What is the axis y^2-16x=0 ?

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axis y^2-16x=0

axis

y^{2} - 16 x = 0

y = 0

Solution steps

y^{2} - 16 x = 0

Rewrite

y^{2} - 16 x = 0

in the standard form:

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

Therefore parabola properties are:

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

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 + 4)^{2} - 9

x = - \frac{( y + 3 ) ^{2}}{20} + 1

y = (x + 8)^{2} + 4 (x + 9) + 1

y = x^{2} - 11 x + 10

y^{2} + x - 10 y + 29 = 0