What is the axis y=x^2+12x+43 ?

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axis y=x^2+12x+43

axis

y = x^{2} + 12 x + 43

x = - 6

Solution steps

y = x^{2} + 12 x + 43

Rewrite

y = x^{2} + 12 x + 43

in the standard form

4 \cdot \frac{1}{4} (y - 7) = (x - (- 6))^{2}

Therefore parabola properties are:

(h, k) = (- 6, 7), 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

y = (0.5 - x)^{2}

y = - x^{2} - 3 x + 4

y = (x - 2 + 5) (x - 2 - 5)

x^{2} + 2 x + 3 y + 22 = 0

y = - \frac{1}{4} x^{2} + x + 3