What is the axis y=x^2-4x+12 ?

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axis y=x^2-4x+12

axis

y = x^{2} - 4 x + 12

x = 2

Solution steps

y = x^{2} - 4 x + 12

Rewrite

y = x^{2} - 4 x + 12

in the standard form

4 \cdot \frac{1}{4} (y - 8) = (x - 2)^{2}

Therefore parabola properties are:

(h, k) = (2, 8), 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 = 2

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

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

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

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

y = (x + 12) (x - 12)