What is the vertices y^2-8x+2y+17=0 ?

The vertices y^2-8x+2y+17=0 is (h,k)=(2,-1),p=2

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vertices y^2-8x+2y+17=0

vertices

y^{2} - 8 x + 2 y + 17 = 0

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

Solution steps

y^{2} - 8 x + 2 y + 17 = 0

Rewrite

y^{2} - 8 x + 2 y + 17 = 0

in the standard form

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

Therefore parabola properties are:

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

y = x^{2} + 4 x + 3

y = x^{2} - 10 x + 7

y = 92 x^{2}

y = x^{2} - 2 x - 35

x^{2} + 7 x + 12

What is the vertices y^2-8x+2y+17=0 ?
The vertices y^2-8x+2y+17=0 is (h,k)=(2,-1),p=2