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Math Help - Please help me with this problem, it is urgent, thanks!

  1. #1
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    Please help me with this problem, it is urgent, thanks!

    The graph of the function y = ax≤ + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.

    I know this is only a basic question, but I really would appreciate your help.

    Thanks
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  2. #2
    MHF Contributor kalagota's Avatar
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    y = ax^2 + bx + c = x^2 + \frac{b}{a}x + \frac{c}{a}

    = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{4ac- b^2}{4a^2} = \left({x +\frac{b}{2a}}\right)^2 + \frac{4ac- b^2}{4a^2}

    so, the turning point is at \left({-\frac{b}{2a}, \frac{4ac-b^2}{4a^2}}\right)

    plug in and you will have 2 equations..

    for the third equation:
    note:
    a point (x_0,y_0) is on the curve y = ax^2 + bx + c if and only if y_0 = ax_0^2 + bx_0 + c..
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  3. #3
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    Quote Originally Posted by tim_mannire View Post
    The graph of the function y = ax≤ + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.

    I know this is only a basic question, but I really would appreciate your help.

    Thanks
    This aproach should give you the same answer but you will only have to solve a symoltaneous equation in two variables.

    You want the equation to pass through (0,5) so substitute x=0, y=5 into the equation to get y = 5 = c

    You want the equation to pass through (-3,2) so substitute x=-3, y=2 into the equation to get 2 = 9*a-3b+5

    Then by knowing that at the turning point the differential of y with respect to x is zero you can write:

     y'=0=2*a*x + b

    You have two equations in two variables which can be solved by symoltaneous equations.
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  4. #4
    Senior Member nikhil's Avatar
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    Lightbulb Here it is

    y=ax^2+bx+c
    since it passes through (0,5) this eq must satisfy it. So putting the values in the eq we get 5=a(0)^2+b(0)+c.which gives us c=5.
    Now at turning point x=-b/2a
    (if u wanna know why is it that so u may ask it in a new thread).since turning point is (-3,2) therefor x=-b/2a=-3 or b=6a. Substituting this value in eq we get y=ax^2+6ax+5 now substituting (-3,2) we get 2=9a-18a+5 or a=1/3 since b=6a therfor b=2.so finaly a=1/3,b=2,c=5
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  5. #5
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    Quote Originally Posted by tim_mannire View Post
    The graph of the function y = ax≤ + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.


    I would really appreciate any help on this problem, thanks.
    First use the turning point form y = a (x - h)^2 + k. Obviously h = 0, k = 5. Substitute (-3, 2) to get a.

    Now expand the turning point form to get the standard form.

    Edit: Don't double post! It wastes peoples time! http://www.mathhelpforum.com/math-he...nt-thanks.html
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by tim_mannire View Post
    The graph of the function y = ax≤ + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.


    I would really appreciate any help on this problem, thanks.
    Use the following formula:

    y = a(x - p)^2 + q

    y = a(x + 3)^2 + 2

    Passes through the point (0;5)

    5 = a(0 + 3)^2 + 2

    3 = 9a

    a = \frac{1}{3}

    y = \frac{1}{3} (x+3)^2 + 2


    EDIT: Oops Mr F beat me to it... by 2 mins
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  7. #7
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    Quote Originally Posted by nikhil View Post
    y=ax^2+bx+c
    since it passes through (0,5) this eq must satisfy it. So putting the values in the eq we get 5=a(0)^2+b(0)+c.which gives us c=5.
    Now at turning point x=-b/2a
    (if u wanna know why is it that so u may ask it in a new thread).since turning point is (-3,2) therefor x=-b/2a=-3 or b=6a. Substituting this value in eq we get y=ax^2+6ax+5 now substituting (-3,2) we get 2=9a-18a+5 or a=1/3 since b=6a therfor b=2.so finaly a=1/3,b=2,c=5

    Thank you very much. very well answered. much appreciated!
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