What is the axis y=x^2-4x+7 ?

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axis y=x^2-4x+7

axis

y = x^{2} - 4 x + 7

x = 2

Solution steps

y = x^{2} - 4 x + 7

Rewrite

y = x^{2} - 4 x + 7

in the standard form

4 \cdot \frac{1}{4} (y - 3) = (x - 2)^{2}

Therefore parabola properties are:

(h, k) = (2, 3), 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

y = 2 x^{2} - 6 x + 4

y = (x - 5)^{2} - 1

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

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

y = - 3 x^{2} - 2