Answer :
To identify key features and commonalities between the polynomial function [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex] and the exponential function [tex]\( g(x) = 2^x \)[/tex], let's examine each function critically.
Firstly, consider the polynomial function [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
1. Finding the y-intercept of [tex]\( f(x) \)[/tex]:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]. Plugging [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = -(0)^2 + 2(0) + 1 = 1. \][/tex]
Therefore, the y-intercept of [tex]\( f(x) \)[/tex] is [tex]\( (0, 1) \)[/tex].
2. Finding the y-intercept of [tex]\( g(x) \)[/tex]:
Similarly, the y-intercept occurs when [tex]\( x = 0 \)[/tex]. Plugging [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = 2^0 = 1. \][/tex]
Thus, the y-intercept of [tex]\( g(x) \)[/tex] is also [tex]\( (0, 1) \)[/tex].
Based on this analysis:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same y-intercept of [tex]\( (0,1) \)[/tex].
Next, let's evaluate the other statements about the functions:
3. Behavior over the interval [tex]\([1, \infty)\)[/tex]:
For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- This is a downward-opening parabola. To see where it starts decreasing, note that the vertex form of the quadratic function indicates it starts decreasing after the vertex. Completing the square or using the vertex formula, we find the vertex at [tex]\( x = 1 \)[/tex]. Therefore, it decreases for [tex]\( x \geq 1 \)[/tex].
For [tex]\( g(x) = 2^x \)[/tex]:
- This is an exponential growth function. It continuously increases as [tex]\( x \)[/tex] increases.
Therefore, it is not true that both functions decrease over the interval [tex]\([1, \infty)\)[/tex].
4. Range of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The range is determined by the maximum value of the parabola. The vertex form shows the maximum value is at the vertex, [tex]\( (1, 2) \)[/tex]. Hence, the range of [tex]\( f(x) \)[/tex] is [tex]\((-\infty, 2]\)[/tex].
For [tex]\( g(x) = 2^x \)[/tex]:
- This grows without bound as [tex]\( x \)[/tex] increases, and it approaches zero as [tex]\( x \)[/tex] decreases without bound. So its range is [tex]\((0, \infty)\)[/tex].
Therefore, the statement about both having the same range of [tex]\((-\infty, 2)\)[/tex] is incorrect.
5. x-intercepts of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- We solve [tex]\( -x^2 + 2x + 1 = 0 \)[/tex] to find the x-intercepts.
[tex]\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(-1)} = \frac{-2 \pm \sqrt{4 + 4}}{-2} = \frac{-2 \pm \sqrt{8}}{-2} = \frac{-2 \pm 2\sqrt{2}}{-2}. \][/tex]
This simplifies to [tex]\( x = -1 \)[/tex] and [tex]\( x = 3 \)[/tex]. So the x-intercepts are [tex]\( (-1, 0) \)[/tex] and [tex]\( (3, 0) \)[/tex].
For [tex]\( g(x) = 2^x \)[/tex]:
- This is always positive for all real [tex]\( x \)[/tex], hence it has no x-intercepts.
Therefore, the statement about both having the same x-intercept of [tex]\( (-1,0) \)[/tex] is false.
In conclusion, the correct statement about the key features the two functions share is:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same y-intercept of [tex]\( (0,1) \)[/tex].
Firstly, consider the polynomial function [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
1. Finding the y-intercept of [tex]\( f(x) \)[/tex]:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]. Plugging [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = -(0)^2 + 2(0) + 1 = 1. \][/tex]
Therefore, the y-intercept of [tex]\( f(x) \)[/tex] is [tex]\( (0, 1) \)[/tex].
2. Finding the y-intercept of [tex]\( g(x) \)[/tex]:
Similarly, the y-intercept occurs when [tex]\( x = 0 \)[/tex]. Plugging [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = 2^0 = 1. \][/tex]
Thus, the y-intercept of [tex]\( g(x) \)[/tex] is also [tex]\( (0, 1) \)[/tex].
Based on this analysis:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same y-intercept of [tex]\( (0,1) \)[/tex].
Next, let's evaluate the other statements about the functions:
3. Behavior over the interval [tex]\([1, \infty)\)[/tex]:
For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- This is a downward-opening parabola. To see where it starts decreasing, note that the vertex form of the quadratic function indicates it starts decreasing after the vertex. Completing the square or using the vertex formula, we find the vertex at [tex]\( x = 1 \)[/tex]. Therefore, it decreases for [tex]\( x \geq 1 \)[/tex].
For [tex]\( g(x) = 2^x \)[/tex]:
- This is an exponential growth function. It continuously increases as [tex]\( x \)[/tex] increases.
Therefore, it is not true that both functions decrease over the interval [tex]\([1, \infty)\)[/tex].
4. Range of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The range is determined by the maximum value of the parabola. The vertex form shows the maximum value is at the vertex, [tex]\( (1, 2) \)[/tex]. Hence, the range of [tex]\( f(x) \)[/tex] is [tex]\((-\infty, 2]\)[/tex].
For [tex]\( g(x) = 2^x \)[/tex]:
- This grows without bound as [tex]\( x \)[/tex] increases, and it approaches zero as [tex]\( x \)[/tex] decreases without bound. So its range is [tex]\((0, \infty)\)[/tex].
Therefore, the statement about both having the same range of [tex]\((-\infty, 2)\)[/tex] is incorrect.
5. x-intercepts of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- We solve [tex]\( -x^2 + 2x + 1 = 0 \)[/tex] to find the x-intercepts.
[tex]\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(-1)} = \frac{-2 \pm \sqrt{4 + 4}}{-2} = \frac{-2 \pm \sqrt{8}}{-2} = \frac{-2 \pm 2\sqrt{2}}{-2}. \][/tex]
This simplifies to [tex]\( x = -1 \)[/tex] and [tex]\( x = 3 \)[/tex]. So the x-intercepts are [tex]\( (-1, 0) \)[/tex] and [tex]\( (3, 0) \)[/tex].
For [tex]\( g(x) = 2^x \)[/tex]:
- This is always positive for all real [tex]\( x \)[/tex], hence it has no x-intercepts.
Therefore, the statement about both having the same x-intercept of [tex]\( (-1,0) \)[/tex] is false.
In conclusion, the correct statement about the key features the two functions share is:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same y-intercept of [tex]\( (0,1) \)[/tex].
