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Thread: Quadratic Functions Question - Find in Form of y = ax2 + bx + c

  1. #1
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    Quadratic Functions Question - Find in Form of y = ax2 + bx + c

    Hey All

    Got a test tomorrow, I'd appreciate some help. I'm pretty good in math but I just don't get these questions here.


    Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph:

    q1) touches the x-axis at 4 and passes through (2,12)
    q2) has vertex (-4, 1) and passes through (1, 11)

    You need to have 2 equations for the 2 different situations. It's the same technique but I just don't know it. Any help appreciated!!
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  2. #2
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    Re: Quadratic Functions Question - Find in Form of y = ax2 + bx + c

    Quote Originally Posted by JemDkyl View Post
    Hey All

    Got a test tomorrow, I'd appreciate some help. I'm pretty good in math but I just don't get these questions here.


    Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph:

    q1) touches the x-axis at 4 and passes through (2,12)
    q2) has vertex (-4, 1) and passes through (1, 11)

    You need to have 2 equations for the 2 different situations. It's the same technique but I just don't know it. Any help appreciated!!
    Vertex point is given by $\displaystyle V\left(\frac{-b}{2a} , \frac{-b^2}{4a}+c\right)$

    If graph passes through point $\displaystyle (x_A,y_A)$ then $\displaystyle y_A=ax^{2}_A+bx_A+c$

    So you have to solve system of three equations with three unknowns in each case .
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  3. #3
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    Re: Quadratic Functions Question - Find in Form of y = ax2 + bx + c

    That wasn't very helpful :/ Could you solve one for me and I'll do the other one myself? I need to see an example, that's how I get things :P
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  4. #4
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    Re: Quadratic Functions Question - Find in Form of y = ax2 + bx + c

    Quote Originally Posted by JemDkyl View Post
    That wasn't very helpful :/ Could you solve one for me and I'll do the other one myself? I need to see an example, that's how I get things :P
    You really are a master of tact. Surely you can substitute $\displaystyle (4,0)$ and $\displaystyle (2,12)$ into $\displaystyle ax^2+bx+c=y$

    Knowing the $\displaystyle x$ coordinate of the vertex is $\displaystyle \frac{-b}{2a}$ gives you another equation to substitute a value in for. This should, in theory, be enough.

    Please have an attempt and show how much you can do for yourself.
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  5. #5
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    Re: Quadratic Functions Question - Find in Form of y = ax2 + bx + c

    -b/2a = 4

    b = -8a

    Then I replace 2 with x and y with 12?

    4a + 2b + c = 12
    16a + 4b + c = 0 (x intercept replaced)
    -----------------

    12a + 2b = -12

    Since b is -8a --> 12a + 2(-8a) = -12
    -4a = - 12
    a = 3

    Afterwards I just replace 3 with a in the previous 12 + 2b = -12 equation.
    b = -24

    I find c the same way, it's 48.

    So I guess the final equation should be y = 3x2 - 24x + 48

    I still don't understand the theory though.. why is -b/2a = 4??

    Isn't 4 the x-intercept, but not the minimum/maximum point of the graph?
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  6. #6
    Super Member Quacky's Avatar
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    Re: Quadratic Functions Question - Find in Form of y = ax2 + bx + c

    It's both.

    As for the vertex, we have $\displaystyle y=ax^2+bx+c$

    Completing the square would give:

    $\displaystyle y=a(x+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c$.

    You should know that the vertex of a quadratic of the form $\displaystyle y=a(x-h)^2+k$ (known as the vertex form of a quadratic) is $\displaystyle (h,k)$ and this is what we have here. $\displaystyle h=\frac{-b}{2a}$ and $\displaystyle k=-a(\frac{b}{2a})^2+c$

    If you're still unsure as to why this is the case, you should probably consult an instructor.
    Last edited by Quacky; Dec 1st 2011 at 05:30 AM.
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    Re: Quadratic Functions Question - Find in Form of y = ax2 + bx + c

    Alright I get it then. I'll solve a few more problems like this one, the thread can be closed now. Thanks!
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