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Math Help - Turning Point of a Parabola

  1. #1
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    Turning Point of a Parabola

    Hello,

    How do I find the turning point of a parabola?

    Thanks in advance
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  2. #2
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    Quote Originally Posted by J4553 View Post
    Hello,

    How do I find the turning point of a parabola?

    Thanks in advance
    Complete the square.

    Alternatively, use calculus.

    Post a more specific question (and state your mathematical background).
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  3. #3
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    Quote Originally Posted by J4553 View Post
    Hello,

    How do I find the turning point of a parabola?

    Thanks in advance
    The turning point of a parabola is the vertex of the parabola.

    For the given equation of parabola, you can find the vertex by completing the square in the form

    y = a(x-h)^2+k

    where (h, k) is vertex.

    If, suppose equation of a parabola is

    y = 3x^2-6x+5

    Then, complete the square, so, eqn becomes,

    y = 3(x-1)^2+2

    so, vertex = (1, 2)

    so, turning point = (1, 2)

    Did you get it now???
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  4. #4
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    Shyam - no, I'm afraid that just confuses me even more.
    I think my lecturer gave the formula: -b/2a to find the x-component of the vertex, then may have said to substitute that value into another equation to find the y-component.
    Does this sound right or am I way off?
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  5. #5
    Senior Member euclid2's Avatar
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    Quote Originally Posted by J4553 View Post
    Shyam - no, I'm afraid that just confuses me even more.
    I think my lecturer gave the formula: -b/2a to find the x-component of the vertex, then may have said to substitute that value into another equation to find the y-component.
    Does this sound right or am I way off?
    that formula is correct, for example, take this function;

    y=x^2+6x+8

    b=6, a=1

    -b/2a=-6/2(1)=-3

    sub -3 into the function to get the value of y

    y=(-3)^2+6(-3)+8=-1

    therefore, the vertex for this example is (-3,-1)

    Shyam is simply using another method then the one you are instructed to use
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  6. #6
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    Perfect - thanks!
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  7. #7
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    Quote Originally Posted by J4553 View Post
    Shyam - no, I'm afraid that just confuses me even more.
    I think my lecturer gave the formula: -b/2a to find the x-component of the vertex, then may have said to substitute that value into another equation to find the y-component.
    Does this sound right or am I way off?
    Yes, your formula is correct.

    If, suppose equation of a parabola is

    y = 3x^2-6x+5

    x = \frac{-b}{2a}=\frac{-(-6)}{(2)(3)}=1

    y = 3(1)^2-6(1)+5

    y = 2

    so, you have (1, 2) as the vertex or turning point of parabola
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