What is the axis y=-1/2 x^2-4x+10 ?

The axis y=-1/2 x^2-4x+10 is x=-4

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axis y=-1/2 x^2-4x+10

axis

y = - \frac{1}{2} x^{2} - 4 x + 10

x = - 4

Solution steps

y = - \frac{1}{2} x^{2} - 4 x + 10

Rewrite

y = - \frac{1}{2} x^{2} - 4 x + 10

in the standard form

4 (- \frac{1}{2}) (y - 18) = (x - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (- 4, 18), p = - \frac{1}{2}

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 = x^{2} - x - 20

x^{2} + y = 5

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

y = 1 - x^{2}

y = 2 x^{2} - 1 - (5 x - 6)