What is the vertices y^2-6y-8x+17=0 ?

The vertices y^2-6y-8x+17=0 is (h,k)=(1,3),p=2

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vertices y^2-6y-8x+17=0

vertices

y^{2} - 6 y - 8 x + 17 = 0

(h, k) = (1, 3), p = 2

Solution steps

y^{2} - 6 y - 8 x + 17 = 0

Rewrite

y^{2} - 6 y - 8 x + 17 = 0

in the standard form

4 \cdot 2 (x - 1) = (y - 3)^{2}

Therefore parabola properties are:

(h, k) = (1, 3), p = 2

y = x^{2} + 3 x - 10

f (x) = x^{2} + 2 x - 15

8 y = x^{2}

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

y = 2 x^{2} - 5

What is the vertices y^2-6y-8x+17=0 ?
The vertices y^2-6y-8x+17=0 is (h,k)=(1,3),p=2