Let's go through the solution step-by-step.
### Part A: Checking for Zeros
To determine if [tex]\(\-5, \-3\)[/tex], and [tex]\(\1\)[/tex] are zeros of the function [tex]\(f(x) = -x^3 - 7x^2 - 7x + 15\)[/tex], we need to substitute each value into the equation and see if it equals zero.
Step 1: Checking [tex]\( x = -5 \)[/tex]
[tex]\[
f(-5) = -(-5)^3 - 7(-5)^2 - 7(-5) + 15
\][/tex]
[tex]\[
= -( -125 ) - 7( 25 ) + 35 + 15
\][/tex]
[tex]\[
= 125 - 175 + 35 + 15
\][/tex]
[tex]\[
= 125 - 175 + 50
\][/tex]
[tex]\[
= 0
\][/tex]
So, [tex]\( -5 \)[/tex] is a zero of [tex]\( f(x) \)[/tex].
Step 2: Checking [tex]\( x = -3 \)[/tex]
[tex]\[
f(-3) = -(-3)^3 - 7(-3)^2 - 7(-3) + 15
\][/tex]
[tex]\[
= -( -27 ) - 7(9) + 21 + 15
\][/tex]
[tex]\[
= 27 - 63 + 21 + 15
\][/tex]
[tex]\[
= 27 - 63 + 36
\][/tex]
[tex]\[
= 0
\][/tex]
So, [tex]\( -3 \)[/tex] is a zero of [tex]\( f(x) \)[/tex].
Step 3: Checking [tex]\( x = 1 \)[/tex]
[tex]\[
f(1) = -(1)^3 - 7(1)^2 - 7(1) + 15
\][/tex]
[tex]\[
= -(1) - 7(1) - 7 + 15
\][/tex]
[tex]\[
= -1 - 7 - 7 + 15
\][/tex]
[tex]\[
= -15 + 15
\][/tex]
[tex]\[
= 0
\][/tex]
So, [tex]\( 1 \)[/tex] is a zero of [tex]\( f(x) \)[/tex].
Thus, [tex]\(-5\)[/tex], [tex]\(-3\)[/tex], and [tex]\(\1\)[/tex] are indeed zeros of [tex]\(f(x)\)[/tex].
### Part B: End Behavior of [tex]\( f(x) \)[/tex]
The end behavior of a polynomial function is determined by its leading term. For the polynomial [tex]\( f(x) = -x^3 - 7x^2 - 7x + 15 \)[/tex], the leading term is [tex]\( -x^3 \)[/tex].
As [tex]\( x \to \infty \)[/tex]:
Consider the term [tex]\( -x^3 \)[/tex]:
- Because of the negative sign, as [tex]\( x \)[/tex] becomes increasingly positive, [tex]\( -x^3 \)[/tex] will become increasingly negative.
Therefore, as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
As [tex]\( x \to -\infty \)[/tex]:
- Similarly, for [tex]\( -x^3 \)[/tex], as [tex]\( x \)[/tex] becomes increasingly negative, [tex]\( -x^3 \)[/tex] will become increasingly positive because the cube of a negative number is negative, and then multiplied by another negative (from the coefficient), it becomes positive.
Therefore, as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
In summary:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]
This is the end behavior of the function [tex]\( f(x) = -x^3 - 7x^2 - 7x + 15 \)[/tex].
