Change this equation to vertex form and find the vertex and axis of symmetry:

y=x^2-8x+2

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We have the equation of the parabola as y = x^2 - 8x + 2

Now let's write it in the form y = (x - h)^2 + k

y = x^2 - 8x + 2

=> y = x^2 - 8x + 16 - 14

=> y = (x - 4)^2 - 14

The equation of a parabola with vertex (h, k) is y = (x - h)^2 + k

Therefore for y = (x - 4)^2 - 14

**the vertex is ( 4, -14)**

**The axis of symmetry is x = 4.**

The vertex form of parabola is y = a(x-h)^2+c, where (h, k) are the coordinates of the vertex. (1/a)/4 = 1/4a is focal distance from the vertex (h,k), and x = h is the axis of of symmetry

The given parabola is y = x^2-8x+2. To convert this to vertex form we have to complete the x^2-8x into a perfect square by adding 4^2 so that x^2-8x+4^2 = (x-4)^2.

Therefore we add and subtract 4^2:

y = (x^2-8x+4^2) - 4^2+2.

y = 1(x-8)^2 -18 is in the required form.

The coordinates of the vertex = (8, -18)

The focal length = 1/4*1 = 1/4 . So the ocus is 1/4 units above the vertex (8,-18).

The axis of symmetry is x= 8, a || line to y axis.

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