What is the axis y^2=16-4x ?

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axis y^2=16-4x

axis

y^{2} = 16 - 4 x

y = 0

Solution steps

y^{2} = 16 - 4 x

Rewrite

y^{2} = 16 - 4 x

in the standard form:

4 (- 1) (x - 4) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (4, 0), 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 = 0

y = + (x + 5)^{2} + 3

y^{2} - 8 x - 6 y - 3 = 0

y = 6 x^{2} + 10 x - 4

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

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