Question

The function f(x) = –x^3 + 7x^2 – 7x – 15 has zeros located at 5, 3, –1. Verify the zeros of f(x) and explain how you verified them. Describe the end behavior of the function.

Answer 1

Answer:

See below for answers and explanations

Step-by-step explanation:

If we check x=5:

$f(5)=-(5)^3+7(5)^2-7(5)-15\\f(5)=-125+7(25)-35-15\\f(5)=-125+175-50\\f(5)=50-50\\f(5)=0$

So $x=5$ is indeed a zero

If we check x=3:

$f(3)=-(3)^3+7(3)^2-7(3)-15\\f(3)=-27+7(9)-21-15\\f(3)=-27+63-36\\f(3)=36-36\\f(3)=0$

So $x=3$ is also indeed a zero

If we check x=-1:

$f(-1)=-(-1)^3+7(-1)^2-7(-1)-15\\f(-1)=1+7+7-15\\f(-1)=8+7-15\\f(-1)=15-15\\f(-1)=0$

So $x=-1$ is also indeed a zero

End Behavior

Since the degree of $-x^3$ is odd $(3)$ and the leading coefficient is negative $(-1)$, the end behavior of function $f$ is as follows:

• As $x\rightarrow-\infty,f(x)\rightarrow+\infty$
• As $x\rightarrow+\infty,f(x)\rightarrow-\infty$