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Thread: Find expressions for the quadratic functions whose graphs are shown.

  1. #1
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    Find expressions for the quadratic functions whose graphs are shown.

    Find expressions for the quadratic functions whose graphs are shown.



    I know how to do f(x) but I'm stuck on g(x).
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  2. #2
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    Quote Originally Posted by zweevu View Post
    Find expressions for the quadratic functions whose graphs are shown.



    I know how to do f(x) but I'm stuck on g(x).
    First, note that the y-intercept is 1.

    So $\displaystyle g(x) = ax^2 + bx + 1$.


    Using the other two points, we will have two equations in two unknowns to solve simultanously.

    Equation 1:
    $\displaystyle 2 = a(-2)^2 + b(-2) + 1$

    $\displaystyle 4a - 2b + 1 = 2$

    $\displaystyle 4a - 2b = 1$.


    Equation 2:
    $\displaystyle -2.5 = a(1)^2 + b(1) + 1$

    $\displaystyle a + b + 1 = -2.5$

    $\displaystyle a + b = -1.5$.


    So your system of equations is...

    $\displaystyle 4a - 2b = 1$

    $\displaystyle a + b = -1.5$.


    Solve for a and b.
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  3. #3
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    The second function has the form

    $\displaystyle y = ax^2+bx+c$

    The y-intercept c is 1 as shown in the picture so

    $\displaystyle y = ax^2+bx+1$

    Now use points (-2,2)

    $\displaystyle y = ax^2+bx+1$

    $\displaystyle 2 = a(-2)^2+b(-2)+1$ ...(1)

    and (1,-2.5)

    $\displaystyle y = ax^2+bx+1$

    $\displaystyle -2.5 = a(1)^2+b(1)+1$ ...(2)

    and solve the sytem
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  4. #4
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    The question is looking g(x) = ... what would I type in for the answer. I got +/- 1/6 for a and b though.

    Edit: NVM You plug them into the original equation. Thanks!
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    You have $\displaystyle g(x) = y = ax^2+bx+c$

    With $\displaystyle c = 1 , b = \frac{-7}{6}$ and $\displaystyle a = \frac{-1}{3}$

    can you find $\displaystyle g(x)$ ?
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  6. #6
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    We assume that:
    $\displaystyle g(x) = ax^2 + bx + c$
    and the graph of the function got through 3 points, they are:
    $\displaystyle (-2, 2), (0, 1), (1, -2.5)$

    $\displaystyle g(-2) = 4a - 2b + c = 2;$
    $\displaystyle g(0) = c = 1$
    $\displaystyle g(1) = a + b +c = -2.5$

    So:
    $\displaystyle a = -1; b = -2.5; c = 1;$
    $\displaystyle g(x) = -x^2 - 2.5x + 1$
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  7. #7
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    can anyone figure out f(x) ???
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  8. #8
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    $\displaystyle f(x) = 2(x-3)^2$
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