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    Hi, I'm really struggling with this question. Can't seem to figure it out. Any help appreciated. Thanks!

    Question:

    A parabola has a vertex with coordinates (-4,1) and goes through the points with coordinates (2,-17).

    Find the equation of the parabola in the form y=ax^2+bx+c

    ----

    P.s. Please explain how you got your answer if you can. I want to understand it.

    Thanks!
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    Funny, this exact question was posted a couple of days ago.

    First notice that you can find the vertex (i.e. stationary point) of a quadratic equation by completing the square on it. If you complete the square on ax^2+bx+c=0 you end up with something in the form a(x+p)^2+q=0. Notice that this is always greater than or equal to q, and attains its minimum when a(x+p)^2=0 (i.e. when x=-p). This means that the vertex of the graph is at (-p,q).

    You know where the vertex in, so you can sub in to find the values of p and q; and then you can expand the brackets. Then you just need to sub in another point on the graph (i.e. (2,-17)) to find the value of a.
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    (Original post by nuodai)
    Funny, this exact question was posted a couple of days ago.

    First notice that you can find the vertex (i.e. stationary point) of a quadratic equation by completing the square on it. If you complete the square on ax^2+bx+c=0 you end up with something in the form a(x+p)^2+q=0. Notice that this is always greater than or equal to q, and attains its minimum when a(x+p)^2=0 (i.e. when x=-p). This means that the vertex of the graph is at (-p,q).

    You know where the vertex in, so you can sub in to find the values of p and q; and then you can expand the brackets. Then you just need to sub in another point on the graph (i.e. (2,-17)) to find the value of a.
    :confused:

    Interesting, although that's not how I did it. I don't want to post my values here since OP's clearly still working on it, but would differentiation help to eliminate c? That's the approach I took, followed by painstakingly forming simultaneous equations. :sigh: I'm sure your way is A LOT faster! :p:
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    Thanks, very much.
    Understand it now
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    (Original post by Tortious)
    :confused:

    Interesting, although that's not how I did it. I don't want to post my values here since OP's clearly still working on it, but would differentiation help to eliminate c? That's the approach I took, followed by painstakingly forming simultaneous equations. :sigh: I'm sure your way is A LOT faster! :p:
    Well you could start from y=ax^2+bx+c, differentiate to find the x-coordinate of the vertex in terms of a and b, sub back in to find the y-value in terms of a,b,c, then sub in the values of the coordinates and solve the resulting simultaneous equations in a,b,c... but it seems like a wasted effort when writing a(x+p)^2+q=0 tells you what p,q (and hence b,c) are immediately, so you just have to sub in to find a.
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    Hey,

    I tried the method. I guess I don't quite understand it yet =S

    I got a(x+4)^2+1 and then did all the subbing in but got a wrong answer (Answers are in back of book but don't explain anything). Please if you could show me all the steps- sorry, i'm slow- it would be very appreciated.

    Apprently the right answer is y=-1/2(x^2) -4x -7
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    (Original post by jammysmt)
    Hey,

    I tried the method. I guess I don't quite understand it yet =S

    I got a(x+4)^2+1 and then did all the subbing in but got a wrong answer (Answers are in back of book but don't explain anything). Please if you could show me all the steps- sorry, i'm slow- it would be very appreciated.

    Apprently the right answer is y=-1/2(x^2) -4x -7
    The a(x+4)^2+1 bit is right. I won't show you all the steps though -- you show me the steps, and then I'll show you where you're going wrong.
 
 
 
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