What is the axis y^2-12x+20=0 ?

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axis y^2-12x+20=0

axis

y^{2} - 12 x + 20 = 0

y = 0

Solution steps

y^{2} - 12 x + 20 = 0

Rewrite

y^{2} - 12 x + 20 = 0

in the standard form:

4 \cdot 3 (x - \frac{5}{3}) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{3}, 0), p = 3

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 = - 3 x^{2} - 2 x

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

(y + 9)^{2} = - 44 (x - 1)

y = 5 x^{2} + 4

y = 5 x^{2} + 3