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Math Help - Quadratic Equations of the form Ax^2 + BX + C

  1. #1
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    Quadratic Equations of the form Ax^2 + BX + C

    I am trying to get to understand this subject but can't seem to grasp it from the books?

    I have; y = X^2 + 2x - 3.

    This is what I have tried so far;

    (x + 2) (x - 1.5)

    x = - 2 or x = 1.5

    - 2 + 1.5 = - 0.25
    2

    y = - 0.25^2 + 2 ( -0.25) - 3

    ( -4, -1)

    I know it's wrong but can't see why?

    Any help appreciated

    Thanks
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  2. #2
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    What are you trying to do? Are you trying so solve F(x)=0 ? If so, then (x+3)(x-1)=0
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Do you want to calculate the zero's? ...
    What have you done to factorize the quadratic equation? Have you used the quadratic formula? Or? ...

    Because:
    (x+2)(x-1,5)=x^2+2x-1,5x-3=x^2+0,5x-3 (and this one is not the given equation).

    Use:
    D=b^2-4.a.c=4-4.(1.(-3))=4+12=16
    Can you now calculate the zero's? ...
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Quote Originally Posted by Siron View Post
    Do you want to calculate the zero's? ...
    What have you done to factorize the quadratic equation? Have you used the quadratic formula? Or? ...

    Because:
    (x+2)(x-1,5)=x^2+2x-1,5x-3=x^2+0,5x-3 (and this one is not the given equation).

    Use:
    D=b^2-4.a.c=4-4.(1.(-3))=4+12=16
    Can you now calculate the zero's? ...
    What I am trying to do is find the co-ordinates of its vertex
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    You can use the calculate the vertex of the parabola:
    x=(-b)/2a

    You get the x-coordinat of the vertex and you just have to enter the value of the x-coordinate in the equation and you'll find the y-coordinate of the vertex.

    Or you can calculate the first derivative and take f'(x)=0.
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Quote Originally Posted by Siron View Post
    You can use the calculate the vertex of the parabola:
    x=(-b)/2a

    You get the x-coordinat of the vertex and you just have to enter the value of the x-coordinate in the equation and you'll find the y-coordinate of the vertex.

    Or you can calculate the first derivative and take f'(x)=0.
    Thanks for the info, but I can't use that method, the reason being is; y = x^2 + 2x - 3.

    I have no values for - b / 2a

    ?????
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  7. #7
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    You have the values, in general a quadratic equation looks like: y=ax^2+bx+c.
    In this case the given equation: y=x^2+2x-3
    That means: a=1,b=2,c=-3

    If you want to use the first derivative, you'll get:
    f'(x)=2x+2
    and f'(x)=0 so
    2x+2=0 <-> 2x=-2 <->x=-1

    Now you've the x-coordinate of the vertex, by entering in the equation you'll get the y-coordinate:
    y=(-1)^2+2(-1)-3=1-2-3=-4

    So the vertex of the parabola: (-1,-4)

    Try it with the other method.
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Quote Originally Posted by Siron View Post
    You have the values, in general a quadratic equation looks like: y=ax^2+bx+c.
    In this case the given equation: y=x^2+2x-3
    That means: a=1,b=2,c=-3

    If you want to use the first derivative, you'll get:
    f'(x)=2x+2
    and f'(x)=0 so
    2x+2=0 <-> 2x=-2 <->x=-1

    Now you've the x-coordinate of the vertex, by entering in the equation you'll get the y-coordinate:
    y=(-1)^2+2(-1)-3=1-2-3=-4

    So the vertex of the parabola: (-1,-4)

    Try it with the other method.
    Thanks for your help, the method you have used i.e. f(x) = 2x +2 is a bit confusing to me at the moment, I can see mistakes I am making following your work and I know your method is right, but it is complex for a beginner to understand.

    Thanks

    David
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    MHF Contributor Siron's Avatar
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    You're welcome! Haven't you learned derivatives yet? ...

    In that case, do you see now you can just read a,b,c out of the equation and calculate the vertex that way?
    Another way is to wright your equation in this form:
    y=(x-p)^2+q
    Now the coordinates of the top are: (p,q).
    In this case:
    y=x^2+2x-3
    you can write it as: y=(x+1)^2-4
    So the coordinates of the vertex are (-1,-4).


    It isn't a problem you haven't learned derivatives yet, but they can be very useful in this case. With derivatives you've also the ability (if it's possible offcourse) to calculate the vertex of many other functions.
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  10. #10
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Quote Originally Posted by Siron View Post
    You're welcome! Haven't you learned derivatives yet? ...

    In that case, do you see now you can just read a,b,c out of the equation and calculate the vertex that way?
    Another way is to wright your equation in this form:
    y=(x-p)^2+q
    Now the coordinates of the top are: (p,q).
    In this case:
    y=x^2+2x-3
    you can write it as: y=(x+1)^2-4
    So the coordinates of the vertex are (-1,-4).


    It isn't a problem you haven't learned derivatives yet, but they can be very useful in this case. With derivatives you've also the ability (if it's possible offcourse) to calculate the vertex of many other functions.
    Sorry not that advanced at the moment!

    Not entiry happy with the course book, its not written with plenty of worked examples, but plenty of history in stead!

    Will have to find some good maths books I think and get reading.

    Thanks again for your help.
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Quote Originally Posted by David Green View Post
    Sorry not that advanced at the moment!

    Not entiry happy with the course book, its not written with plenty of worked examples, but plenty of history in stead!

    Will have to find some good maths books I think and get reading.

    Thanks again for your help.
    Post #2 tells you how to find the x-intercepts. For a parabola, you should know that the turning point lies on the axis of symmetry, which is a vertical line halfway between the x-intercepts.

    So you should be able to get the x-coordinate easily. Then substitute that value into the given parabola rule to get the y-coordinate.

    I'm sure your class notes and textbook would have examples to follow - have you refered to them?
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    Re: Quadratic Equations of the form Ax^2 + BX + C

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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Quote Originally Posted by David Green View Post
    Thanks for the info, but I can't use that method, the reason being is; y = x^2 + 2x - 3.

    I have no values for - b / 2a

    ?????
    What is wrong with a=1, b=2, c=-3 ?

    CB
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  14. #14
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    You can also complete the square to turn the parabola into turning point form then read the coordinates of the vertex from there
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  15. #15
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    Re: Quadratic Equations of the form Ax^2 + BX + C

    Well, we can tell him a thousand ways to do this,
    but 1st he needs to get exposed to the quadratic formula;
    as it is, he doesn't seem to know what d'heck A,B,C are!
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