What is the vertices x=y^2+2y-1-(2y^2-3y+6) ?

The vertices x=y^2+2y-1-(2y^2-3y+6) is (h,k)=(-3/4 , 5/2),p=-1/4

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vertices x=y^2+2y-1-(2y^2-3y+6)

vertices

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

(h, k) = (- \frac{3}{4}, \frac{5}{2}), p = - \frac{1}{4}

Solution steps

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

Rewrite

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

in the standard form

4 (- \frac{1}{4}) (x - (- \frac{3}{4})) = (y - \frac{5}{2})^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{4}, \frac{5}{2}), p = - \frac{1}{4}

f (x) = x^{2} + 6 x - 8

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

f (x) = x^{2} + 6 x + 4

f (x) = 2 x^{2} - 10 x + 12

\frac{x ^{3} - 1}{x - 1}

What is the vertices x=y^2+2y-1-(2y^2-3y+6) ?
The vertices x=y^2+2y-1-(2y^2-3y+6) is (h,k)=(-3/4 , 5/2),p=-1/4