How do you find the slope of the tangent line #y=(x+1)(x-2)# at x=1?

1 Answer
Mar 27, 2017

The slope of the tangent line to #y# is #1#

Explanation:

The slope of the tangent line is evaluated by computing #y'_(x=1)#
#" "#
Let us differentiate #y#:
#" "#
Applying the product rule of differentiation :

#" "#
#color(blue)((uv)'=u'v + v'u#
#" "#
#color(blue)(y'=(x+1)'(x-2)+(x-2)'(x+1)#
#" "#
#y'=1(x-2)+1(x+1)#
#" "#
#y'=x-2+x+1#
#" "#
#y'=2x-1#
#" "#
The slope is #y'_(x=1)#, so let us substitute #x=1# in #y'#

#" "#
#y'_(x=1) =2(1)-1#

#" "#
#y'_(x=1) =2-1=1#
#" "#
Therefore, the slope of the tangent line at #x=1" "# is #1#.