Question

consider function f:$f(x) = x^3 + 2x^2 - 5x - 6$select the locations of the zeros of function $f$ on the coordinate plane. then select the end behavior of its graph.as $x$ approaches positive infinity, $f(x)$ approaches negative infinityas $x$ approaches negative infinity, $f(x)$ approaches negative infinityas $x$ approaches negative infinity, $f(x)$ approaches positive infinityas $x$ approaches positive infinity, $f(x)$ approaches positive infinity

Explanation:

Step1: Factor the cubic function

First, use rational root theorem to test possible roots. Test $x=1$:
$f(1)=1^3 + 2(1)^2 -5(1)-6=1+2-5-6=-8
eq0$
Test $x=-1$:
$f(-1)=(-1)^3 + 2(-1)^2 -5(-1)-6=-1+2+5-6=0$
So $(x+1)$ is a factor. Perform polynomial division or use synthetic division:
$$\frac{x^3+2x^2-5x-6}{x+1}=x^2+x-6$$
Factor the quadratic: $x^2+x-6=(x+3)(x-2)$
Thus, $f(x)=(x+3)(x+1)(x-2)$

Step2: Find zeros of the function

Set $f(x)=0$, solve for $x$:
$(x+3)(x+1)(x-2)=0$
$x+3=0 \implies x=-3$
$x+1=0 \implies x=-1$
$x-2=0 \implies x=2$
The zeros are at $(-3,0)$, $(-1,0)$, $(2,0)$

Step3: Determine end behavior

For cubic function $f(x)=x^3+2x^2-5x-6$, the leading term is $x^3$ with positive coefficient.

• As $x\to+\infty$, $x^3\to+\infty$, so $f(x)\to+\infty$
• As $x\to-\infty$, $x^3\to-\infty$, so $f(x)\to-\infty$

Answer:

1. Zeros locations: $(-3, 0)$, $(-1, 0)$, $(2, 0)$
2. End behavior:

• As $x$ approaches positive infinity, $f(x)$ approaches positive infinity
• As $x$ approaches negative infinity, $f(x)$ approaches negative infinity