What is the axis y=(x-7)^2+10 ?

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axis y=(x-7)^2+10

axis

y = (x - 7)^{2} + 10

x = 7

Solution steps

y = (x - 7)^{2} + 10

Rewrite

y = (x - 7)^{2} + 10

in the standard form:

4 \cdot \frac{1}{4} (y - 10) = (x - 7)^{2}

Therefore parabola properties are:

(h, k) = (7, 10), p = \frac{1}{4}

Parabola is of the form

4 p (y - k) = (x - h)^{2}

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = 7

6 x^{2} - 8

y = x^{2} - 6 x - 12

(x + 9)^{2} + 2 (y - 4) = 0

x^{2} = 36 y

y = - 0.0000848 x^{2} - 0.006005 x + 1.004