What is the axis y=-1/4 x^2+x+6 ?

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axis y=-1/4 x^2+x+6

axis

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

x = 2

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, 7), p = - 1

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 = 2

+ y = x^{2} - 6 x + 8

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

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

33 x^{2} - 3 {x - 2 x [x - 7 (x - 2) - 3] + 2} - 60 x

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