\( f \)[/tex] vertically up by 2 units.
- If the original [tex]\( f(x) \)[/tex] had a y-intercept at (0,2), then [tex]\( h(x) = f(x) + 2 \)[/tex] will have a y-intercept 2 units higher.
- Therefore, the y-intercept will be at (0,4).
Matching:
- [tex]\( h(x) = f(x) + 2 \)[/tex] matches with [tex]\( y \)[/tex]-intercept at (0,4).
2. Understanding [tex]\( j(x) = f(x + 2) \)[/tex]:
- This transformation shifts the function [tex]\( f \)[/tex] horizontally to the left by 2 units.
- The behavior of the function as [tex]\( x \)[/tex] changes could be shifted similarly. If the function [tex]\( f \)[/tex] decreases as [tex]\( x \)[/tex] increases, the shift to the left won't change the decreasing nature of the function.
Matching:
- [tex]\( j(x) = f(x + 2 \)[/tex] matches with [tex]\( \)[/tex]function decreases as [tex]\( x \)[/tex] increases.
3. Understanding [tex]\( m(x) = -f(x) \)[/tex]:
- This transformation reflects the function [tex]\( f \)[/tex] over the x-axis.
- If [tex]\( f(x) \)[/tex] has an asymptote, reflecting it over the x-axis won't affect the x-position of the asymptote, but its y-value may change based on the nature of [tex]\( f \)[/tex].
Matching:
- [tex]\( m(x) = -f(x) \)[/tex] matches with [tex]\( \)[/tex]asymptote of [tex]\( y = 2 \)[/tex].
4. Understanding [tex]\( g(x) = 2 f(x) \)[/tex]:
- This transformation stretches the function [tex]\( f \)[/tex] vertically by a factor of 2.
- If the original [tex]\( f(x) \)[/tex] had a y-intercept at (0,1), then [tex]\( g(x) = 2 f(x) \)[/tex] will double the y-intercept value to (0,2).
Matching:
- [tex]\( g(x) = 2 f(x) \)[/tex] matches with [tex]\( y \)[/tex]-intercept at (0,2).
Putting it all together, the correct matches are:
[tex]\[ \begin{align*} \boxed{h(x) = f(x) + 2} \ &\longrightarrow \ y \text{-intercept at } (0,4), \\ \boxed{j(x) = f(x + 2)} \ &\longrightarrow \ \text{function decreases as } x \text{ increases}, \\ \boxed{m(x) = -f(x)} \ &\longrightarrow \ \text{asymptote of } y = 2, \\ \boxed{g(x) = 2 f(x)} \ &\longrightarrow \ y \text{-intercept at } (0,2). \end{align*} \][/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.
