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

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

axis

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

x = - 1

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 1, - 2), 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 = 3 x^{2} (1.3)

(y - 4)^{2} = - 16 (x - 2)

36 y^{2} + 72 x - 132 y - 239 = 0

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

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