What is the axis y=(x+4)^2-9 ?

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axis y=(x+4)^2-9

axis

y = (x + 4)^{2} - 9

x = - 4

Solution steps

y = (x + 4)^{2} - 9

Rewrite

y = (x + 4)^{2} - 9

in the standard form:

4 \cdot \frac{1}{4} (y - (- 9)) = (x - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (- 4, - 9), 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

x = - \frac{( y + 3 ) ^{2}}{20} + 1

y = (x + 8)^{2} + 4 (x + 9) + 1

y = x^{2} - 11 x + 10

y^{2} + x - 10 y + 29 = 0

2 x^{2} + 8 x - 4 y + 12 = 0