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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lfryerda | High School Teacher | (Level 2) Educator

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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)`




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: