What is the axis y=4(x+1)^2-3 ?

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axis y=4(x+1)^2-3

axis

y = 4 (x + 1)^{2} - 3

x = - 1

Solution steps

y = 4 (x + 1)^{2} - 3

Rewrite

y = 4 (x + 1)^{2} - 3

in the standard form:

4 \cdot \frac{1}{16} (y - (- 3)) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, - 3), p = \frac{1}{16}

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 = x^{2} + 8 x + 12

x^{2} + 8 y + 2 x - 23 = 0

y = x^{2} - 4 x + 5

3 y^{2} = 24 x

y = x^{2} - 14 x + 45