What is the axis x=((y-3)^2)/8+5 ?

The axis x=((y-3)^2)/8+5 is y=3

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axis x=((y-3)^2)/8+5

axis

x = \frac{( y - 3 ) ^{2}}{8} + 5

y = 3

Solution steps

x = \frac{( y - 3 ) ^{2}}{8} + 5

Rewrite

x = \frac{( y - 3 ) ^{2}}{8} + 5

in the standard form:

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

Therefore parabola properties are:

(h, k) = (5, 3), p = 2

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 = 3

y = x^{2} - 42 + 10 x

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

y = - 2 x^{2} + 12 x - 14

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

y = - 6 (x - 5)^{2} + 12