Change this equation to vertex form and find the vertex and axis of symmetry: y=x^2-8x+2
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.