we have the equation
![f\mleft(x\mright)=2x^2−4x+2](https://img.qammunity.org/2023/formulas/mathematics/college/8y0rsj974a60z9c7luvfcx72ht968jjray.png)
The domain of any quadratic equation is all real numbers
so
The domain is the interval (-infinite, infinite)
To find out the range, we need the vertex
Convert the given equation into vertex form
![\begin{gathered} f\mleft(x\mright)=2x^2−4x+2 \\ f(x)=2(x^2-2x)+2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iofuroq0cvf2e9kw03niq9f4ah8f91d6y2.png)
Complete the square
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/vjembll3h484c6x0epc1k22ap97j56l571.png)
The vertex is the point (1,0)
The vertical parabola opens upward (the leading coefficient is positive)
The vertex is a minimum
therefore
The range is the interval [0, infinite)
All real numbers greater than or equal to zero