Question

test review #3, chapter 5
score: 6.5/37 answered: 6/37
question 7
given the function $p(x) = x^3 - x^2 - 12x$.
the $y$-intercept is
the $x$-intercepts is/are
when $x \to \infty$, $y \to$?
when $x \to -\infty$, $y \to$?
question help: video

Explanation:

Step1: Find y-intercept (set $x=0$)

$P(0) = 0^3 - 0^2 - 12(0) = 0$

Step2: Factor for x-intercepts (set $P(x)=0$)

$x^3 - x^2 - 12x = 0$
$x(x^2 - x - 12) = 0$
$x(x-4)(x+3) = 0$

Step3: Solve for x-intercept values

$x=0$, $x-4=0 \implies x=4$, $x+3=0 \implies x=-3$

Step4: Analyze end behavior (leading term $x^3$)

As $x \to \infty$, $x^3 \to \infty$, so $y \to \infty$

Step5: Analyze end behavior (negative x)

As $x \to -\infty$, $x^3 \to -\infty$, so $y \to -\infty$

Answer:

The y-intercept is $0$
The x-intercepts is/are $-3, 0, 4$
When $x \to \infty$, $y \to \infty$
When $x \to -\infty$, $y \to -\infty$