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Math Help - Need help solving for:

  1. #1
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    Need help solving for:

    the axis of symmetry, the vertex, any x-and y-intercepts and graphing the parabolas.

    here's an example question

    y=-4x^2-9x-2

    also y=x^2-6x+25
    Last edited by Vuong; January 3rd 2010 at 02:31 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Vuong View Post
    the axis of symmetry, the vertex, any x-and y-intercepts and graphing the parabolas.

    here's an example question

    y=-4x^2-9x-2
    Let f(x)=ax^2+bx+c. A little completing the square shows that f(x)=a\left(a+\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a}. From this we see that the vertex is at \left(\tfrac{-b}{2a},c-\tfrac{b^2}{4a}\right), the axis of symmetry at x=\frac{-b}{2a}, the x-intercepts are at x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} given \Delta\geqslant0, and the y intercepts are f(0)=c.
    Last edited by mr fantastic; January 3rd 2010 at 09:33 PM. Reason: Added a math tag
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  3. #3
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    I never understood what the f(x) meant...
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Vuong View Post
    I never understood what the f(x) meant...
    Same thing as y
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  5. #5
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    Quote Originally Posted by Vuong View Post
    y=-4x^2-9x-2
    Start by factoring -4x^2 - 9x - 2
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  6. #6
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    Quote Originally Posted by Wilmer View Post
    Start by factoring -4x^2 - 9x - 2
    What do I do after?
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  7. #7
    Super Member Quacky's Avatar
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    y=-4x^2-9x-2

    Ok, the line will cut the y axis where x=0.
    Substitute x=0 into the equation, and you get y=-2. So point of interception is at:
    (0,-2)
    Now, the line will cut the x axis where y=0. To find this, we can try to factorise:

    -4x^2-9x-2=0
    4x^2+9x+2=0
    (4x+1)(x+2)=0
    x=-2 and x=-1/4
    x intercepts are
    (-2,0) and (-1/4,0)

    Parabolas are symmetrical, so if you take the x intercepts, the line of symmetry will be halfway between them.
    -2.25/2 = -1.125 (wow, nasty numbers)
    x=-1.125 is the line of symmetry.

    A question about the vertex: Have you covered differentiation?
    Last edited by Quacky; February 5th 2010 at 05:26 PM. Reason: Added \math' tags for clarification
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  8. #8
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    Quote Originally Posted by Quacky View Post
    y=-4x^2-9x-2

    Ok, the line will cut the y axis where x=0.
    Substitute x=0 into the equation, and you get y=-2. So point of interception is at:
    (0,-2)
    Now, the line will cut the x axis where y=0. To find this, we can try to factorise:

    -4x^2-9x-2=0
    4x^2+9x+2=0
    (4x+1)(x+2)=0
    x=-2 and x=-1/4
    x intercepts are
    (-2,0) and (-1/4,0)

    Parabolas are symmetrical, so if you take the x intercepts, the line of symmetry will be halfway between them.
    -2.25/2 = -1.125 (wow, nasty numbers)
    x=-1.125 is the line of symmetry.

    A question about the vertex: Have you covered differentiation?
    Why are they minus?
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  9. #9
    Super Member Quacky's Avatar
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    (4x+1)(x+2)=0
    Substitute x=-2 in and see what happens.
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  10. #10
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    Quote Originally Posted by Quacky View Post
    (4x+1)(x+2)=0
    Substitute x=-2 in and see what happens.
    Uhh (-7)(0) = 0?
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  11. #11
    Super Member Quacky's Avatar
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    Yes.
    If you substitute 2 into the equation, you get
    (9)(4) = 36.
    As you want the solutions to
    (4x+1)(x+2)=0 (not 36)
    x=-2 is the correct solution, not x=2.
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  12. #12
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    Quote Originally Posted by Quacky View Post
    Yes.
    If you substitute 2 into the equation, you get
    (9)(4) = 36.
    As you want the solutions to
    (4x+1)(x+2)=0 (not 36)
    x=-2 is the correct solution, not x=2.
    Great, thanks! Big help. Do you know how to factor a problem like 6x^2-5x-6? I've been trying a bunch of combinations, still unsuccessful
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  13. #13
    Super Member Quacky's Avatar
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    Think of it this way:

    For that to be true, either
    (4x+1)=0 or (x+2)=0
    If 4x+1=0
    4x=-1
    x=-1/4

    If x+2=0
    x=-2

    Edit: I'll look at the above problem now.
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  14. #14
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Vuong View Post
    Great, thanks! Big help. Do you know how to factor a problem like 6x^2-5x-6? I've been trying a bunch of combinations, still unsuccessful
    6x^2-5x-6=0\implies x=\frac{5\pm\sqrt{25-4(6)(-6)}}{12}={{{\tfrac{18}{12}}\brace{\tfrac{-8}{12}}}} so that 6x^2-5x-6=\left(x-\tfrac{18}{12}\right)\cdot\left(x+\tfrac{8}{12}\ri  ght)
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  15. #15
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    Quote Originally Posted by Drexel28 View Post
    6x^2-5x-6=0\implies x=\frac{5\pm\sqrt{25-4(6)(-6)}}{12}={{{\tfrac{18}{12}}\brace{\tfrac{-8}{12}}}} so that 6x^2-5x-6=\left(x-\tfrac{18}{12}\right)\cdot\left(x+\tfrac{8}{12}\ri  ght)
    You could use quadratic to factor?
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