For #f(x)=x-x^2# what is the equation of the tangent line at #x=1#?

1 Answer
May 23, 2018

#y=-x+1#

Explanation:

#"to obtain the equation we require slope and a point on"#
#"the tangent"#

#"the slope is the value of "f'(1)#

#f'(x)=1-2xrArrf'(1)=-1#

#"and "f(1)=0rArr(x_1,y_1)=(1,0)#

#y=-1(x-1)#

#y=-x+1larrcolor(red)"equation of tangent"#