What is the axis y=-(x-6)^2+1 ?

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axis y=-(x-6)^2+1

axis

y = - (x - 6)^{2} + 1

x = 6

Solution steps

y = - (x - 6)^{2} + 1

Rewrite

y = - (x - 6)^{2} + 1

in the standard form:

4 (- \frac{1}{4}) (y - 1) = (x - 6)^{2}

Therefore parabola properties are:

(h, k) = (6, 1), 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 = 6

y = \frac{3}{2} x^{2}

y = \frac{1}{5} x^{2} - \frac{8}{5} x + \frac{1}{5}

4 x^{2} - \frac{1}{2} x^{2} + 8 x - 2 y = - 2

y = x^{2} - 8 x + 16

y = 4 x^{2} + 3 x - 1