What is the vertices y^2=6x-32 ?

The vertices y^2=6x-32 is (h,k)=(16/3 ,0),p= 3/2

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vertices y^2=6x-32

vertices

y^{2} = 6 x - 32

(h, k) = (\frac{16}{3}, 0), p = \frac{3}{2}

Solution steps

y^{2} = 6 x - 32

Rewrite

y^{2} = 6 x - 32

in the standard form:

4 \cdot \frac{3}{2} (x - \frac{16}{3}) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (\frac{16}{3}, 0), p = \frac{3}{2}

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

2 x^{2} - 4 x

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

y = - 2 x^{2} - 8 x + 3

x^{2} - 2 x + 1

What is the vertices y^2=6x-32 ?
The vertices y^2=6x-32 is (h,k)=(16/3 ,0),p= 3/2