91Ó°ÊÓ

The given function is:

$$\begin{gathered} f\left(x\right)=2x^2-4x+1 \\ f\left(x\right)=2(x^2-2x)+1 \end{gathered}$$

Add and subtract the square of half of the coefficient of x inside the parenthesis.

$$\begin{gathered} f\left(x\right)=2(x^2-2x+1-1)+1 \\ f\left(x\right)=2(x^2-2x+1)-2+1 \\ f\left(x\right)=2(x-1{)}^2-1 \end{gathered}$$

In the function $f\left(x\right)=a(x-h{)}^2+k$, a is a constant and $(h,k)$is the vertex.

In the given function $a=2,h=1,k=-1$. It means the graph of the given function is a parabola that opens up and has its vertex at $(1,-1)$and its axis of symmetry is the line $x=1$.

The graph of $y=x^2$vertically stretched by factor 2, shifts 1 units right and 1 units down.

First, plot the graph of $y=x^2$then multiply the values by 2 to get the graph of $y=2x^2$, then shift it 1 unit right to get the graph of the function localid="1646042810491" $y=2(x-1{)}^2$. After that shift the resulting graph 1 unit down to get the graph of the function $f\left(x\right)=2(x-1{)}^2-1$.

The required graph is shown below:

Using a graphing utility, we get a graph that is the same as the above graph.