Answer :
Let's analyze the function [tex]\( f(x) = -2(3)^x + 6 \)[/tex] to determine its key features:
1. [tex]\( y \)[/tex]-intercept at [tex]\((0,4)\)[/tex]:
- To find the [tex]\( y \)[/tex]-intercept, substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -2(3)^0 + 6 = -2 \cdot 1 + 6 = -2 + 6 = 4 \][/tex]
- Therefore, [tex]\( f(x) \)[/tex] has a [tex]\( y \)[/tex]-intercept at [tex]\( (0, 4) \)[/tex].
2. Function approaches an integer as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\(-\infty \)[/tex], [tex]\( 3^x \)[/tex] approaches 0 because [tex]\( 3^x \)[/tex] exponentially decays to 0 for negative [tex]\( x \)[/tex].
- Thus,
[tex]\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (-2(3)^x + 6) = -2 \cdot 0 + 6 = 6 \][/tex]
- Therefore, [tex]\( f(x) \)[/tex] approaches the integer 6 as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex].
3. Increasing or decreasing on all intervals of [tex]\( x \)[/tex]:
- The function [tex]\( f(x) = -2(3)^x + 6 \)[/tex] is an exponential function where the base [tex]\( 3 \)[/tex] raised to the power of [tex]\( x \)[/tex] makes [tex]\( 3^x \)[/tex] an increasing function.
- Since it is multiplied by [tex]\(-2\)[/tex], [tex]\( -2(3)^x \)[/tex] is a decreasing function.
- Adding 6 does not change the fact that the function is decreasing.
- Thus, [tex]\( f(x) \)[/tex] is decreasing on all intervals of [tex]\( x \)[/tex].
4. [tex]\( x \)[/tex]-intercept at [tex]\((1,0)\)[/tex]:
- To find the [tex]\( x \)[/tex]-intercept, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 0 = -2(3)^x + 6 \implies -2(3)^x = -6 \implies (3)^x = 3 \implies x = 1 \][/tex]
- Therefore, [tex]\( f(x) \)[/tex] has an [tex]\( x \)[/tex]-intercept at [tex]\( (1,0) \)[/tex].
Considering these features, we answer the matching problem as follows:
- [tex]\( y \)[/tex]-intercept at [tex]\((0,4)\)[/tex]: only [tex]\( f(x) \)[/tex]
- Approaches an integer as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]: [tex]\( f(x) \)[/tex] and [tex]\( h(x) \)[/tex]
- Increasing on all intervals of [tex]\( x \)[/tex]: [tex]\( g(x) \)[/tex] (assumed given function)
- [tex]\( x \)[/tex]-intercept at [tex]\((1,0)\)[/tex]: only [tex]\( f(x) \)[/tex]
[tex]\[ \begin{align*} y-\text{intercept at } (0,4) & \quad \text{only } f(x) \\ \text{approaches an integer as } x \text{ approaches } -\infty & \quad f(x) \text{ and } h(x) \\ \text{increasing on all intervals of } x & \quad \text{only } g(x) \\ x-\text{intercept at } (1,0) & \quad \text{only } f(x) \\ \end{align*} \][/tex]
1. [tex]\( y \)[/tex]-intercept at [tex]\((0,4)\)[/tex]:
- To find the [tex]\( y \)[/tex]-intercept, substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -2(3)^0 + 6 = -2 \cdot 1 + 6 = -2 + 6 = 4 \][/tex]
- Therefore, [tex]\( f(x) \)[/tex] has a [tex]\( y \)[/tex]-intercept at [tex]\( (0, 4) \)[/tex].
2. Function approaches an integer as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\(-\infty \)[/tex], [tex]\( 3^x \)[/tex] approaches 0 because [tex]\( 3^x \)[/tex] exponentially decays to 0 for negative [tex]\( x \)[/tex].
- Thus,
[tex]\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (-2(3)^x + 6) = -2 \cdot 0 + 6 = 6 \][/tex]
- Therefore, [tex]\( f(x) \)[/tex] approaches the integer 6 as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex].
3. Increasing or decreasing on all intervals of [tex]\( x \)[/tex]:
- The function [tex]\( f(x) = -2(3)^x + 6 \)[/tex] is an exponential function where the base [tex]\( 3 \)[/tex] raised to the power of [tex]\( x \)[/tex] makes [tex]\( 3^x \)[/tex] an increasing function.
- Since it is multiplied by [tex]\(-2\)[/tex], [tex]\( -2(3)^x \)[/tex] is a decreasing function.
- Adding 6 does not change the fact that the function is decreasing.
- Thus, [tex]\( f(x) \)[/tex] is decreasing on all intervals of [tex]\( x \)[/tex].
4. [tex]\( x \)[/tex]-intercept at [tex]\((1,0)\)[/tex]:
- To find the [tex]\( x \)[/tex]-intercept, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 0 = -2(3)^x + 6 \implies -2(3)^x = -6 \implies (3)^x = 3 \implies x = 1 \][/tex]
- Therefore, [tex]\( f(x) \)[/tex] has an [tex]\( x \)[/tex]-intercept at [tex]\( (1,0) \)[/tex].
Considering these features, we answer the matching problem as follows:
- [tex]\( y \)[/tex]-intercept at [tex]\((0,4)\)[/tex]: only [tex]\( f(x) \)[/tex]
- Approaches an integer as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]: [tex]\( f(x) \)[/tex] and [tex]\( h(x) \)[/tex]
- Increasing on all intervals of [tex]\( x \)[/tex]: [tex]\( g(x) \)[/tex] (assumed given function)
- [tex]\( x \)[/tex]-intercept at [tex]\((1,0)\)[/tex]: only [tex]\( f(x) \)[/tex]
[tex]\[ \begin{align*} y-\text{intercept at } (0,4) & \quad \text{only } f(x) \\ \text{approaches an integer as } x \text{ approaches } -\infty & \quad f(x) \text{ and } h(x) \\ \text{increasing on all intervals of } x & \quad \text{only } g(x) \\ x-\text{intercept at } (1,0) & \quad \text{only } f(x) \\ \end{align*} \][/tex]
