Answer :
Sure, let’s go through the problem step by step.
### Part (a): Determine the end behavior of the graph of the function
Given the function [tex]\( f(x) = (x + 4)^2 (1 - x) \)[/tex], the term with the highest degree, which is [tex]\(-x^3\)[/tex], will dominate as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex]. Therefore, the end behavior of the function is similar to that of the function [tex]\( y = -x^3 \)[/tex].
End behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \approx -x^3 \to -\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \approx -x^3 \to \infty \)[/tex].
So, the end behavior is:
The graph of [tex]\( f \)[/tex] behaves like [tex]\( y = -x^3 \)[/tex] for large values of [tex]\( |x| \)[/tex].
### Part (b): Find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-intercepts of the graph of the function
To find the intercepts:
[tex]\( x \)[/tex]-intercepts:
Set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[ f(x) = (x + 4)^2 (1 - x) = 0 \][/tex]
The product is zero if any factor is zero:
1. [tex]\( (x + 4)^2 = 0 \Rightarrow x + 4 = 0 \Rightarrow x = -4 \)[/tex]
2. [tex]\( 1 - x = 0 \Rightarrow x = 1 \)[/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are [tex]\( (-4, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
[tex]\( y \)[/tex]-intercept:
Evaluate [tex]\( f(0) \)[/tex].
[tex]\[ f(0) = (0 + 4)^2 (1 - 0) = 16 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 16) \)[/tex].
### Summary of intercepts:
- [tex]\( x \)[/tex]-intercepts: [tex]\( (-4, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( (0, 16) \)[/tex]
Therefore:
The graph of [tex]\( f \)[/tex] behaves like [tex]\( y = -x^3 \)[/tex] for large values of [tex]\( |x| \)[/tex].
The [tex]\( x \)[/tex]-intercepts are [tex]\((-4, 0)\)[/tex] and [tex]\((1, 0)\)[/tex].
The [tex]\( y \)[/tex]-intercept is [tex]\((0, 16)\)[/tex].
### Part (a): Determine the end behavior of the graph of the function
Given the function [tex]\( f(x) = (x + 4)^2 (1 - x) \)[/tex], the term with the highest degree, which is [tex]\(-x^3\)[/tex], will dominate as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex]. Therefore, the end behavior of the function is similar to that of the function [tex]\( y = -x^3 \)[/tex].
End behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \approx -x^3 \to -\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \approx -x^3 \to \infty \)[/tex].
So, the end behavior is:
The graph of [tex]\( f \)[/tex] behaves like [tex]\( y = -x^3 \)[/tex] for large values of [tex]\( |x| \)[/tex].
### Part (b): Find the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-intercepts of the graph of the function
To find the intercepts:
[tex]\( x \)[/tex]-intercepts:
Set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[ f(x) = (x + 4)^2 (1 - x) = 0 \][/tex]
The product is zero if any factor is zero:
1. [tex]\( (x + 4)^2 = 0 \Rightarrow x + 4 = 0 \Rightarrow x = -4 \)[/tex]
2. [tex]\( 1 - x = 0 \Rightarrow x = 1 \)[/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are [tex]\( (-4, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
[tex]\( y \)[/tex]-intercept:
Evaluate [tex]\( f(0) \)[/tex].
[tex]\[ f(0) = (0 + 4)^2 (1 - 0) = 16 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 16) \)[/tex].
### Summary of intercepts:
- [tex]\( x \)[/tex]-intercepts: [tex]\( (-4, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( (0, 16) \)[/tex]
Therefore:
The graph of [tex]\( f \)[/tex] behaves like [tex]\( y = -x^3 \)[/tex] for large values of [tex]\( |x| \)[/tex].
The [tex]\( x \)[/tex]-intercepts are [tex]\((-4, 0)\)[/tex] and [tex]\((1, 0)\)[/tex].
The [tex]\( y \)[/tex]-intercept is [tex]\((0, 16)\)[/tex].
