Question

y = (2x^2-x+1)/(x-1)

a) Determine the points where the tangent line is horizontal

b) Determine the equation of the tangent to the point determined in question (a)

c) Determine at what x value is the slope of the tangent undefined

Answer 1

Using the quotient rule, ...
dy/dx = ( (x-1)(4x - 1) - (2x^2 - x + 1)(1) )/(x-1)^2
= (4x^2 - 5x + 1 - 2x^2 + x - 1)/(x-1)^2
= (2x^2 - 4x)/(x-1)^2
= 0 for a horizontal slop

2x^2 - 4x = 0
2x(x - 2) = 0
x = 0 or x = 2

if x = 0, y = 1/-1 = -1
if x = 2, y = (8-2+1)/(2-1) = 7

so the tangent line is horizontal at (0,-1) and (2,7)

horizontal lines have the form y = c, where c is a constant
So, looking at our points, that would be y = -1 and y = 7

for a tangent to be undefined, we must have divided by zero, that is,
in (2x^2 - 4x)/(x-1)^2, x-1 = 0 or x = 1

if x = 1, then we would have an asymptote, (a tangent would be vertical)

check:

www.wolframalpha.com/input/?i=graph+y+%3D+%282x%5E2-x%2B1%29%2F%28x-1%29