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Math Help - quadratic equations

  1. #1
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    quadratic equations

    hi all, i need some help on finding quadratic equations from points on graphs, could someone please please explain how i would determine which general equation to use to solve.

    1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

    a.) someone please explain how to pick a formula to use

    2.) a parabola passes through points (-3,9) find the rule dictating the parabola.

    a.) someone please explain how to pick a formula to use

    also at the start of the exercise there is an example like this: a parabola passes through points (0,3) and (-3,1) it says this equation is in the from y =ax^2 + c... What the hell? i nvr learnt any ax^2 +c

    pls help i have a test this coming thursday
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  2. #2
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    Quote Originally Posted by andrew2322 View Post
    hi all, i need some help on finding quadratic equations from points on graphs, could someone please please explain how i would determine which general equation to use to solve.

    1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

    Mr F says: Use the model y = A(x - B)(x - C). B = 1, C = 3 (why?). To get A substitute the given y-intercept, that is, the point (0, 3).

    a.) someone please explain how to pick a formula to use

    2.) a parabola passes through points (-3,9) find the rule dictating the parabola. Mr F says: Insufficient data to get a unique rule.

    a.) someone please explain how to pick a formula to use

    also at the start of the exercise there is an example like this: a parabola passes through points (0,3) and (-3,1) it says this equation is in the from y =ax^2 + c... What the hell? i nvr learnt any ax^2 +c

    Mr F says: They're giving you the heads-up that b = 0 in the standard form! Run with the ball!! Substitute the two given points to get two equations in a and c. Solve these two equations simultaneously.

    pls help i have a test this coming thursday
    ..
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  3. #3
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    sorry

    huh? im like so lost... how can y = A(x-c)(x-b) work?gah pls go over it again with the actual answers pls? nd how to select which formula to use pls
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    Quote Originally Posted by andrew2322 View Post
    huh? im like so lost... how can y = A(x-c)(x-b) work?gah pls go over it again with the actual answers pls? nd how to select which formula to use pls
    Consider ....... y = A(x - B)(x - C) has x-intercepts found from the equation

    A(x - B)(x - C) = 0

    => x - B = 0 or x - C = 0

    => x = B or x = C.

    So if you know the x-intercepts, you use the model y = A(x - B)(x - C) because knowing the x-intercepts means you know the value of B and C .....

    So for the question:

    Quote Originally Posted by andrew2322
    1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.
    you would use this model and you know that B = 1 and C = 3.

    So y = A(x - 1)(x - 3).

    Now substitute the y-inetercept, that is, the point (0, 3), that is, x = 0 and y = 3:

    3 = A(0 - 1)(0 - 3) = A(-1)(-3) = 3A => A = 1.

    So y = (x - 1)(x - 3).
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    woah

    wow that just hit me in the face, how cud i have not seen the obvious... wow ur really smart. and one last thing, cud u pls explain the values B and C and how they are different from X? and also, there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.
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    Quote Originally Posted by andrew2322 View Post
    wow that just hit me in the face, how cud i have not seen the obvious... wow ur really smart. and one last thing, cud u pls explain the values B and C and how they are different from X?
    They are the x-intercepts. In other words, the value of x where the parabola cuts the x-axis. In other words, when y = 0, x = B or x = C.

    Try working backwards from the answer y = (x - 1)(x - 3)to see how it all fits together.
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    sorry

    there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.
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    Quote Originally Posted by andrew2322 View Post
    there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.
    They give you the model - on a platter. Don't question your good luck, just use it!!!!

    Substitute the given point, that is x = 2 and y = 5:

    5 = a (2)^2 = 4a => a = 5/4.

    Therefore y = (5/4) x^2.
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    hey

    could you please tell me how, if i didnt have good luck, know what equation to use?
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    Quote Originally Posted by andrew2322 View Post
    could you please tell me how, if i didnt have good luck, know what equation to use?
    You wouldn't. Without being given the model, a single point is insufficient data to get a unique equation. That's why you were given the equation (so it wasn't really good luck .... it was actually absolutely essential).
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  11. #11
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    one last

    ok just one FINAL question

    Y intercept: (0,5)
    X intercept: (4,0)

    Find the equation, please give me step by step working out with the final answer, thanks
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  12. #12
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    Quote Originally Posted by andrew2322 View Post
    ok just one FINAL question

    Y intercept: (0,5)
    X intercept: (4,0)

    Find the equation, please give me step by step working out with the final answer, thanks
    Insufficient data to find a unique equation. Is the x-intercept also a turning point?
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