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Math Help - TP of quadratic when given y & x ints

  1. #1
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    Exclamation TP of quadratic when given y & x ints

    A parabola has a turning point at (z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

    Find the value of z
    Find the equation of the parabola

    It's seemingly easy but I can't seem to do it.
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  2. #2
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    Quote Originally Posted by freswood
    A parabola has a turning point at (z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

    Find the value of z
    Find the equation of the parabola

    It's seemingly easy but I can't seem to do it.
    Hello,

    if the vertex of the parabola has the coordinates (z, v) then the equation of the parabola, which you should use, is:
    y=a \cdot(x-z)^2+v

    You've got 3 points, which belong to the parabola:
    P(0, 10)
    V(z, -8)
    N(5, 0)

    Put in the coordinates of these points into the equation and you'll get:
    10=az^2
    0=a\cdot(5-z)^2-8

    Solve the 1rst equation for a: a=\frac{18}{z^2}
    and put it into the 2nd equation. You'll get:
    0=\frac{18}{z^2} \cdot (5-z)^2-8

    Solve for z. The results are: z=3 or z=15

    I've attached a graph to demonstrate these results.

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails TP of quadratic when given y & x ints-zwei_par.gif  
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  3. #3
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    Quote Originally Posted by freswood
    A parabola has a turning point at (z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

    Find the value of z
    Find the equation of the parabola

    It's seemingly easy but I can't seem to do it.
    Here is one way.

    The turning point---the vertex--- is at (z,-8) and the parabola intersects the x-axis in two points, and it intersects the y-axis at one point only. That means the parabola is a vertical parabola---it opens upward or downward.

    The standard form of the equation of a vertical parabola whose vertex is at (h,k) is:
    (y-k) = a(x-h)^2 ----***

    Given: two points (0,10) and (5,0), and vertex (z,-8)

    At point (0,10),
    (10 -(-8)) = a(0 -z)^2
    18 = a(z^2) ----------------(1)

    At point (5,0),
    (0 -(-8)) = a(5 -z)^2
    8 = a(5 -z)^2 --------------(2)

    From (1), a = 18/(z^2), substitute that into (2),
    8 = [18/(z^2)][25 -10z +z^2]
    Clear the fractions, multiply both sides by z^2,
    8z^2 = 18[25 -10z +z^2]
    Divide both sides by 2,
    4z^2 = 9[25 -10z +z^2]
    4z^2 = 225 -90z +9z^2
    0 = 225 -90z +9z^2 -4z^2
    0 = 225 -90z +5z^2
    Divide both sides by 5,
    0 = 45 -18z +z^2
    Or,
    z^2 -18z +45 = 0
    (z -15)(z -3) = 0
    z = 15 or 3 -----------------answer.

    Since there are two possible z's, then there are two possible parabolas.


    One is using the vertex (15,-8).
    First we find the "a" here.
    Substitute z=15 into (1),
    18 = a(15^2)
    a = 18/225
    So,
    (y -(-8)) = (18/225)(x -15)^2
    y +8 = (18/225)(x -15)^2 -----------answer.
    If you want to expand that, and get the general form---y = ax^2 +bx +c---of the parabola,
    y +8 = (18/225)(x^2 -30x +225)
    y +8 = (18/225)x^2 -(540/225)x +18
    y = (18/225)x^2 -(12/5)x +18 -8
    y = 0.08x^2 -2.4x +10 ------------------answer.

    The other is using the vertex (3,-8).
    a = 18/(3^2) = 2
    So,
    (y -(-8)) = 2(x -3)^2
    y +8 = 2(x -3)^2 -------------answer.
    Or,
    y +8 = 2(x^2 -6x +9)
    y +8 = 2x^2 -12x +18
    y = 2x^2 -12x +18 -8
    y = 2x^2 -12x +10 -----------answer.
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  4. #4
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    Thanks heaps for your help! It makes sense now. Stupid me - I always rely on elimination, and never think to use substitution
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