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Math Help - graph

  1. #1
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    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
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  2. #2
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    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
                    |                           *
         *          |
                    |                            *
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