What is the vertices x=-2(y+3)^2+4 ?

The vertices x=-2(y+3)^2+4 is (h,k)=(4,-3),p=-1/8

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vertices x=-2(y+3)^2+4

vertices

x = - 2 (y + 3)^{2} + 4

(h, k) = (4, - 3), p = - \frac{1}{8}

Solution steps

x = - 2 (y + 3)^{2} + 4

Rewrite

x = - 2 (y + 3)^{2} + 4

in the standard form:

4 (- \frac{1}{8}) (x - 4) = (y - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (4, - 3), p = - \frac{1}{8}

x^{2} + 5 x - 14

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

3 x^{2} + 3 x + 2 y = 0

f (x) = 2 x^{2} + 8 x + 3

y = 2 x^{2} - 7

What is the vertices x=-2(y+3)^2+4 ?
The vertices x=-2(y+3)^2+4 is (h,k)=(4,-3),p=-1/8