Sketch the graph of `f(x)=4x^3+6x^2-24x-2` :

(1) The y-intercept is -2

(2) `f'(x)=12x^2+12x-24`

Any extrema occur at critical points. Since f is a polynomial, we need only find where `f'(x)=0` :

`12x^2+12x-24=0`

`12(x+2)(x-1)=0`

`=>x=-2,1` are the critical points.

`f'(-3)=48 => f'(x)>0` on `(-oo,-2)` thus the function increases on this interval.

`f'(-1)=-24` `=>f'(x)<0` on `(-2,1)` thus the function decreases on this interval.

`f'(2)=48=>f'(x)>0` on `(1,oo)` thus the function increases on this interval.

Applying the first derivative test we have (-2,38) is a local maximum, and (1,-16) is a local minimum.

(3) `f''(x)=24x+12` . Any inflection points occur where the second derivative is zero.

`f''(x)=0 => 24x+12=0 => x=-1/2` . Testing values of the second derivative and applying the second derivative test we find the function is concave down on `(-oo,-1/2)` and concave up on `(-1/2,oo)`

**(4) Thus we have the function increasing on `(-oo,-2)` and concave down, achieving a local max at (-2,38), decreasing on (-2,1), going through an inflection point at `(-1/2,11)` changing to concave up , achieving a local minimum at (1,-16) an increasing on `(1,oo)` . We also have the y-intercept as -2. Putting it all together we get a sketch:**

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From precalculus we know the general shape of a third degree polynomial with positive leading coefficient. Finding just the local max and min (the turning points) and a few values would get a suitable graph.

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