Answer
The equation of the given graph is \[j\left( x \right)=-{{x}^{2}}-1\]
Work Step by Step
A quadratic function can be expressed as $f\left( x \right)=a{{x}^{2}}+bx+c$ corresponding to which the graph is a parabola whose vertex is the point $\left( h,k \right)=\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$. If $a>0$, it opens upward and if $a<0$ then it opens downward.
Now, it can be observed that the given parabola opens downward so $a<0$. So, the vertex of the parabola in the graph is on $\left( 0,-1 \right)$. Thus, in comparison with the standard equation, the value of $\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$ is:
$\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)=\left( 0,-1 \right)$
This implies, $b=0$ and $f\left( 0 \right)=-1$.
Putting in the values in the standard form we get:
$\begin{align}
& f\left( 0 \right)=a{{\left( 0 \right)}^{2}}+b\left( 0 \right)+c \\
& -1=c
\end{align}$
Substituting the values of b and c obtained in the standard equation, we obtain:
$\begin{align}
& f\left( x \right)=a{{x}^{2}}+0x-1 \\
& =a{{x}^{2}}-1
\end{align}$
Where, $a<0$, the parabola opens downwards. This equation is similar to $j\left( x \right)=-{{x}^{2}}-1$
Therefore, the required equation of the graph is $j\left( x \right)=-{{x}^{2}}-1$.
