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Math Help - Solving A Quadratic problem

  1. #1
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    Solving A Quadratic problem

    I was wanting some help with this one please. It's a word problem for which I have to figure out the quadratic equation. I have the answer but am insure how to arrive at it. The answer id y= -1.75(x-2)^2 -1

    Find the equation of the quadratic function with; axis of symmetry x=2, maximum value -1 and point (4, -8) .


    So my thinking is to use the Turning Point references into a Vertex form function and use the x,y point reference to solve. But I'm not well practiced and I got stuck.
    Could someone show me the working on this one please?
    Thanks from Melody2
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  2. #2
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    Re: Solving A Quadratic problem

    Hi Tren 301,
    I am very happy to answer your question. I get to practice writing in Latex so we both win.

    Find the equation of the quadratic function with; axis of symmetry x=2, maximum value -1 and point (4, -8) .

    The vertex is (2,-1)
    I am assuming that this is a standard parabola, not a sideways one so the formula will be

    \begin{align*}(y+1) = &k(x-2)^2\\&Now\;sub\; in (4,-8)\\(-8+1)=&k(4-2)^2\\-7=&k*4\\-1.75=&k\\&Hence\\(y+1)=&-1.75(x-2)^2\\y=&-1.75(x-2)^{2}-1\end{align*}

    And now the answer is in the form that you wanted.
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    Re: Solving A Quadratic problem

    Hi Melody2,
    Thanks for your reply. That Latex idea sounds interesting (assuming your female).
    I had the equation like this -8= (4-2)^2-1. As my textbook shows the TP go in the places for b & c in: y=a(x-b)^2+c
    So why is the -1 with the y value in your equation and not in the place of 'c' where I had it?
    I did not allow a value for 'k' either.

    Thanks
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  4. #4
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    Re: Solving A Quadratic problem

    Quote Originally Posted by Melody2 View Post
    Hi Tren 301,
    I am very happy to answer your question. I get to practice writing in Latex so we both win.

    Find the equation of the quadratic function with; axis of symmetry x=2, maximum value -1 and point (4, -8) .

    The vertex is (2,-1)
    I am assuming that this is a standard parabola, not a sideways one so the formula will be
    You don't need to "assume" it. If the axis of symmetry is "x= " then it must be a vertical "standard" parabola.

    \begin{align*}(y+1) = &k(x-2)^2\\&Now\;sub\; in (4,-8)\\(-8+1)=&k(4-2)^2\\-7=&k*4\\-1.75=&k\\&Hence\\(y+1)=&-1.75(x-2)^2\\y=&-1.75(x-2)^{2}-1\end{align*}

    And now the answer is in the form that you wanted.
    Last edited by HallsofIvy; November 15th 2013 at 03:24 PM.
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    Re: Solving A Quadratic problem

    Yes of course you don't have to assume it, how silly of me.
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  6. #6
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    Re: Solving A Quadratic problem

    Okay, My explaination will not be very formal but I'll try and explain how I think of it.
    I'll start with a very basic parabola
    y=ax^2
    This is a parabola with a vertex at (0,0)
    If a is positive it is concave up
    If a is negative it is concave down.
    |a| determines the 'width' of the parabola.
    The smaller |a| is the skinnier the parabola will be
    The larger |a| is the wider the parabola will be.

    If I want to translate the parabola 10 units left, so that the vertex is (-10,0) I would replace the x with x+10
    and the equation would be y=a(x+10)^2

    If I want to translate the parabola 8 units up, so that the vertex is (0,8) I would replace the y with y-8
    and the equation would be (y-8)=ax^2

    If I want to translate the parabola left and up so the new vertex is (-10,8) I would replace the y with y-8 and the x with x+10
    and the equation would be (y-8)=a(x+10)^2

    You may be much more familiar with this when dealing with circles
    The equation of a circle centred at the origin is x^2+y^2=r^2
    If you translate it so that the new centre is (h,k)
    the new formula is (x-h)^2+(y-k)^2=r^2

    Anyway, I hope i have explained a lot of my logic, the k determines the width and I just plugged in the extra point you had to determine the value of k.

    I think your text book does it in a common way. I prefer my way because I can use it when translating any equation.

    Does that explain things better?

    And yes, I am female. My real life name is Melody.
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  7. #7
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    Re: Solving A Quadratic problem

    oh, 'k' and 'a' are the same thing. I suppose it would have been less confusing if I had called it 'a' all the way through.
    I am NOT referring to the circle 'k' just the initial parabola 'k'

    Don't you hate it when teachers use the same letter to stand for different things and different letters to stand for the same thing all in the space of 1 question!
    Sorry.
    Last edited by Melody2; November 15th 2013 at 10:47 PM.
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    Re: Solving A Quadratic problem

    Hi Melody,
    Thanks for your excellent and thoughtful help. I was able to make sense of your original answer when I looked at it again and tried it myself. The quadratic reference to the circle lost me a bit. I'm not well practice enough in Maths. I have my Maths 2C/D exam this week 21/11 (Perth WA), so I'm practicing up.
    Yes, I see that 'k' and 'a' occupy the same position. I just need more practice.
    If it's alright I might ask you again when I need help. Would that be OK?

    Thanks Again
    Peter.
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  9. #9
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    Re: Solving A Quadratic problem

    That would be absolutely fine Peter. I love helping people, and practicing Latex. But keep in mind that there are other very capable people on this forum that also love to help people.

    The circle is not a quadratic, I was just trying to show you that this method works when any graph is translated to some new position on the xy plane (cartesian plane). I won't say more, maybe you aren't there yet, but when you learn about circles try to keep what I have said in your mind.
    Translate just means slide from one spot to another, no reflecting or rotating, just sliding. It occurred to me that you may not know this word.

    Melody.
    Last edited by Melody2; November 15th 2013 at 11:56 PM.
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