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#