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Popular >

vertices y=s^2+2s+2

Solution

vertices

y = s^{2} + 2 s + 2

Solution

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

Solution steps

y = s^{2} + 2 s + 2

Rewrite

y = s^{2} + 2 s + 2

in the standard form

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

Therefore parabola properties are:

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

Popular Examples

vertices f(x)=3x^2+2x-1 vertices

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

vertices x^2-5y=0 vertices

x^{2} - 5 y = 0

vertices y=2x^2-8x vertices

y = 2 x^{2} - 8 x

vertices f(x)=-4x^2+2x vertices

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

vertices y=2x^2-6x vertices

y = 2 x^{2} - 6 x

Frequently Asked Questions (FAQ)

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