What is the axis y^2-8y+4x=0 ?

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axis y^2-8y+4x=0

axis

y^{2} - 8 y + 4 x = 0

y = 4

Solution steps

y^{2} - 8 y + 4 x = 0

Rewrite

y^{2} - 8 y + 4 x = 0

in the standard form

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

Therefore parabola properties are:

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

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = 4

y = 3 x^{2} + 2 x - 1

y = 2 x^{2} + 20 x + 53

y = x^{2} + 2 x + 7

y = 3 x^{2} - 27

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