What is the vertices x=4y^2 ?

The vertices x=4y^2 is (h,k)=(0,0),p= 1/16

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vertices x=4y^2

vertices

x = 4 y^{2}

(h, k) = (0, 0), p = \frac{1}{16}

Solution steps

x = 4 y^{2}

Rewrite

x = 4 y^{2}

in the standard form:

4 \cdot \frac{1}{16} (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{16}

2 x^{2} - 3

x^{2} = - 12 y

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

x^{2} + 8 y + 2 x - 23 = 0

y = x^{2} - 10 x + 24