What is the axis y=x^2+12x+45 ?

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axis y=x^2+12x+45

axis

y = x^{2} + 12 x + 45

x = - 6

Solution steps

y = x^{2} + 12 x + 45

Rewrite

y = x^{2} + 12 x + 45

in the standard form

4 \cdot \frac{1}{4} (y - 9) = (x - (- 6))^{2}

Therefore parabola properties are:

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

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

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

y = (8 - 2 x)^{2}

y = 7 - 6 x^{2}

x^{2} - y = 4