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Math Help - Graphs, parabolas, hyperbolas...

  1. #1
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    Graphs, parabolas, hyperbolas...

    Hi.

    I'm really stuck on a few questions I have to do. :/
    Help would be greatly appreciated.


    1. Sketch the graph of: y = (x-2)2 (2x+5) showing all points of interest. The secondary stationary point occurs when x=-1


    2. The Parabola y = 2x2 + 5x + 8 and the line y = ax - 10 intersect exactly at one point. Determine value(s) of a and the point of intersection.


    3. A truncas has intercepts of (2,0), (6,0), and (0,6). Determine its equation, domain, and range.


    4. The remainder when f(x) = 4x4 - 4x3 + jx2 + x + 6 is divided by x-2 is -60. Demonstrate that j must be equal to -25.
    Use long division to then factorise f(x)
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  2. #2
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    Re: Graphs, parabolas, hyperbolas...

    Quote Originally Posted by QueenLuv View Post
    1. Sketch the graph of: y = (x-2)2 (2x+5) showing all points of interest. The secondary stationary point occurs when x=-1
    To find the zeros make y=0 and solve for x. To find the stationary pts make y'=0 and solve for x

    Quote Originally Posted by QueenLuv View Post

    2. The Parabola y = 2x2 + 5x + 8 and the line y = ax - 10 intersect exactly at one point. Determine value(s) of a and the point of intersection.
    Make the equations equal, then solve the discriminant equalling zero


    Quote Originally Posted by QueenLuv View Post

    3. A truncas has intercepts of (2,0), (6,0), and (0,6). Determine its equation, domain, and range.
    You have y = \frac{a}{(x-b)^2}+c

    Subsitute your values one at a time and solve the system

    Quote Originally Posted by QueenLuv View Post

    4. The remainder when f(x) = 4x4 - 4x3 + jx2 + x + 6 is divided by x-2 is -60. Demonstrate that j must be equal to -25.
    Use long division to then factorise f(x)
    Note that f(2) = -60
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  3. #3
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    Re: Graphs, parabolas, hyperbolas...

    Thanks. But I'm still stuck on 2 and 4. The wording of them really confuse me. :/

    So sorry, but could you (or anyone else) in layman's terms, please tell me the steps on how to get the answers?
    Last edited by QueenLuv; March 13th 2012 at 01:54 AM.
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  4. #4
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    Re: Graphs, parabolas, hyperbolas...

    Hello, QueenLuv!

    1. Sketch the graph of: y \:=\: (x-2)^2(2x+5) showing all points of interest.
    The secondary stationary point occurs when x=-1.

    The graph is a cubic with a positive leading coefficient.
    . . Its shape is: . \begin{array}{c}\cap \\ [-3mm]\;\;\;\cup \end{array}

    The graph has x-intercepts 2 and -\tfrac{5}{2}
    The graph is tangent to the x-axis at x = 2. (minimum)
    The y-intercept is: . y \,=\,(-2)^2(5) \,=\,20

    When x = -1,\:y \:=\:(-3)^2(3) \:=\:27 (maximum)

    Can you sketch the graph now?

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  5. #5
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    Re: Graphs, parabolas, hyperbolas...

    Hello again, QueenLuv!

    4. The remainder when f(x) \:=\: 4x^4 - 4x^3+ jx^2 + x + 6 is divided by x-2 is -60.
    Demonstrate that j must be equal to -25.

    Use long division to then factorise f(x).

    We are told that: . f(2) \,=\,-60

    . . 4(2^4) - 4(2^3) + j(2^2) + 2 + 6 \:=\:-60 \quad\Rightarrow\quad 4j + 40 \:=\:-60

    . . 4j \:=\:-100 \quad\Rightarrow\quad \boxed{j \:=\:-25}



    \begin{array}{cccccccccccc} &&&& 4x^3 & - & 12x^2 & - & x & + & 3 \\ && -- & -- & -- & -- & -- & -- & -- & -- & -- \\ x+2 & | & 4x^4 &-& 4x^3 &-& 25x^2 &+& x &+& 6 \\ && 4x^4 &+& 8x^3 \\ && -- & -- & -- \\ &&& - & 12x^3 &-& 25x^2 \\ &&& - & 12x^3 &-& 24x^2 \\ &&& -- & -- & -- & -- \\ &&&&& - & x^2 &+& x \\ &&&&& - & x^2 &-& 2x \\ &&&&& -- & -- & -- & -- \\ &&&&&&&& 3x &+& 6 \\ &&&&&&&& 3x &+& 6 \\ &&&&&&&& -- & -- & -- \end{array}



    \begin{array}{ccccccccccc} &&&& 4x^2 &&& - & 1 \\ && -- & -- & -- & -- & -- & -- & -- \\ x-3 & | & 4x^3 &-& 12x^2 &-& x &+& 3 \\ && 4x^3 &-& 12x^2 \\ && -- & -- & -- \\ &&&&& - & x &+& 3 \\ &&&&& - & x &+& 3 \\ &&&&& -- & -- & -- & -- \end{array}



    Therefore: . f(x) \;=\;(x+2)(x-3)(4x^2-1) \;=\;(x+2)(x-3)(2x-1)(2x+1)

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  6. #6
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    Re: Graphs, parabolas, hyperbolas...

    Oh, now I get it. Thank you so much!!!
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