What is the vertices 4y^2-48x-20y=71 ?

The vertices 4y^2-48x-20y=71 is (h,k)=(-2, 5/2),p=3

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vertices 4y^2-48x-20y=71

vertices

4 y^{2} - 48 x - 20 y = 71

(h, k) = (- 2, \frac{5}{2}), p = 3

Solution steps

4 y^{2} - 48 x - 20 y = 71

Rewrite

4 y^{2} - 48 x - 20 y = 71

in the standard form

4 \cdot 3 (x - (- 2)) = (y - \frac{5}{2})^{2}

Therefore parabola properties are:

(h, k) = (- 2, \frac{5}{2}), p = 3

y = x^{2} + 2 x - 15

f (x) = - 4 (x + 1)^{2} - 5

x^{2} - 4

2 x^{2} + 3 x - 20

y = x^{2} - 2 x^{1}

What is the vertices 4y^2-48x-20y=71 ?
The vertices 4y^2-48x-20y=71 is (h,k)=(-2, 5/2),p=3