# 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

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### 1 Answer

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:

`f'(x)=2(x+6)(x-2)^2+(x+6)^2 2(x-2)`

`=2(x+6)(x-2)(x-2+x+6)`

`=2(x+6)(x-2)(2x+4)`

`=4(x+6)(x-2)(x+2)`

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.

`f''(x)=4((x-2)(x+2)+(x+6)(x+2)+(x+6)(x-2))`

`=4(x^2-4+x^2+8x+12+x^2+4x-12)`

`=4(3x^2+12x-4)`

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:**