What is the axis x=2(y-5)^2+3 ?

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axis x=2(y-5)^2+3

axis

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

y = 5

Solution steps

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

Rewrite

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

in the standard form:

4 \cdot \frac{1}{8} (x - 3) = (y - 5)^{2}

Therefore parabola properties are:

(h, k) = (3, 5), p = \frac{1}{8}

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 = 5

x^{2} + 10 x = - 12 y - 1

2 y^{2} - 8 x - 24 y + 56 = 0

y = \frac{2 x ^{2} + 7 x + 12}{2}

- x^{2} + 4 x - 11

y = 5 x^{2} - 9