Question

Given the quadratic function y=x^2-x+k-1 and line y=x+1, find the number of points that their graphs have in common. (Assume k>4)

Answer 1

x^2 - x + k - 1 = x + 1

x^2 -2x +(k-2) = 0

x = [ 2 +/- sqrt(4 -4(k-2)) ] /2

x = [ 2 +/- 2 sqrt ( 3-k) ]/2

so two points assuming k>3

Answer 2

suppose k = -1
then y = x^2 - x - 2
= (x-2)(x+1)
since the line y = x+1 is to intersect it ...
(x-2)(x+1) = x+1
(x-2)(x+1) - (x+1) = 0
(x+1)(x-3) = 0
x= -1 or x = 3
y = 0, or y = 4
they would intersect twice at (3,4) and (-1,0)
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2-x-2,+y%3Dx%2B1

now as we increase k we are doing a vertical shift of the parabola, so at first the two intersection points move closer to each other.
e.g. let k = 2
x^2 - x + 2-1 with y = x+1
x^2 - 2x = 0
x(x-2)) = 0
x = 0 or x = 2
y = 1, y = 3
http://www.wolframalpha.com/input/?i=plot+x%5E2+-+x+%2B+2-1+,+y+%3D+x%2B1

let k = 4, then we have y = x^2 - x + 3 , with y = x+1
algebraically .... x^2 - x + 3 = x+1
x^2 - 2x + 2 = 0

http://www.wolframalpha.com/input/?i=plot+x%5E2+-+x+%2B+4-1+,+y+%3D+x%2B1

notice now the parabola has moved vertically to the point it no longer intersect the straight line, so clearly any value of k greater will just make things worse.

No intersection points!!