What is the axis x^2-8x+14 ?

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axis x^2-8x+14

axis

x^{2} - 8 x + 14

x = 4

Solution steps

y = x^{2} - 8 x + 14

Rewrite

y = x^{2} - 8 x + 14

in the standard form

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

Therefore parabola properties are:

(h, k) = (4, - 2), 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 = 4

f (x) = x^{2} - 18 x + 80

y = lo g_{5} (2) x^{2}

y = 25 x^{2} - 100

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

y = 3 x^{2} - 8 x - 3