# f(x)=(x+5)^2(x-3)^2 Need power function y= and the Behavior of the x intercept at 3 and -5?I got the x intercepts (-5,0)(3,0) and y of (0,225)

embizze | Certified Educator

Given `f(x)=(x+5)^2(x-3)^2` :

(1) The x-intercepts are x=-5,x=3 using the zero product property

(2) The y-intercept is 225 (Substitute x=0)

(3) If you expand you will not get a power function; you will get a quartic polynomial:

`(x+5)^2(x-3)^2`

`=(x^2+10x+25)(x^2-6x+9)`

`=x^4+4x^3-26x^2-60x+225`

(4) Both x=-5 and x=3 are minimums of the function. Note that the function is always nonnegative (the product of two nonnegative numbers) so 0 will be a minimum.

Using calculus:

(a) Use the product rule

`f'(x)=2(x+5)(x-3)^2+2(x+5)^2(x-3)`

`=(2x+10)(x^2-6x+9)+(2x-6)(x^2+10x+25)`

`=4x^3+12x^2-52x-60`

(b) Differentiating the expanded product gives

`f'(x)=4x^3+12x^2-52x-60`

Setting the derivative equal to zero we get:

`4x^3+12x^2-52x-60=0`

`4(x+5)(x+1)(x-3)=0`

So the critical points are -5,-1,3

Using the first derivative test we find minimums at x=-5,x=3 and a local maximum at x=-1 (The function is symmetric so the maximum occurs at the avereage of the minimums)

The graph: