What is the axis y=x^2-10x+12 ?

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axis y=x^2-10x+12

axis

y = x^{2} - 10 x + 12

x = 5

Solution steps

y = x^{2} - 10 x + 12

Rewrite

y = x^{2} - 10 x + 12

in the standard form

4 \cdot \frac{1}{4} (y - (- 13)) = (x - 5)^{2}

Therefore parabola properties are:

(h, k) = (5, - 13), 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 = 5

y = 3 x^{2} + 6

y^{2} - 6 y - 4 x + 5 = 0

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

y = 7 x^{2} + 14 x

f (x) = x^{2} + 6 x + 8