Answer
The function of the given graph is\[g\left( x \right)={{x}^{2}}-2x+1\]
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 has a vertex on $\left( 1,0 \right)$ and opens upwards. Thus, in comparison with the standard equation, the value of $\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$ is:
$\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)=\left( 1,0 \right)$
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 we get:
$\begin{align}
& f\left( 1 \right)=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\
& 0=a+b+c \\
& 0=a-2a+c \\
& a=c
\end{align}$
Substituting the values of b and c obtained in the standard equation, we obtain:
$f\left( x \right)=a{{x}^{2}}-2ax+a$
Where, $a>0$ the parabola opens upwards. The coefficients of ${{x}^{2}}$ and constant $c$ are equal and have opposite in sign to x. This equation is similar to $g\left( x \right)={{x}^{2}}-2x+1$.
Therefore, the required equation of the graph is $g\left( x \right)={{x}^{2}}-2x+1$.
