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Math Help - Equation of function

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    Post Equation of function

    Can u help me by explaining and show working out 2 this question? I can understand when some1 explains and shows me.....I've attempted it but cant

    a) Determine the equation of the function whose graph is shown, given that it has the form y=ax^2 + bx + c, for -6<x<2.

    points on graph are; (-6,13) (2,5)
    crosses y axis at (0,1)
    b) Determine the turning point of the curve.

    THIS ISNT FOR AN ASSIGNMENT OR EXAM THING. Just normal work.
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    Member arpitagarwal82's Avatar
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    where is the graph?
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    Quote Originally Posted by wandering_lost View Post
    a) Determine the equation of the function whose graph is shown, given that it has the form y=ax^2 + bx + c, for -6<x<2.

    points on graph are; (-6,13) (2,5)
    crosses y axis at (0,1)
    Substitute the three points to get a system of linear equations,

    \left\{\begin{array}{rcrcrcr}<br />
13&=&36a&-&6b&+&c\\<br />
5&=&4a&+&2b&+&c\\<br />
1&=&&&&&c<br />
\end{array}\right.

    Solve for a,\;b, and c. You should get (a,\,b,\,c)=\left(\frac12,\,1,\,1\right).

    b) Determine the turning point of the curve.
    Set the derivative equal to zero, solve, and test the intervals to see where the derivative changes sign.
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    Quote Originally Posted by wandering_lost View Post
    Can u help me by explaining and show working out 2 this question? I can understand when some1 explains and shows me.....I've attempted it but cant

    a) Determine the equation of the function whose graph is shown, given that it has the form y=ax^2 + bx + c, for -6<x<2.

    points on graph are; (-6,13) (2,5)
    crosses y axis at (0,1)
    b) Determine the turning point of the curve.

    THIS ISNT FOR AN ASSIGNMENT OR EXAM THING. Just normal work.
    Part A.
    Use the y-intercept (0,1) to find c.

    y = ax^2 + bx + c
    1 = a*0 + b * 0 + c
    c = 1

    Now use one of the other points to find a relation between a and b. I'll use (2,5)

    y = ax^2 + bx + 1
    5 = a * 2^2 + b * 2 + 1
    4 = 4a + 2b
    2b = 4 - 4a
    b = (4 - 4a)/2
    b = 2 - 2a

    Now replace b in the original formula with the relation you just found.

    y = ax^2 + bx + 1
    y = ax^2 + (2 - 2a) * x + 1

    Now use that equation and the last point (-6,13) to solve for a.

    y = ax^2 + (2 - 2a) * x + 1
    13 = a * -6^2 + (2 - 2a) * -6 + 1
    13 = a * 36 - 12 + 12a + 1
    24 = 36a + 12a
    48a = 24
    a = 1/2 = 0.5

    Now plug a back into the original equation and use any point to find b. I'll use (2,5)

    y = ax^2 + bx + 1
    y = 0.5 * x^2 + bx + 1
    5 = 0.5 * 2^2 + b * 2 + 1
    5 = 0.5 * 4 + 2b + 1
    2 = 2b
    b = 1

    Now plug a, b, and c into the original equation

    y = ax^2 + bx + c
    y = 0.5x^2 + x + 1

    Part B
    Divide by 0.5
    y = 0.5x^2 + x + 1
    2y = x^2 + 2x +2

    Complete the square
    2y = x^2 + 2x + 2
    2y = x^2 + 2x + 2 + 1 - 1
    2y = x^2 + 2x + 1 + 1
    2y = (x + 1)^2 + 1

    Divide by 2
    2y = (x + 1)^2 + 1
    y = 0.5(x + 1)^2 + 0.5

    From this equation the x-value at the turning point is -1 and the y-value at the turning point is 0.5. Therefore, (-1,0.5)
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