Question 353350
The slope of the tangent line is the value of the derivative of the function.
{{{y=(x-1)^(-1)}}}
{{{dy/dx=-(x-1)^(-2)=-1}}}
{{{(x-1)^2=1}}}
{{{x-1=0 +- 1}}}
{{{x=0}}} and {{{x=2}}}
There are two points where the slope is equal to {{{-1}}}.
Find the corresponding y values using the function.
When {{{x=0}}}, {{{y=1/(0-1)=-1}}}
When {{{x=2}}}, {{{y=1/(2-1)=1}}}
Then using the point slope form of a line with the slope and the point,{{{y-yp=m(x-xp)}}} for both points.
{{{y-(-1)=-1(x-0)}}}
{{{y+1=-x}}}
{{{highlight(y1=-x-1)}}}
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{{{y-1=-1(x-2))}}}
{{{y-1=-x+2}}}
{{{highlight_green(y2=-x+3)}}}
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{{{drawing(300,300,-6,6,-6,6,circle(0,-1,0.2),circle(2,1,0.2),grid(1),graph(300,300,-6,6,-6,6,1/(x-1),-x-1,-x+3))}}}