What is the vertices y^2=6x-4y-12 ?

The vertices y^2=6x-4y-12 is (h,k)=(4/3 ,-2),p= 3/2

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vertices y^2=6x-4y-12

vertices

y^{2} = 6 x - 4 y - 12

(h, k) = (\frac{4}{3}, - 2), p = \frac{3}{2}

Solution steps

y^{2} = 6 x - 4 y - 12

Rewrite

y^{2} = 6 x - 4 y - 12

in the standard form

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

Therefore parabola properties are:

(h, k) = (\frac{4}{3}, - 2), p = \frac{3}{2}

y = + x^{2} + 8

- 3 x^{2} + 30 x - 8

f (x) = 4 x^{2} - 8 x + 7

f (x) = 2 x^{2} + 5 x + 2

f (x) = 8 x^{2}

What is the vertices y^2=6x-4y-12 ?
The vertices y^2=6x-4y-12 is (h,k)=(4/3 ,-2),p= 3/2