What is the axis y=x^2+14x+48 ?

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axis y=x^2+14x+48

axis

y = x^{2} + 14 x + 48

x = - 7

Solution steps

y = x^{2} + 14 x + 48

Rewrite

y = x^{2} + 14 x + 48

in the standard form

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

Therefore parabola properties are:

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

y = x^{2} - 3 x^{1}

(y - 7)^{2} = - 12 (x + 1)

x = (y + 5)^{2}

y = 4 x^{2} + 6 x + 1

y = 0.3 x^{2} - 66 x + 23114