What is the axis y=+(x+4)^2-6 ?

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axis y=+(x+4)^2-6

axis

y = + (x + 4)^{2} - 6

x = - 4

Solution steps

y = (x + 4)^{2} - 6

Rewrite

y = (x + 4)^{2} - 6

in the standard form:

4 \cdot \frac{1}{4} (y - (- 6)) = (x - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (- 4, - 6), 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 = - 4

y = (x + 8)^{2}

(x + 3)^{2} = 24 (y - 1)

y = (z - 2)^{2}

3 x^{2} - 9 x - 2 = 5 y

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