Question 593813: how do I find the equation of the line with a slope -1 that is tangent to the curve y=1/x-1?
I don't think there is a solution to this since there are two curves and the y-axis is a vertical asymptote.
Any help is greatly appreciated.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! how do I find the equation of the line with a slope -1 that is tangent to the curve y=1/x-1?
Find the derivative of:
y=1/x-1
y=x^(-1)-1
y' = -x^(-2)
y' = -1/x^2
.
Set y' to -1 and solve for x:
y' = -1/x^2
-1 = -1/x^2
x^2 = 1
x = {-1, +1}
.
Using x=1, find y:
y=1/x-1
y=1/1-1
y=1-1
y=0
so, the point at which the slope is -1 is at (1,0)
.
equation of line, plug into the "point-slope" form:
y - y1 = m(x - x1)
y - 0 = -1(x - 1)
y = -x + 1 (this is what they're looking for)
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