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)
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
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!!
To find the number of points that the quadratic function and the line have in common, we need to find the values of x for which the two equations are equal and graphically represent their intersection points.
First, let's set the quadratic function y = x^2 - x + k - 1 equal to the line y = x + 1:
x^2 - x + k - 1 = x + 1
Rearranging the equation, we get:
x^2 - 2x + (k - 2) = 0
Now, we can use the quadratic formula to find the values of x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -2, and c = (k - 2). Plugging in these values, we get:
x = (2 ± √((-2)^2 - 4(1)(k - 2))) / 2(1)
x = (2 ± √(4 - 4(k - 2))) / 2
x = (2 ± √(4 - 4k + 8)) / 2
x = (2 ± √(12 - 4k)) / 2
x = 1 ± √(3 - k)
Now, we know that for the quadratic and line to intersect, the discriminant (3 - k) must be non-negative. Since k > 4 (as given), the discriminant (3 - k) will be negative, and we won't have any real solutions. Therefore, the quadratic function and the line have 0 points in common.
In summary, when k > 4, the quadratic function y = x^2 - x + k - 1 and the line y = x + 1 have no intersection points.