What is the axis y=+x^2+14x+45 ?

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axis y=+x^2+14x+45

axis

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

x = - 7

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (y - (- 4)) = (x - (- 7))^{2}

Therefore parabola properties are:

(h, k) = (- 7, - 4), 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 = - 7

y = 1.4 x^{2} + 2 x + 1

y = 2 x^{2} - 12 x - 5

y = 18 x^{2} - 6 x + 4

y = x (x + 3)

x^{2} + 16 y = 0