# How do you find the equation of a line tangent to the function #y=1/(2-x)# at x=-1?

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To find the equation of a line tangent to the function y=1/(2-x) at x=-1, we need to find the slope of the tangent line at that point and then use the point-slope form of a line to write the equation.

To find the slope, we can take the derivative of the function y=1/(2-x) with respect to x.

The derivative of y=1/(2-x) is dy/dx = 1/(2-x)^2.

Substituting x=-1 into the derivative, we get dy/dx = 1/(2-(-1))^2 = 1/9.

So, the slope of the tangent line at x=-1 is 1/9.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute x1=-1, y1=1/(2-(-1)) = 1/3, and m=1/9 into the equation.

Therefore, the equation of the line tangent to the function y=1/(2-x) at x=-1 is y - 1/3 = (1/9)(x - (-1)).

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