Question

Find the coordinates of the point on the curve y = x^2 -x -1 where the tangent is perpendicular to the tangent at the point (-1,1)

Answer 1

@redpellet
\[y'(x) = \frac 1 3\\
x =\frac 23\\
\left(\frac 23,-\frac {11}9\right)
\]
is the point

Answer 2

For the slope of the tangent at the point (-1, 1):

y = x^2 - x - 1
y' = 2x - 1

y' = 2(-1) - 1
= -3
therefore its slope is -3.

(a perpendicular slope is the negative reciprocal):
so the slope of line perpendicular to the tangent is 1/3.

1/3 = 2x - 1
x = 2/3

and then sub that into y = x^2 -x -1 for the y component of the coordinate

the point on the graph at that x value is where the slope of its tangent is perpendicular to the tangent at (-1, 1)