Question

f(x)=(x+6)^2(x-2)^2
what is the y-intercept of f?
Power function of f?
Turning points?
Behavior of xintercept (2)
" " -6

Answer 1

y-intercept
(0,f(0)) = (0,144)

Answer 2

The degree of f as a polynomial is 4

Answer 3

\[
f'(x)=4 x^3+24 x^2-16 x-96\\
f''(x)=12 x^2+48 x-16
f''(x) = 0\\

\begin{array}{c}
x=\frac{2}{3} \left(-3-2
\sqrt{3}\right) \\
x=\frac{2}{3} \left(-3+2
\sqrt{3}\right) \\
\end{array}
\]
These are the x's where f has inflection (turning point)

Answer 4

The function is tangent to the x-axis at x=-6 and x=2
because these are double roots.

Answer 5

So my behavior of the x-intercept for 2 ?

Answer 6

The graph is tangent to the x-axis.

Answer 7

See the attached graph of your function.

Answer 8

ok, I see so for my behavior of the x-intercept of 2 is y=-6 and 2?

and my maxium number of turning points

Answer 9

Are turning points inflection points?

Answer 10

f has a local minimum at x =-6 and x =2
it has a local maximum at x=-2

Answer 11

Did you get it?

Answer 12

I think...
so for the x-intercept of (2) It the -6 and 2
for x-intercept(-6) I put -2

Answer 13

x-intercepts are at x=-6 and x =2

Answer 14

I have to go bye.

Answer 15

see my problem says what s he behavior of grahp near the xintercept of 2