What is the equation of the parabola that has a vertex at (14, -9) (14,9) and passes through point (0, 2) (0,2)?

1 Answer
Aug 25, 2016

y=11/196(x-14)^2-9y=11196(x14)29

Explanation:

The equation of a parabola in color(blue)"vertex form"vertex form is

color(red)(|bar(ul(color(white)(a/a)color(black)(y=a(x-h)^2+k)color(white)(a/a)|)))
where (h ,k) are the coordinates of the vertex and a, is a constant.

here h = 14 and k = - 9, so we can write a partial equation

y=a(x-14)^2-9

To find a, substitute the coordinates of (0 ,2) a point on the parabola, into the partial equation.

rArra(0-14)^2-9=2rArr196a=11rArra=11/196

rArry=11/196(x-14)^2-9" is equation in vertex form"

The equation may be expressed in color(blue)"standard form"

That is y=ax^2+bx+c by distributing the bracket and simplifying.

rArry=11/196(x^2-28x+196)-9=11/196x^2-11/7x+2

graph{11/196(x-14)^2-9 [-20, 20, -10, 10]}