Answer
The y-intercept is 0
The x-intercepts are -2, -1, 0, and 1
$\lim\limits_{x \to \infty} x^2(x^2-1)^2(x+2) = \infty$
$\lim\limits_{x \to -\infty} x^2(x^2-1)^2(x+2) = -\infty$
Work Step by Step
$y = x^2(x^2-1)^2(x+2)$
When $x=0$, then $~~y = (0)^2(0^2-1)^2(0+2) = 0$
When $y=0$:
$x^2(x^2-1)^2(x+2)= 0$
$x^2(x-1)^2(x+1)^2(x+2)= 0$
$x = 0, 1, -1, -2$
$\lim\limits_{x \to \infty} x^2(x^2-1)^2(x+2) = \infty$
This limit is the product of a large magnitude positive number, a large magnitude positive number, and a large magnitude positive number.
$\lim\limits_{x \to -\infty} x^2(x^2-1)^2(x+2) = -\infty$
This limit is the product of a large magnitude positive number, a large magnitude positive number, and a large magnitude negative number.
Note that the graph does not cross the x-axis at $x = -1$, $x=0,$ or $x=1$ because the terms $(x+1)^2$, $x^2$, and $(x-1)^2$ have even exponents.
