What is the axis x^2+2x-8y+33=0 ?

The axis x^2+2x-8y+33=0 is x=-1

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axis x^2+2x-8y+33=0

axis

x^{2} + 2 x - 8 y + 33 = 0

x = - 1

Solution steps

x^{2} + 2 x - 8 y + 33 = 0

Rewrite

x^{2} + 2 x - 8 y + 33 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 1, 4), p = 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 = - 1

y = (8 - 2 x)^{2}

y = 7 - 6 x^{2}

x^{2} - y = 4

y = - x^{2} + 18 x - 79

y = 12 x^{2} + 8 x - 15