How do you find the quadratic function with vertex $\left( 3,4 \right)$ and point (1, 2)?

Hint: The general form of the quadratic function with vertex (h, k) and “a” as the constant multiplier is as follows: $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$. Then we are going to substitute the values of vertex (h, k) given in the above problem i.e. (3, 4) in this function in x. And then we will substitute the point (1, 2) in place of x and f(x) in this function. And hence, we will get the required quadratic function.

Complete step by step answer:
In the above problem, we are asked to find the quadratic function having vertex (3, 4) and the point (1, 2).
We know that the general form of a quadratic function is:
$f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ …………… Eq. (1)
In the above formula, (h, k) is the vertex of the quadratic function and the “a” is some constant. Now, substituting 3 in place of h and 4 in place of k in the above equation we get,
$f\left( x \right)=a{{\left( x-3 \right)}^{2}}+4$ …………… Eq. (2)
Substituting the point (1, 2) in the above equation by replacing x by 1 and f(x) by 2 we get,
$\begin{align}
  & 2=a{{\left( 1-3 \right)}^{2}}+4 \\
 & \Rightarrow 2=a{{\left( -2 \right)}^{2}}+4 \\
 & \Rightarrow 2=4a+4 \\
\end{align}$
Subtracting 4 on both the sides we get,
$\begin{align}
  & 2-4=4a+4-4 \\
 & \Rightarrow -2=4a \\
\end{align}$
Dividing 4 on both the sides we get,
$-\dfrac{2}{4}=a$
Simplifying the L.H.S of the above equation we get,
$-\dfrac{1}{2}=a$
Now, substituting the above value of $''a''$ in eq. (2) we get,
$f\left( x \right)=-\dfrac{1}{2}{{\left( x-3 \right)}^{2}}+4$
The above equation is the quadratic function in vertex form.

Hence, we have found the quadratic function with vertex (3, 4) and point (1, 2) as $f\left( x \right)=-\dfrac{1}{2}{{\left( x-3 \right)}^{2}}+4$.

Note: The quadratic function which we have solved above is in the vertex form $f\left( x \right)=-\dfrac{1}{2}{{\left( x-3 \right)}^{2}}+4$. The other way of writing this quadratic function form is by simplifying the R.H.S of this quadratic function which is done by expanding ${{\left( x-3 \right)}^{2}}$ so expanding this quadratic expression in f(x) we get,
$\begin{align}
  & f\left( x \right)=-\dfrac{1}{2}\left( {{x}^{2}}+9-6x \right)+4 \\
 & \Rightarrow f\left( x \right)=-\dfrac{1}{2}\left( {{x}^{2}}-6x \right)+4-\dfrac{9}{2} \\
 & \Rightarrow f\left( x \right)=-\dfrac{1}{2}\left( {{x}^{2}}-6x \right)+\dfrac{8-9}{2} \\
 & \Rightarrow f\left( x \right)=-\dfrac{1}{2}\left( {{x}^{2}}-6x \right)-\dfrac{1}{2} \\
\end{align}$
From the above function, we can take $-\dfrac{1}{2}$ as common then the above function will look like:
$f\left( x \right)=-\dfrac{1}{2}\left( {{x}^{2}}-6x+1 \right)$