Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Find the vertex of the parabola?

how do you find the vertex

of this parabola?

y = x^2-8x+7

please and thank you

best answer will be rewarded

9 Answers

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  • cidyah
    Lv 7
    1 decade ago
    Best Answer

    a=1 b=-8 c=7 if the parabola is y=ax^2+bx+c

    x-coordinate of the vertex = -b / 2a

    =-(-8) /2 =4

    y-coordinate of the vertex is

    (4)^2-8(4)+7

    =16-32+7 = -9

    (4,-9) is the vertex

  • 1 decade ago

    well since the coefficient of the x^2 is positive (in this case it's 1), then you know your parabola will be opening upward. So the vertex is on the bottom, making it a minimum. I don't know what math level you are on. But I am going to use some precal/cal.

    Find the derivative of y.

    y = x^2 - 8x +7

    dy/dx = 2x - 8

    set that equal to 0 and solve:

    2x -8= 0

    2x=8

    x=4

    plug 4 into the original equation for x.

    y = 4^2 - 8(4) +7

    y = 16-32+7

    y = -9

    vertex = (4, -9)

    A simpler way is just using -b/2a to give you the x coordinate:

    8/2 = 4

    Then plug 4 in the original equation to get y = -9.

    Source(s): I'm a math major
  • 4 years ago

    I can find the vertex of the parabola, but I am unaware of any curve named 'vertex parabola !' f(x) = x^2 - 4x + 3 = (x - 2)^2 - 1 The vertex is at: (2, - 1) QED

  • T
    Lv 5
    1 decade ago

    Take the derivative, set that equal to 0 and solve for x. Put that value into the original equation and solve for y.

    dy/dx = 2x - 8 = 0

    2x = 8

    x = 4

    y = 4^2 - 8(4) + 7

    y = 16 - 32 + 7

    y = -9

    (4,-9)

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  • This has the form ax^2+bx+x

    y=1x^2-8x+7

    a=1

    b=-8

    c=7

    x intercept=7,1

    y intercept=7

    vertex=(4,-9)

    x coordinate of the vertex=-b/2a

    =-(-8)/2(1)

    =8/2

    =4

    now plug this into the quadratic equation and find the y point

    y=4^2-8(4)+7

    y=16-32+7

    y=-16+7

    y=-9

  • 1 decade ago

    the vertex of a parabola is -b/2a

    so this is -(-8 / 2)

    which is 4

  • 6 years ago

    Here a=1, b=-8andc=7

    Now for any parabola ,

    Vertex (x, y) is given as

    x=-b/2a= 8/2=4

    y=-D/4a

    where D = b^2 -4ac

    Therefore,

    y=-[(-8)^2-(4*1*7)]/(4*1)

    =-(64-28)/4=36/4=9

    Hence, vertex is (4, 9)

  • Como
    Lv 7
    1 decade ago

    Method 1

    f (x) = (x ² - 8 x + 16) - 16 + 7

    f (x) = (x - 4)² - 9

    Vertex is V (4 , - 9)

    Method 2

    f `(x) = 2x - 8 = 0 for turning point

    x = 4

    f(4) = - 9

    Vertex is V (4 , - 9)

  • 1 decade ago

    y = x^2-8x+7 {we will try to make a perfect square involving x}

    y = x^2-8x+16 -9

    y+9 = (x -4)^2

    Vertex is (4, -9)

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