Question

Use the x-intercepts to find the intervals on which the graph of f is above and below the x-axis.
F(x)=(x-2)^2(x+4)^2

Answer 1

solve for x

Answer 2

ok, then what?

Answer 3

You have to show me your solutions for x first.

Answer 4

2x^2+20

Answer 5

Nope. \((x-2)^2 (x+4)^2 =0\) has only 2 real solutions.
\((x-2)(x-2)(x+4)(x+4) =0\) hence x =2 and x =4 are solutions.
But (x-2)^2 >0 for all x, (x+4) >0 for all x also
Hence f(x) is >0 for all x , no negative value,

Answer 6

I don't understand how to find the answer though...