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

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Answer 1 · 2017-10-07T13:05:02

Zeros at: $4 , -4 , 0$

We need to find roots, so:

Notice that we can rewrite the expression as a quadratic:

Let $z = x^2$

Then:

$3z^2-48z=0$

Solving for $z$ :

$z(3z-48)=0 => z = 0, and z=48/3 = 16$

But $z= x^2$

So:

$x^2=0=> x=0$

$x^2 = 16=> x=sqrt16=> x= +-4$

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

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