What is the vertices x=3y^2 ?

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

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vertices x=3y^2

vertices

x = 3 y^{2}

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

Solution steps

x = 3 y^{2}

Rewrite

x = 3 y^{2}

in the standard form:

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

Therefore parabola properties are:

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

y = x^{2} + 4 x - 12

2 (x + 1)^{2} - 3

2 y^{2} - 8 x = 0

x^{2} + 3 x + 1

x^{2} + 3 x - 1