Answer
The required parabola is as shown below.
Work Step by Step
We know that for the provided quadratic equation $f\left( x \right)=2x-{{x}^{2}}+3$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ one gets:
$\begin{align}
& a=-1, \\
& b=2, \\
& c=3
\end{align}$
And use the steps shown below to determine the graph of the quadratic equation.
Step 1: Determine how the parabola opens:
Note that a, the coefficient of ${{x}^{2}}$ , is -1. If $a>0$ , the parabola opens in the upward direction and if $a<0$ then the parabola opens downward. Also, if $\left| a \right|$ is small, the parabola opens more flatly than if $\left| a \right|$ is large.
Now, from the provided equation of the function, it is observed that the graph opens downward as $a<0$.
Step 2: Evaluate the vertex
The x-coordinate can be calculated as:
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{2}{2\times \left( -1 \right)} \\
& =1
\end{align}$
And, the y-coordinate can be calculated as:
$\begin{align}
& y=2x-{{x}^{2}}+3 \\
& =2\left( 1 \right)-{{\left( 1 \right)}^{2}}+3 \\
& =2-1+3 \\
& =4
\end{align}$
Thus, the vertex is $\left( 1,4 \right)$
Step 3:
Above steps lead to the parabola that is open downwards and has a vertex at $\left( 1,4 \right)$ and it intersects x-axis at $3$ and $-1$ , and y-axis at 3.
So, the required parabola is as shown above.
