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Popular Geometry >

vertices y^2-12y-4x+28=0

Solution

vertices

y^{2} - 12 y - 4 x + 28 = 0

Solution

(h, k) = (- 2, 6), p = 1

Solution steps

y^{2} - 12 y - 4 x + 28 = 0

Rewrite

y^{2} - 12 y - 4 x + 28 = 0

in the standard form

4 \cdot 1 \cdot (x - (- 2)) = (y - 6)^{2}

Therefore parabola properties are:

(h, k) = (- 2, 6), p = 1

Graph

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Frequently Asked Questions (FAQ)

What is the vertices y^2-12y-4x+28=0 ?
The vertices y^2-12y-4x+28=0 is (h,k)=(-2,6),p=1