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let f(x)=(3x)/(x-q) ...

q cannot be equal to x.
a) find the equations of the vertical and horizontal asymptotes of the graph of f.
the vertical and horizontal asymptotes to the graph of f intersect at the point Q(1,3)
b) find the value of q
c) the point P(x,y) lies on the graph of f. show that PQ = sqrt( ((x-1)^2) + ( (3)/(x-1) )^2 )
d) Find the coordinates of the points on the graph of f that are closest to (1,3)
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1 Answer

a)  The vertical asymptote is the value of x such that x-q = 0(i.e. when the f(x)-> ∞.
     Therefore, x = q.
     The horizontal asymptote is the value of f(x) when x-> ∞.
      Therefore, y = 3 (the ratio of the coefficients of x in f(x))
b)  Since the these two lines intersect at Q(1, 3), the value of q = 1.  (the value of x is the vertical asymptote and the value of y is the horizontal asymptote.
c)  PQ = √((x-1)2 + (f(x)-3)).     Replace f(x) by (3x)/(x-1) and simplify.
d)  Minimize  (x-1)2 + ((3x)/(x-1) - 3)).
I hope this helps.