What is the vertices x=-3y^2+5 ?

The vertices x=-3y^2+5 is (h,k)=(5,0),p=-1/12

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vertices x=-3y^2+5

vertices

x = - 3 y^{2} + 5

(h, k) = (5, 0), p = - \frac{1}{12}

Solution steps

x = - 3 y^{2} + 5

Rewrite

x = - 3 y^{2} + 5

in the standard form:

4 (- \frac{1}{12}) (x - 5) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (5, 0), p = - \frac{1}{12}

f (x) = x^{2} + 10 x + 6

- 3 x^{2} - 12 x - 9

y = x^{2} - 2 x 5

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

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