To find the vertex and axis of simmetry we need to complete the squared on the function:
[tex]\begin{gathered} y=2x^2-8x+6 \\ y=2(x^2-4x)+6 \\ y=2(x^2-4x+(-2)^2)+6-2(-2)^2 \\ y=2(x-2)^2+6-8 \\ y=2(x-2)^2-2 \end{gathered}[/tex]Now the function is written in the form:
[tex]y=a(x-h)^2+k[/tex]in this form the vertex is (h,k). Comparing the equations we conclude that the vertex is at the point (2,-2).
Now, the axis of symmetry on a vertical parabola has the form:
[tex]x=k[/tex]Therefore, the axis of symmetry is:
[tex]x=2[/tex]To graph the function we need to find points on it:
Now we plot this points on the plane and join them with a smooth line:
From the graph we notice that the domain is:
[tex]D=(-\infty,\infty)[/tex]and the range is:
[tex]R=\lbrack-2,\infty)[/tex]
