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Sagot :
Sure, let's match each transformation of function [tex]\( f \)[/tex] to its relevant feature step-by-step.
1. Feature: [tex]\( y \)[/tex]-intercept at [tex]\((0, 2)\)[/tex]
- We need to find the function for which the [tex]\( y \)[/tex]-intercept is at the point [tex]\((0, 2)\)[/tex].
- The [tex]\( y \)[/tex]-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex].
- Given the function [tex]\( h(x) = f(x) + 2 \)[/tex], when [tex]\( x = 0 \)[/tex],
[tex]\[ h(0) = f(0) + 2 \][/tex]
If [tex]\( f(0) = 0 \)[/tex], then [tex]\( h(0) = 2 \)[/tex]. Thus, [tex]\( h(x) = f(x) + 2 \)[/tex] shifts the original function [tex]\( f \)[/tex] vertically up by 2 units, making the [tex]\( y \)[/tex]-intercept [tex]\( (0, 2) \)[/tex].
Therefore, [tex]\( h(x) = f(x) + 2 \)[/tex] corresponds to a [tex]\( y \)[/tex]-intercept at [tex]\((0, 2)\)[/tex].
2. Feature: Asymptote of [tex]\( y = 2 \)[/tex]
- An asymptote is a line that a function approaches as [tex]\( x \)[/tex] tends to infinity or negative infinity.
- For the function [tex]\( h(x) = f(x) + 2 \)[/tex], if [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex], then [tex]\( h(x) \)[/tex] will have a horizontal asymptote at [tex]\( y = 2 \)[/tex].
- The function [tex]\( h(x) = f(x) + 2 \)[/tex] shifts the original graph of [tex]\( f(x) \)[/tex] vertically upwards by 2 units.
Thus, [tex]\( h(x) = f(x) + 2 \)[/tex] also corresponds to an asymptote of [tex]\( y = 2 \)[/tex].
3. Feature: [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex]
- We need to find the function for which the [tex]\( y \)[/tex]-intercept is at the point [tex]\((0, 4)\)[/tex].
- For the function [tex]\( g(x) = 2f(x) \)[/tex], evaluate it at [tex]\( x = 0 \)[/tex],
[tex]\[ g(0) = 2f(0) \][/tex]
If [tex]\( f(0) = 2 \)[/tex], then [tex]\( g(0) = 4 \)[/tex]. Thus, [tex]\( g(x) = 2f(x) \)[/tex] scales the output of the function [tex]\( f \)[/tex] by a factor of 2, making the [tex]\( y \)[/tex]-intercept [tex]\( (0, 4) \)[/tex].
Therefore, [tex]\( g(x) = 2f(x) \)[/tex] corresponds to a [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex].
4. Feature: Function decreases as [tex]\( x \)[/tex] increases
- We need to identify the function that results in the original function [tex]\( f(x) \)[/tex] decreasing as [tex]\( x \)[/tex] increases.
- For the function [tex]\( m(x) = -f(x) \)[/tex], if [tex]\( f(x) \)[/tex] is increasing, [tex]\( -f(x) \)[/tex] will be decreasing, because negating the function inverses its slope.
[tex]\[ \text{If } f(x) \text{ increases with respect to } x, \text{then } -f(x) \text{ decreases with respect to } x. \][/tex]
Therefore, [tex]\( m(x) = -f(x) \)[/tex] corresponds to the feature where the function decreases as [tex]\( x \)[/tex] increases.
Summarizing:
- [tex]\( y \)[/tex]-intercept at [tex]\((0, 2)\)[/tex] matches with [tex]\( h(x) = f(x) + 2 \)[/tex]
- Asymptote of [tex]\( y = 2 \)[/tex] matches with [tex]\( h(x) = f(x) + 2 \)[/tex]
- [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex] matches with [tex]\( g(x) = 2f(x) \)[/tex]
- Function decreases as [tex]\( x \)[/tex] increases matches with [tex]\( m(x) = -f(x) \)[/tex]
1. Feature: [tex]\( y \)[/tex]-intercept at [tex]\((0, 2)\)[/tex]
- We need to find the function for which the [tex]\( y \)[/tex]-intercept is at the point [tex]\((0, 2)\)[/tex].
