What is the vertices y^2=-6x ?

The vertices y^2=-6x is (h,k)=(0,0),p=-3/2

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vertices y^2=-6x

vertices

y^{2} = - 6 x

(h, k) = (0, 0), p = - \frac{3}{2}

Solution steps

y^{2} = - 6 x

Rewrite

y^{2} = - 6 x

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, 0), p = - \frac{3}{2}

4 x - 2 x^{2}

y = \frac{95 x ^{2} - 10 x + 26}{10}

(x - 1)^{2} = - 2 y - 11

f (x) = x^{2} - 6 x + 10

- x^{2} + 4 x