91Ó°ÊÓ

The given function is:

$$\begin{gathered} f\left(x\right)=-2x^2+6x+2 \\ f\left(x\right)=-2(x^2-3x)+2 \end{gathered}$$

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

$$\begin{gathered} f\left(x\right)=-2(x^2-3x+(1.5{)}^2-(1.5{)}^2)+2 \\ f\left(x\right)=-2(x^2-3x+(1.5{)}^2)+4.5+2 \\ f(x)=-2(x-1.5{)}^2+6.5 \end{gathered}$$

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 $a=-2,h=1.5,k=6.5$. It means the graph of the given function is a parabola that opens down and has its vertex at $(1.5,6.5)$and its axis of symmetry is the line $x=1.5$.

The graph of $y=x^2$reflects along the x-axis and vertically stretched by factor 2, shifts 1.5 units right and 6.5 units up.

First, plot the graph of $y=x^2$then reflect it along the x-axis vertically stretched by factor 2 to get the graph of $y=-2x^2$, then shift it 1.5 units right to get the graph of the function $y=-2(x-1.5{)}^2$. After that shift the resulting graph 6.5 units up to get the graph of the function $f\left(x\right)=-2(x-1.5{)}^2+6.5$.

The required graph is shown below:

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