# How do you graph by using the zeros for f(x)=-48x^2+3x^4?

Oct 7, 2017

Zeros at: $4 , - 4 , 0$

#### Explanation:

We need to find roots, so:

Notice that we can rewrite the expression as a quadratic:

Let $z = {x}^{2}$

Then:

$3 {z}^{2} - 48 z = 0$

Solving for $z$:

$z \left(3 z - 48\right) = 0 \implies z = 0 , \mathmr{and} z = \frac{48}{3} = 16$

But $z = {x}^{2}$

So:

${x}^{2} = 0 \implies x = 0$

${x}^{2} = 16 \implies x = \sqrt{16} \implies x = \pm 4$

Other values can be found by putting in value for $x$ and calculating corresponding values of $y$