What is the vertices y^2=x+1 ?

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

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vertices y^2=x+1

vertices

y^{2} = x + 1

(h, k) = (- 1, 0), p = \frac{1}{4}

Solution steps

y^{2} = x + 1

Rewrite

y^{2} = x + 1

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 1, 0), p = \frac{1}{4}

f (x) = - x^{2} + 8 x - 12

x^{2} - 6 x + 8

y = - 4 x^{2} - 24 x - 40

f (x) = x^{2} + 6 x + 9

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