Answer :
Let's analyze the given functions [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex] and [tex]\( g(x) = 2^x \)[/tex] to identify their key features step by step.
### 1. Analyze Decrease Over the Interval [tex]\([1, \infty)\)[/tex]:
- For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The function is a downward-opening parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative).
- To confirm decreasing behavior, complete the square or find the vertex:
[tex]\[ f(x) = -(x^2 - 2x - 1) = -(x^2 - 2x + 1 - 1) = -(x - 1)^2 + 2. \][/tex]
- The vertex is at [tex]\(x = 1\)[/tex] and the maximum value is at [tex]\( f(1) = 2 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x \geq 1 \)[/tex].
- For [tex]\( g(x) = 2^x \)[/tex]:
- [tex]\( g(x) \)[/tex] is an exponential function with base 2.
- The function [tex]\( g(x) \)[/tex] is increasing for all [tex]\( x \)[/tex], so it doesn't decrease for any interval.
Thus, the claim that both functions decrease over the interval [tex]\([1, \infty)\)[/tex] is false.
### 2. Compare the Range:
- Range of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The maximum value occurs at [tex]\( x = 1 \)[/tex], [tex]\( f(1) = 2 \)[/tex].
- Since it opens downwards, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, 2]. \][/tex]
- Range of [tex]\( g(x) = 2^x \)[/tex]:
- Since [tex]\( g(x) \)[/tex] is an exponential function, the range is:
[tex]\[ (0, \infty). \][/tex]
Thus, the claim that both functions have the same range of [tex]\((- \infty, 2)\)[/tex] is false.
### 3. Compare the [tex]\( x \)[/tex]-Intercepts:
- [tex]\( x \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^2 + 2x + 1 = 0. \][/tex]
- Solving the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 4}}{-2}. \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{8}}{-2} = 1 \pm \sqrt{2}. \][/tex]
- The [tex]\( x \)[/tex]-intercepts are [tex]\( x = 1 + \sqrt{2} \)[/tex] and [tex]\( x = 1 - \sqrt{2} \)[/tex].
- [tex]\( x \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( g(x) = 0 \)[/tex]:
- The exponential function [tex]\( 2^x \)[/tex] never crosses the x-axis (it never equals zero), so there is no [tex]\( x \)[/tex]-intercept.
Thus, the claim that both functions have the same [tex]\( x \)[/tex]-intercept of [tex]\((-1, 0)\)[/tex] is false.
### 4. Compare the [tex]\( y \)[/tex]-Intercepts:
- [tex]\( y \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 + 2 \cdot 0 + 1 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
- [tex]\( y \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 2^0 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
Thus, the claim that both functions have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex] is true.
### Conclusion:
The only correct statement is:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex].
### 1. Analyze Decrease Over the Interval [tex]\([1, \infty)\)[/tex]:
- For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The function is a downward-opening parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative).
- To confirm decreasing behavior, complete the square or find the vertex:
[tex]\[ f(x) = -(x^2 - 2x - 1) = -(x^2 - 2x + 1 - 1) = -(x - 1)^2 + 2. \][/tex]
- The vertex is at [tex]\(x = 1\)[/tex] and the maximum value is at [tex]\( f(1) = 2 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x \geq 1 \)[/tex].
- For [tex]\( g(x) = 2^x \)[/tex]:
- [tex]\( g(x) \)[/tex] is an exponential function with base 2.
- The function [tex]\( g(x) \)[/tex] is increasing for all [tex]\( x \)[/tex], so it doesn't decrease for any interval.
Thus, the claim that both functions decrease over the interval [tex]\([1, \infty)\)[/tex] is false.
### 2. Compare the Range:
- Range of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The maximum value occurs at [tex]\( x = 1 \)[/tex], [tex]\( f(1) = 2 \)[/tex].
- Since it opens downwards, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, 2]. \][/tex]
- Range of [tex]\( g(x) = 2^x \)[/tex]:
- Since [tex]\( g(x) \)[/tex] is an exponential function, the range is:
[tex]\[ (0, \infty). \][/tex]
Thus, the claim that both functions have the same range of [tex]\((- \infty, 2)\)[/tex] is false.
### 3. Compare the [tex]\( x \)[/tex]-Intercepts:
- [tex]\( x \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^2 + 2x + 1 = 0. \][/tex]
- Solving the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 4}}{-2}. \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{8}}{-2} = 1 \pm \sqrt{2}. \][/tex]
- The [tex]\( x \)[/tex]-intercepts are [tex]\( x = 1 + \sqrt{2} \)[/tex] and [tex]\( x = 1 - \sqrt{2} \)[/tex].
- [tex]\( x \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( g(x) = 0 \)[/tex]:
- The exponential function [tex]\( 2^x \)[/tex] never crosses the x-axis (it never equals zero), so there is no [tex]\( x \)[/tex]-intercept.
Thus, the claim that both functions have the same [tex]\( x \)[/tex]-intercept of [tex]\((-1, 0)\)[/tex] is false.
### 4. Compare the [tex]\( y \)[/tex]-Intercepts:
- [tex]\( y \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 + 2 \cdot 0 + 1 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
- [tex]\( y \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 2^0 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
Thus, the claim that both functions have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex] is true.
### Conclusion:
The only correct statement is:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex].
