What is the axis y=+2(x+1)^2-3 ?

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axis y=+2(x+1)^2-3

axis

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

x = - 1

Solution steps

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

Rewrite

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

in the standard form:

4 \cdot \frac{1}{8} (y - (- 3)) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, - 3), 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 = \frac{x ^{2} - 4 x}{11}

\frac{a + 1}{b} = \frac{a - 1}{b} + \frac{b + 1}{a}

y^{2} - 8 y + 4 x = 0

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

y = 2 x^{2} + 20 x + 53