What is the axis y=-2x^2+16x-31 ?

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axis y=-2x^2+16x-31

axis

y = - 2 x^{2} + 16 x - 31

x = 4

Solution steps

y = - 2 x^{2} + 16 x - 31

Rewrite

y = - 2 x^{2} + 16 x - 31

in the standard form

4 (- \frac{1}{8}) (y - 1) = (x - 4)^{2}

Therefore parabola properties are:

(h, k) = (4, 1), p = - \frac{1}{8}

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 = 4

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

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

x = \frac{y ^{2}}{8}

y = 18 x^{2} - 14 x + 9

y = - (- 12 x + 13)^{2} - 1