Answer
$f(x)$ has the minimum value $f(x)= 49/10$
Work Step by Step
We are given the points $(2,1)$ and $(x,3x+2)$. Recall the distance formula between two points:
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
We are told that the function is the square of the distance, so we square the formula above and plug in the given points to obtain:
\begin{align*}
f(x)&=(x-2)^{2}+(3 x+2-1)^{2}\\
&=x^{2}-4 x+4+9 x^{2}+6 x+1\\
&=10 x^{2}+2 x+5\\
&=10\left(x^{2}+\frac{1}{5} x+\frac{1}{100}\right)+5-\frac{1}{10}\\
&=10\left(x+\frac{1}{10}\right)^{2}+\frac{49}{10}
\end{align*}
Compare with the quadratic function
$$f(x)=a(x-h)^{2}+k$$
Then the vertex is: $(h,k)= ( 1/10, 49/10)$
Since $a>0 $, then $f(x)$ has a minimum value of $f(x)= 49/10$