- The [tex]\( y \)[/tex]-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex].
- Given the function [tex]\( h(x) = f(x) + 2 \)[/tex], when [tex]\( x = 0 \)[/tex],
[tex]\[ h(0) = f(0) + 2 \][/tex]
If [tex]\( f(0) = 0 \)[/tex], then [tex]\( h(0) = 2 \)[/tex]. Thus, [tex]\( h(x) = f(x) + 2 \)[/tex] shifts the original function [tex]\( f \)[/tex] vertically up by 2 units, making the [tex]\( y \)[/tex]-intercept [tex]\( (0, 2) \)[/tex].
Therefore, [tex]\( h(x) = f(x) + 2 \)[/tex] corresponds to a [tex]\( y \)[/tex]-intercept at [tex]\((0, 2)\)[/tex].
2. Feature: Asymptote of [tex]\( y = 2 \)[/tex]
- An asymptote is a line that a function approaches as [tex]\( x \)[/tex] tends to infinity or negative infinity.
- For the function [tex]\( h(x) = f(x) + 2 \)[/tex], if [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex], then [tex]\( h(x) \)[/tex] will have a horizontal asymptote at [tex]\( y = 2 \)[/tex].
- The function [tex]\( h(x) = f(x) + 2 \)[/tex] shifts the original graph of [tex]\( f(x) \)[/tex] vertically upwards by 2 units.
Thus, [tex]\( h(x) = f(x) + 2 \)[/tex] also corresponds to an asymptote of [tex]\( y = 2 \)[/tex].
3. Feature: [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex]
- We need to find the function for which the [tex]\( y \)[/tex]-intercept is at the point [tex]\((0, 4)\)[/tex].
- For the function [tex]\( g(x) = 2f(x) \)[/tex], evaluate it at [tex]\( x = 0 \)[/tex],
[tex]\[ g(0) = 2f(0) \][/tex]
If [tex]\( f(0) = 2 \)[/tex], then [tex]\( g(0) = 4 \)[/tex]. Thus, [tex]\( g(x) = 2f(x) \)[/tex] scales the output of the function [tex]\( f \)[/tex] by a factor of 2, making the [tex]\( y \)[/tex]-intercept [tex]\( (0, 4) \)[/tex].
Therefore, [tex]\( g(x) = 2f(x) \)[/tex] corresponds to a [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex].
4. Feature: Function decreases as [tex]\( x \)[/tex] increases
- We need to identify the function that results in the original function [tex]\( f(x) \)[/tex] decreasing as [tex]\( x \)[/tex] increases.
- For the function [tex]\( m(x) = -f(x) \)[/tex], if [tex]\( f(x) \)[/tex] is increasing, [tex]\( -f(x) \)[/tex] will be decreasing, because negating the function inverses its slope.
[tex]\[ \text{If } f(x) \text{ increases with respect to } x, \text{then } -f(x) \text{ decreases with respect to } x. \][/tex]
Therefore, [tex]\( m(x) = -f(x) \)[/tex] corresponds to the feature where the function decreases as [tex]\( x \)[/tex] increases.
Summarizing:
- [tex]\( y \)[/tex]-intercept at [tex]\((0, 2)\)[/tex] matches with [tex]\( h(x) = f(x) + 2 \)[/tex]
- Asymptote of [tex]\( y = 2 \)[/tex] matches with [tex]\( h(x) = f(x) + 2 \)[/tex]
- [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex] matches with [tex]\( g(x) = 2f(x) \)[/tex]
- Function decreases as [tex]\( x \)[/tex] increases matches with [tex]\( m(x) = -f(x) \)[/tex]
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