Question 844183
Some help on the analysis:


The denominator contains factors, (x-1) and (x-5).  These factors do not occur in the numerator.  Between x=1 and x=5, the denominator becomes negative, but to the left of x=1 and to the right of x=5, the denominator is positive.  You want a way for the rational expression to stay positive as described, "graph does not have an x-intercept" and contains "point (3,1)".  A way to do this is have both binomial factors in the denominator, squared.  This way, the rational expression is always positive; and the extreme left and right behavior will approach zero.


The function can be something, {{{k/((x-1)^2(x-5)^2)}}}.
You want to find k using the included point of (3,1).
At x=3, the expression must be 1.  Find k.