1. ## graph

1.describe the graph of $f(x)=-(x+3)(x-4)(x-3)^2$ under

a)x and y intercepts
b)end behaviors
c)local maximum points
d)local minimum points

2. Hello, William!

It seems straight-forward ... Where is your difficulty?

1. Describe the graph of $f(x)\:=\:-(x+3)(x-4)(x-3)^2$ under:

a) $x$- and $y$-intercepts
b) end behaviors
c) local maximum points
d) local minimum points

(a) For $x$-intercepts, let $y = 0\!:\;\;-(x+3)(x-3)^2(x-4) \:=\:0$

. . . . $x\text{-intercepts:} \;(\text{-}3,0),\;(3,0),\;(4,0)$

. . For $y$-intercepts, let $x = 0\!:\;\;y \:=\:-(3)(\text{-}3)^2(\text{-}4)$

. . . . $y\text{-intercept:}\;(0,108)$

(b) When $x\to \text{-}\infty,\;y \to \text{-}\infty$
. . .When $x\to+\infty,\;y\to\text{-}\infty$

. . .The graph "goes down" at both ends.
. . .It is shaped like an $\wedge\!\wedge$

(c) (d) For extreme points, solve $f'(x) = 0$

We have: . $f'(x) \:=\:-(x+3)(x-4)2(x-3) - (x+3)(x-3)^2 - (x-4)(x-3)^2 \;=\;0$

Factor: . $-(x-3)\bigg[2(x+3)(x-4) + (x+3)(x-3) + (x-4)(x-3)\bigg] \:=\:0$

. . and we have: . $(x-3)(4x^2-9x-21) \:=\:0$

There is a minimum at $x \,=\, 3$ and maximums at $x\,=\,\frac{9\pm\sqrt{417}}{8}$

The graph looks like this:
Code:
                |
*  |
*     |*
*       |  *                *
*        |   *           *       *
|      *      *           *
---*---------+----------*---------------*----
-3         |          3               4
|                           *
*          |
|                            *