Answer :
Certainly! Let's analyze both functions to determine their maximum [tex]\( y \)[/tex]-values.
### Function [tex]\( f(x) = -5^x + 2 \)[/tex]
This is an exponential function of the form [tex]\( f(x) = -5^x + 2 \)[/tex].
1. Behavior as [tex]\( x \)[/tex] increases:
- As [tex]\( x \)[/tex] becomes very large (moving towards [tex]\( +\infty \)[/tex]):
[tex]\[ -5^x \][/tex] becomes very large (positive) in magnitude and negative in value, causing [tex]\( f(x) \)[/tex] to decrease without bound.
- Thus, [tex]\( f(x) \)[/tex] heads towards negative infinity.
2. Behavior as [tex]\( x \)[/tex] decreases:
- As [tex]\( x \)[/tex] becomes very negative (moving towards [tex]\( -\infty \)[/tex]):
[tex]\[ 5^x \][/tex] approaches 0 because any positive number raised to a large negative power approaches 0.
- Hence,
[tex]\[ f(x) \approx -0 + 2 = 2 \][/tex]
- However, note that [tex]\( f(x) \)[/tex] never actually reaches 2; it only gets arbitrarily close as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex].
Therefore, the maximum [tex]\( y \)[/tex]-value that [tex]\( f(x) \)[/tex] approaches is 2, but it never attains exactly 2.
### Function [tex]\( g(x) = -5x^2 + 2 \)[/tex]
This is a quadratic function of the form [tex]\( g(x) = -5x^2 + 2 \)[/tex].
1. This is a downward-facing parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative).
2. The vertex of this parabola will give us the maximum [tex]\( y \)[/tex]-value:
- The general form of a parabola [tex]\( ax^2 + bx + c \)[/tex] has its vertex at [tex]\( x = -\frac{b}{2a} \)[/tex]
- Here, the function is [tex]\( g(x) = -5x^2 + 2 \)[/tex]
- The [tex]\( x \)[/tex]-coordinate of the vertex [tex]\( x = -\frac{b}{2a} = -\frac{0}{2(-5)} = 0 \)[/tex]
3. Substituting [tex]\( x = 0 \)[/tex] back into the function:
[tex]\[ g(0) = -5(0)^2 + 2 = 2 \][/tex]
Therefore, the maximum [tex]\( y \)[/tex]-value for [tex]\( g(x) \)[/tex] is exactly 2.
### Conclusion
- For [tex]\( f(x) \)[/tex], the maximum [tex]\( y \)[/tex]-value approaches 2 but never quite reaches it.
- For [tex]\( g(x) \)[/tex], the maximum [tex]\( y \)[/tex]-value is exactly 2.
Based on our analysis:
- Answer D ("The maximum [tex]\( y \)[/tex]-value of [tex]\( f(x) \)[/tex] approaches 2.") accurately describes [tex]\( f(x) \)[/tex].
- Answer C ("[tex]\( g(x) \)[/tex] has the largest possible [tex]\( y \)[/tex]-value.") is true since 2 is a maximum value reached by [tex]\( g(x) \)[/tex].
Thus, the best response is:
C. [tex]\( g(x) \)[/tex] has the largest possible [tex]\( y \)[/tex]-value.
### Function [tex]\( f(x) = -5^x + 2 \)[/tex]
This is an exponential function of the form [tex]\( f(x) = -5^x + 2 \)[/tex].
1. Behavior as [tex]\( x \)[/tex] increases:
- As [tex]\( x \)[/tex] becomes very large (moving towards [tex]\( +\infty \)[/tex]):
[tex]\[ -5^x \][/tex] becomes very large (positive) in magnitude and negative in value, causing [tex]\( f(x) \)[/tex] to decrease without bound.
- Thus, [tex]\( f(x) \)[/tex] heads towards negative infinity.
2. Behavior as [tex]\( x \)[/tex] decreases:
- As [tex]\( x \)[/tex] becomes very negative (moving towards [tex]\( -\infty \)[/tex]):
[tex]\[ 5^x \][/tex] approaches 0 because any positive number raised to a large negative power approaches 0.
- Hence,
[tex]\[ f(x) \approx -0 + 2 = 2 \][/tex]
- However, note that [tex]\( f(x) \)[/tex] never actually reaches 2; it only gets arbitrarily close as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex].
Therefore, the maximum [tex]\( y \)[/tex]-value that [tex]\( f(x) \)[/tex] approaches is 2, but it never attains exactly 2.
### Function [tex]\( g(x) = -5x^2 + 2 \)[/tex]
This is a quadratic function of the form [tex]\( g(x) = -5x^2 + 2 \)[/tex].
1. This is a downward-facing parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative).
2. The vertex of this parabola will give us the maximum [tex]\( y \)[/tex]-value:
- The general form of a parabola [tex]\( ax^2 + bx + c \)[/tex] has its vertex at [tex]\( x = -\frac{b}{2a} \)[/tex]
- Here, the function is [tex]\( g(x) = -5x^2 + 2 \)[/tex]
- The [tex]\( x \)[/tex]-coordinate of the vertex [tex]\( x = -\frac{b}{2a} = -\frac{0}{2(-5)} = 0 \)[/tex]
3. Substituting [tex]\( x = 0 \)[/tex] back into the function:
[tex]\[ g(0) = -5(0)^2 + 2 = 2 \][/tex]
Therefore, the maximum [tex]\( y \)[/tex]-value for [tex]\( g(x) \)[/tex] is exactly 2.
### Conclusion
- For [tex]\( f(x) \)[/tex], the maximum [tex]\( y \)[/tex]-value approaches 2 but never quite reaches it.
- For [tex]\( g(x) \)[/tex], the maximum [tex]\( y \)[/tex]-value is exactly 2.
Based on our analysis:
- Answer D ("The maximum [tex]\( y \)[/tex]-value of [tex]\( f(x) \)[/tex] approaches 2.") accurately describes [tex]\( f(x) \)[/tex].
- Answer C ("[tex]\( g(x) \)[/tex] has the largest possible [tex]\( y \)[/tex]-value.") is true since 2 is a maximum value reached by [tex]\( g(x) \)[/tex].
Thus, the best response is:
C. [tex]\( g(x) \)[/tex] has the largest possible [tex]\( y \)[/tex]-value.
