Question 1019504
<pre>
We draw the graph:

{{{drawing(400,800/3,-7,8,-2,8,
graph(400,800/3,-7,8,-2,8), green(line(1,-3,1,9)),


circle(5,3,0.15),circle(5,3,0.13),circle(5,3,0.11),circle(5,3,0.09),circle(5,3,0.07),circle(5,3,0.05),circle(5,3,0.03),circle(5,3,0.01),
locate(5.4,3.2,"U(5,3)") )}}}

The green line is the line x = 1.  If we pretend that the
green line is a mirror, where would it seem as though its
reflection U' in the mirror would be?

U' would be the same distance from the left side of the
green line as U(5,3) is from the right side on the green 
line.  Using the x-axis as a measuring stick we can count 
that there are 4 units between U and the green line, 
indicated by the dotted line:

{{{drawing(400,800/3,-7,8,-2,8,
graph(400,800/3,-7,8,-2,8,

3*(sqrt(sin(13x))/sqrt(sin(13x)))*(sqrt(x-1)/sqrt(x-1))*(sqrt(5-x)/sqrt(5-x))


), green(line(1,-3,1,9)),


circle(5,3,0.15),circle(5,3,0.13),circle(5,3,0.11),circle(5,3,0.09),circle(5,3,0.07),circle(5,3,0.05),circle(5,3,0.03),circle(5,3,0.01),
locate(5.4,3.2,"U(5,3)") )}}}

Therefore, to locate the image point U', we count 4
units from the green line on the left side, and we 
see that the image point U' ends up at U'(-3,3), 
like this:

{{{drawing(400,800/3,-7,8,-2,8,
graph(400,800/3,-7,8,-2,8,

3*(sqrt(sin(13x))/sqrt(sin(13x)))*(sqrt(x+3)/sqrt(x+3))*(sqrt(5-x)/sqrt(5-x))


), green(line(1,-3,1,9)),


circle(5,3,0.15),circle(5,3,0.13),circle(5,3,0.11),circle(5,3,0.09),circle(5,3,0.07),circle(5,3,0.05),circle(5,3,0.03),circle(5,3,0.01),
locate(5.4,3.2,"U(5,3)"),locate(-5.7,3.2,"U'(-3,3)"),




circle(-3,3,0.15),circle(-3,3,0.13),circle(-3,3,0.11),circle(-3,3,0.09),circle(-3,3,0.07),circle(-3,3,0.05),circle(-3,3,0.03),circle(-3,3,0.01) )}}}

Edwin</pre>