Question 182408
The slope of the tangent line is the value of the derivative of the function at that point. 
First, find the derivative of the function.
{{{y=2x-1/x=2x-x^(-1)}}}
{{{dy/dx=2-(-1)x^(-2)=2+1/x^2}}}
At x=0.5,
{{{(dy/dx)=2+1/(0.5)^2=6}}}
Now you have the slope of the line, use the point-slope form of the line,
{{{y-y[p]=m(x-x[p])}}}
{{{y-(-1)=6(x-0.5)}}}
{{{y+1=6x-3}}}
{{{y=6x-4}}}
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{{{drawing(300,300,-3,3,-3,3,grid(1),circle(0.5,-1,.1),graph( 300, 300, -3, 3, -3, 3, 6x-4, 2x-1/x)) }}}