we have the equation
[tex]f\mleft(x\mright)=2x^2−4x+2[/tex]The domain of any quadratic equation is all real numbers
so
To find out the range, we need the vertex
Convert the given equation into vertex form
[tex]\begin{gathered} f\mleft(x\mright)=2x^2−4x+2 \\ f(x)=2(x^2-2x)+2 \end{gathered}[/tex]Complete the square
[tex]\begin{gathered} f(x)=2(x^2-2x+1-1)+2 \\ f(x)=2(x^2-2x+1)+2-2 \\ f(x)=2(x^2-2x+1) \\ f(x)=2(x-1)^2 \end{gathered}[/tex]The vertex is the point (1,0)
The vertical parabola opens upward (the leading coefficient is positive)
The vertex is a minimum
therefore
All real numbers greater than or equal to zero
