Answer
The required parabola is as shown below.
Work Step by Step
So for the provided quadratic equation $f\left( x \right)=5-4x-{{x}^{2}}$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ we get:
$\begin{align}
& a=-1, \\
& b=-4, \\
& c=5
\end{align}$
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:
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\left( \frac{-4}{2\times \left( -1 \right)} \right) \\
& =-2
\end{align}$
Or the y-coordinate can be calculated as:
$\begin{align}
& y=5-4\left( -2 \right)-{{\left( -2 \right)}^{2}} \\
& =5+8-4 \\
& =9
\end{align}$
Therefore, the vertex is $\left( -2,9 \right)$
Step 3:
And the above steps lead to the parabola that is open downwards and has a vertex at $\left( -2,9 \right)$ and it intersects x-axis at $1$ and $-5$ , and y-axis at 5.
So, the required parabola is as shown above.
