Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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]
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]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
