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

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

asked Nov 30, 2013 in ALGEBRA 2 by angel12 Scholar

1 Answer

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Parabola vertex form is y = a(x-h)^2+k

Given (h,k) = (13,-5)

(x,y) = (6,2)

Now find a

2 = a (6-13)^2-5

2 = a(-7)^2-5

2 = 49a-5

Add 5 to each side.

2+5 = 49a-5+5

7 = 49a

Divide to each side by 49.

7/49 = 49a/49

1/7 = a

a = 1/7

Parabol vertex form is y = a(x-h)^2+k

Given parabola Parabol vertex form is y = 1/7(x-13)^2-5

y = 1/7(x^2-26x+169)-5

y = x^2/7-26/7x+169/7-5

y = x^2/7-(26/7)x+(169-35)/7

y = 1/7(x^2)-(26/7)x+134/7

Now it is in quadratic equation form.

answered Dec 13, 2013 by ashokavf Scholar

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