What is the vertices x=-1/4 (y+2)^2 ?

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

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vertices x=-1/4 (y+2)^2

vertices

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

(h, k) = (0, - 2), p = - 1

Solution steps

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

Rewrite

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

in the standard form:

4 (- 1) (x - 0) = (y - (- 2))^{2}

Therefore parabola properties are:

(h, k) = (0, - 2), p = - 1

2 x^{2} - 4 x + 7

x^{2} - 5 x - 84

x^{2} + 4 x - 24 y - 92 = 0

y = x^{2} - 12 x + 2

f (x) = - x^{2} + 7 x + 18

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