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A line is drawn from the origin O to a point P(x,y) in the first quadrant on the graph of y=1/x. From point P, a line is drawn perpendicular to the x-axis, meeting the x-axis at B. Express the perimeter of OPB as a function of x.

I need help setting up the equation for this. Do I just need to determine the equations for each side of the triangle (a, b, c) using the points given. So it would end up looking like x + y + hypotenuse. Then wherever I see a y, I replace it with (1/x) giving me x + (1/x) + x^2 + (1/x)^2.

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up vote 4 down vote accepted

You are essentially right. The two "legs" have length $x$ and $1/x$. The hypotenuse is $\sqrt{x^2+1/x^2}$. You just left out the square root symbol. So the perimeter is $$x+\frac{1}{x}+\sqrt{x^2+\frac{1}{x^2}}.$$

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