Answer
The equation of the given graph is\[f\left( x \right)={{x}^{2}}+2x+1\]
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 $a>0$ , the parabola opens upward and if $a<0$ then the parabola opens downward.
Now, it can be observed that the given parabola opens upward so $a>0$. So, the vertex of the parabola in the graph is on $\left( -1,0 \right)$. Hence, in comparison with the standard equation, the value of $\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$ is:
$\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)=\left( -1,0 \right)$
This implies,
$\begin{align}
& \frac{-b}{2a}=-1 \\
& b=2a
\end{align}$
and,
$f\left( -1 \right)=0$
Putting in the values in the standard form to we get:
$\begin{align}
& f\left( -1 \right)=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\
& 0=a-b+c \\
& b=a+c
\end{align}$
Putting in the values in the standard form we get:
$f\left( x \right)=a{{x}^{2}}+\left( a+c \right)x+c$
Where, $a>0$, the parabola opens upwards. The coefficient of $x$ is the sum of the coefficient of ${{x}^{2}}$ and has the same sign of x. This equation is similar to
$f\left( x \right)={{x}^{2}}+2x+1$.
Therefore, the required equation of the graph is $f\left( x \right)={{x}^{2}}+2x+1$.
