What is the axis y=4x^2+8x+13 ?

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axis y=4x^2+8x+13

axis

y = 4 x^{2} + 8 x + 13

x = - 1

Solution steps

y = 4 x^{2} + 8 x + 13

Rewrite

y = 4 x^{2} + 8 x + 13

in the standard form

4 \cdot \frac{1}{16} (y - 9) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, 9), p = \frac{1}{16}

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 = - 1

16 x^{2} - 72 x - 112 y = - 221

x = (4 y + 3) (2 y - 5)

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

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

x = (y - 2)^{2} + 3