What is the axis y=-2x^2+12x-14 ?

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axis y=-2x^2+12x-14

axis

y = - 2 x^{2} + 12 x - 14

x = 3

Solution steps

y = - 2 x^{2} + 12 x - 14

Rewrite

y = - 2 x^{2} + 12 x - 14

in the standard form

4 (- \frac{1}{8}) (y - 4) = (x - 3)^{2}

Therefore parabola properties are:

(h, k) = (3, 4), 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 = 3

y = x^{2} - 5 x + 3

y = - 6 (x - 5)^{2} + 12

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

- x^{2} - (- y) = 4

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