What is the vertices 2x-y^2=4 ?

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

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vertices 2x-y^2=4

vertices

2 x - y^{2} = 4

(h, k) = (2, 0), p = \frac{1}{2}

Solution steps

2 x - y^{2} = 4

Rewrite

2 x - y^{2} = 4

in the standard form:

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

Therefore parabola properties are:

(h, k) = (2, 0), p = \frac{1}{2}

f (x) = - 2 x^{2} + 5

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

2 x^{2} - 12 x + 3

y = - 3 x^{2} + 6 x - 1

y = (x + 5)^{2}