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

The vertices x=(y-3)^2 is (h,k)=(0,3),p= 1/4

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vertices x=(y-3)^2

vertices

x = (y - 3)^{2}

(h, k) = (0, 3), p = \frac{1}{4}

Solution steps

x = (y - 3)^{2}

Rewrite

x = (y - 3)^{2}

in the standard form:

4 \cdot \frac{1}{4} (x - 0) = (y - 3)^{2}

Therefore parabola properties are:

(h, k) = (0, 3), p = \frac{1}{4}

y^{2} = x + 1

f (x) = - x^{2} + 8 x - 12

x^{2} - 6 x + 8

y = - 4 x^{2} - 24 x - 40

f (x) = x^{2} + 6 x + 9