What is the vertices x=7y^2 ?

The vertices x=7y^2 is (h,k)=(0,0),p= 1/28

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vertices x=7y^2

vertices

x = 7 y^{2}

(h, k) = (0, 0), p = \frac{1}{28}

Solution steps

x = 7 y^{2}

Rewrite

x = 7 y^{2}

in the standard form:

4 \cdot \frac{1}{28} (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{28}

y = - x^{2} + 10 x - 28

+ y^{2} = 28 x

x^{2} - 5 x + 4

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

y = 7 x^{2} - 10 x + 5