What is the axis y=(x+1)^2+1 ?

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axis y=(x+1)^2+1

axis

y = (x + 1)^{2} + 1

x = - 1

Solution steps

y = (x + 1)^{2} + 1

Rewrite

y = (x + 1)^{2} + 1

in the standard form:

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

Therefore parabola properties are:

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

Parabola is of the form

4 p (y - k) = (x - h)^{2}

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = - 1

y = 3 x^{2} - 5 x + 1

y = (x - 3)^{2}

y = 84 x - x^{2}

(y + 6)^{2} = 12 (x + 1)

y^{2} - 4 x = 0