What is the axis y=(x+8)^2+4(x+9)+1 ?

The axis y=(x+8)^2+4(x+9)+1 is x=-10

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

axis

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

x = - 10

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (y - 1) = (x - (- 10))^{2}

Therefore parabola properties are:

(h, k) = (- 10, 1), 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 = - 10

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

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

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

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

y^{2} + 8 x + 2 y + 33 = 0