how to write a quadratic equation with the vertex (13,-5) and the point (6,2)

how to write a quadratic equation with the vertex (13,-5) and the point (6,2)
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y=a(x-13)^2-5 is the standard vertex form for a parabola. To find a we plug in the given point: 2=a(6-13)^2-5.

2=49a-5, 49a=7, a=1/7, so y=(1/7)(x-13)^2-5=(1/7)(x^2-26x+169)-5.

Multiply through by 7: 7y=x^2-26x+169-35=x^2-26x+134.

The quadratic is y=(1/7)(x^2-26x+134).

by Top Rated User (1.2m points)

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