Answer
The required parabola is as shown below.
Work Step by Step
So for the provided quadratic equation $f\left( x \right)={{x}^{2}}+4x-1$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , one gets:
$\begin{align}
& a=1, \\
& b=4, \\
& c=-1
\end{align}$
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 upwards and if $a<0$ then the parabola opens downwards. 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 upwards as $a>0$.
Step 2: Calculate the vertex:
The x-coordinate can be calculated as:
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{4}{2\times 1} \\
& =-2
\end{align}$
Or, the y-coordinate can be calculated as:
$\begin{align}
& y={{\left( -2 \right)}^{2}}+4\left( -2 \right)-1 \\
& =4-8-1 \\
& =-5
\end{align}$
Therefore, the vertex is $\left( -2,-5 \right)$.
Step 3:
So, the above steps lead to the parabola that opens upwards and has a vertex at $\left( -2,-5 \right)$ and it intersects x-axis at $\left( -2+\sqrt{5} \right)$ and $\left( -2-\sqrt{5} \right)$ , and y-axis at -1.
Thus, the required parabola is as shown above.
