What is the vertices y^2-6y-4x+5=0 ?

The vertices y^2-6y-4x+5=0 is (h,k)=(-1,3),p=1

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vertices y^2-6y-4x+5=0

vertices

y^{2} - 6 y - 4 x + 5 = 0

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

Solution steps

y^{2} - 6 y - 4 x + 5 = 0

Rewrite

y^{2} - 6 y - 4 x + 5 = 0

in the standard form

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

Therefore parabola properties are:

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

5 x (2 x + 4) - 3 x

f (x) = - 4 x^{2} - 12 x + 5

f (x) = 3 x^{2} - x + 1

y = - 2 x^{2} + 5

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

What is the vertices y^2-6y-4x+5=0 ?
The vertices y^2-6y-4x+5=0 is (h,k)=(-1,3),p=1