What is the axis x=(y-7)^2+1 ?

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axis x=(y-7)^2+1

axis

x = (y - 7)^{2} + 1

y = 7

Solution steps

x = (y - 7)^{2} + 1

Rewrite

x = (y - 7)^{2} + 1

in the standard form:

4 \cdot \frac{1}{4} (x - 1) = (y - 7)^{2}

Therefore parabola properties are:

(h, k) = (1, 7), p = \frac{1}{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 = 7

y = x^{2} + 8

x^{2} + 10 x - 2 y + 21 = 0

y = - 2 (x - 4)^{2} + 5

x = (y + 1) (y + 2)

4 x^{2} - 128 y = 0