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

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axis y=2(x+1)^2

axis

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

x = - 1

Solution steps

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

Rewrite

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

in the standard form:

4 \cdot \frac{1}{8} (y - 0) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, 0), p = \frac{1}{8}

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 = a x^{2}

y = + x^{2} + 3 x - 4

y = x^{2} + 8 x + 15

y = (2 x - 1) (5 x - 6)

y = 1 - x - x^{2}