Answer
The y-intercept is 3
The x-intercepts are -1, 1, and 3
$\lim\limits_{x \to \infty} (3-x)(1+x)^2(1-x)^4 = -\infty$
$\lim\limits_{x \to -\infty} (3-x)(1+x)^2(1-x)^4 = \infty$
Work Step by Step
$y = (3-x)(1+x)^2(1-x)^4$
When $x=0$, then $~~y = (3-0)(1+0)^2(1-0)^4 = 3$
When $y=0$:
$(3-x)(1+x)^2(1-x)^4 = 0$
$x = 3, -1, 1$
$\lim\limits_{x \to \infty} (3-x)(1+x)^2(1-x)^4 = -\infty$
This limit is the product of a large magnitude negative number, a large magnitude positive number, and a large magnitude positive number.
$\lim\limits_{x \to -\infty} (3-x)(1+x)^2(1-x)^4 = \infty$
This limit is the product of a large magnitude positive number, a large magnitude positive number, and a large magnitude positive number.
Note that the graph does not cross the x-axis at $x = -1$ or $x=1$ because the terms $(1+x)^2$ and $(1-x)^4$ have even exponents.
