Evaluate the graph of `f(x)=-2x^3-5x^2+7x+6` :

We will need the first derivative:

`f'(x)=-6x^2-10x+7`

(1) The y-intercept occurs when x=0 so the y-intercept is 6

(2) The x-intercepts occur when f(x)=0. Using a grapher the zeros are `x~~-3.287,-.640,1.427`

(3) The local extrema can be found using the first derivative. Extrema can...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Evaluate the graph of `f(x)=-2x^3-5x^2+7x+6` :

We will need the first derivative:

`f'(x)=-6x^2-10x+7`

(1) The y-intercept occurs when x=0 so the y-intercept is 6

(2) The x-intercepts occur when f(x)=0. Using a grapher the zeros are `x~~-3.287,-.640,1.427`

(3) The local extrema can be found using the first derivative. Extrema can only occur when `f'(x)=0` or fails to exist.

Using the quadratic formula the zeros of `f'(x)` are `x=-5/6+-sqrt(67)/6` or `x~~0.531,-2.198`

Using the first derivative test we find `x~~-2.198` to be a local minimum, and `x~~.0531` to be the local maximum.

(4) The end behavior of a polynomial is the end behavior of the leading term. Since the degree is 3 with a negative leading coefficient, `lim_(x->-oo)f(x)=oo,lim_(x->oo)f(x)=-oo`

The graph: