# How do you graph by using the zeros for f(x)=x^2(x-4)^2?

Jun 16, 2017

$x$-intercepts: (0, 0), (4, 0); " "y-intercept: $\left(0 , 0\right)$
touch points at $x = 0 \text{ and } x = 4$
When $x \to \infty$, $y = f \left(x\right) \to \infty$.

#### Explanation:

Given: $f \left(x\right) = {x}^{2} {\left(x - 4\right)}^{2}$

The zeros are the values of $x$ of the polynomial where $f \left(x\right) = 0$.

The $x$-intercepts are the points where $f \left(x\right) = 0 : \left(x , 0\right)$

Finding the zeros of the function:

Factor the function completely:

$f \left(x\right) = {x}^{2} {\left(x - 4\right)}^{2} \text{ }$ Already completed

Set $f \left(x\right) = 0 : \text{ } f \left(x\right) = {x}^{2} {\left(x - 4\right)}^{2} = 0$

${x}^{2} = 0 \text{ and } {\left(x - 4\right)}^{2} = 0$

$x = 0 \text{ and } x = 4$

$x$-intercepts: $\left(0 , 0\right) , \left(4 , 0\right)$

Find the $y$-intercept by setting $x = 0$: $y = {0}^{2} {\left(0 - 4\right)}^{2} = 0$

$y$-intercept: $\left(0 , 0\right)$

Find multiplicity : $\text{ } 2$ at $x = 0 \text{ and } 2$ at $x = 4$.

Even multiplicity causes touch points. Instead of crossing the $x$-axis, the function just touches the $x$-axis at the zeros. If the multiplicity is odd, the function will cross the $x$-axis.

Find the limit of $f \left(x\right)$ as $x \to \infty$ :

If you distribute $f \left(x\right) = {x}^{4} + \ldots$. When $x \to \infty$, $y = f \left(x\right) \to {\infty}^{4} = \infty$.

This means the graph starts high and to the right, touches the $x$-axis at $x = 4$, increases in the positive $y$-direction for a short distance, then eventually decreases and touches the $x$-axis at $x = 0$ and then increases without bound.

The graph looks like a W:
graph{x^2(x-4)^2 [-43.17, 36.83, -5.76, 34.24]}