document.write( "Question 353350: Find the equation of the line with slope -1 that is the tangent to the curve y=1/(x-1)
\n" ); document.write( "(5marks: 1mark for graph, for the general equation of the line, for getting the quadratic, for solving for K, for the solution) \n" ); document.write( "

Algebra.Com's Answer #252664 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
The slope of the tangent line is the value of the derivative of the function.
\n" ); document.write( " $\"y=%28x-1%29%5E%28-1%29\"$
\n" ); document.write( " $\"dy%2Fdx=-%28x-1%29%5E%28-2%29=-1\"$
\n" ); document.write( " $\"%28x-1%29%5E2=1\"$
\n" ); document.write( " $\"x-1=0+%2B-+1\"$
\n" ); document.write( " $\"x=0\"$ and $\"x=2\"$
\n" ); document.write( "There are two points where the slope is equal to $\"-1\"$ .
\n" ); document.write( "Find the corresponding y values using the function.
\n" ); document.write( "When $\"x=0\"$ , $\"y=1%2F%280-1%29=-1\"$
\n" ); document.write( "When $\"x=2\"$ , $\"y=1%2F%282-1%29=1\"$
\n" ); document.write( "Then using the point slope form of a line with the slope and the point, $\"y-yp=m%28x-xp%29\"$ for both points.
\n" ); document.write( " $\"y-%28-1%29=-1%28x-0%29\"$
\n" ); document.write( " $\"y%2B1=-x\"$
\n" ); document.write( " $\"highlight%28y1=-x-1%29\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( " $\"y-1=-1%28x-2%29%29\"$
\n" ); document.write( " $\"y-1=-x%2B2\"$
\n" ); document.write( " $\"highlight_green%28y2=-x%2B3%29\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "

\n" ); document.write( " \n" ); document.write( "

\n" );