find the following of f(x)=(x+6)^2(x-2)^2 turning points behavior of the graph of f near each x intercept graph itmust show work
The turning points are found by solving for the derivative equalling zero. That is, solving `f'(x)=0`. Taking the derivative using the product rule and factoring we get:
which means that the turning points are at x=-6, x=2 and x=-2.
To find the behaviour, we can take the second derivative and evaluate using the second derivative test.
So on evaluation we get:
`f''(-6)=128>0` concave up
`f''(2)=128>0` concave up
`f''(-2)=-64<0` concave down
This means that the turning points are x=-6, concave up, x=-2, concave down, x=2, concave up. The graph is: