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Math Help - Stuck with cal graph

  1. #1
    NeF
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    Arrow Stuck with cal graph

    Given  h(x)=x^3-3x^2-9x-5
           =(x+1)^2(x-5)
    1 Calculate the co-ords of the x and y intercepts of the curve of h
    2 Find the co-ords of the turning point(s) of the curve
    3 Use the info above to draw a graph
    4 Use the graph to determine the co-ords of the turning points of the curve of
    y=x^3-3x^2-9x+5

    Some other question:
    p(x)=x^3-5x^2-8x+12 at  (-1:41)

    My answers:
    1 x=-1 x=5
     y=-5

    2  \frac{dy}{dx}= 3x^2-6x-9
     3(x^2-2x-3) = 0
     (x-3)(x-1) = 0
     x=3    x=1
     f(3) = (3)^3 -(3)^2 -9(3) -5 = -32
     f(-1) = (-1)^3 -3(-1)^2-9(-1)-5 = 0

     (3;-32)     (1;0)

    3 I got a wierd one, want to see what u guys get.


    Thanks guys xD
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  2. #2
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    Quote Originally Posted by NeF
    Given  h(x)=x^3-3x^2-9x-5
           =(x+1)^2(x-5)
    1 Calculate the co-ords of the x and y intercepts of the curve of h
    The y-intercept is when x=0 thus,
    (0+1)^2(0-5)=1^2(-5)=-5
    The x-intercept is when y=0 thus,
    (x+1)^2(x-5)^2=0 thus,
    x=-1,5

    2 Find the co-ords of the turning point(s) of the curve
    Find derivative, (product rule)
    y'=[(x+1)^2]'(x-5)+(x+1)^2(x-5)'
    y'=2(x+1)(x-5)+(x+1)^2
    y'=(x+1)[2(x-5)+(x+1)]
    y'=(x+1)(3x-9)
    y'=3(x+1)(x-3)
    Set y'=0 thus,
    3(x+1)(x-3)=0 thus,
    x=-1,3
    Now,
    f(-1)=0,f(3)=-32
    Thus, the turning points are on,
    (-1,0),(3,-32)
    3 Use the info above to draw a graph
    You should be able to do that now.
    4 Use the graph to determine the co-ords of the turning points of the curve of
    y=x^3-3x^2-9x+5
    This curve is the same as the other curve in the sense that it is shifted up by 10 units.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NeF
    Given  h(x)=x^3-3x^2-9x-5
           =(x+1)^2(x-5)
    1 Calculate the co-ords of the x and y intercepts of the curve of h
    2 Find the co-ords of the turning point(s) of the curve
    3 Use the info above to draw a graph
    4 Use the graph to determine the co-ords of the turning points of the curve of
    y=x^3-3x^2-9x+5

    Some other question:
    p(x)=x^3-5x^2-8x+12 at  (-1:41)

    My answers:
    1 x=-1 x=5
     y=-5

    2  \frac{dy}{dx}= 3x^2-6x-9
     3(x^2-2x-3) = 0
     (x-3)(x-1) = 0
     x=3    x=1
     f(3) = (3)^3 -(3)^2 -9(3) -5 = -32
     f(-1) = (-1)^3 -3(-1)^2-9(-1)-5 = 0

     (3;-32)     (1;0)

    3 I got a wierd one, want to see what u guys get.


    Thanks guys xD
    A technical point... You might need this depending on how picky your Math professor is. The x and y intercepts are actually points, so really the x intercepts are (-1, 0) and (5, 0) and the y intercept is (0, -5). Normally professors don't care, but you do run into the occasional one that does.

    -Dan
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  4. #4
    NeF
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    Hectic thanks guys. One more question though, I post my questions in pre-calculus but it always ends up in the uni calculus section.
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