Calculate the coordinates of the intercepting point of the line x-2 and the curve x^2-4x+4.
f(x) = x^2 - 4x + 4
g(x) = x-2
The intercepting point is when f(x) = g(x)
x^2 - 4x + 4 = x-2
Group similar terms:
==> x^2 - 5x + 6 = 0
==> (x-3)(x-2) = 0
==> x1= 3 ==> y1= 3-2 = 1
==> x2= 2 ==> y2= 2-2 = 0
Then the curve f(x) and the line y are intercepting in two point :
Point of intercepting are: A(3, 1) and B(2, 0)
First, we'll note the line y = x-2 and the curve as y = x^2-4x+4.
To determine the coordinates of the intercepting point of the line and curve, we'll have to solve the system:
y = x-2 (1)
y = x^2-4x+4 (2)
x-2 = x^2-4x+4 (3)
We notice that the equation (2) is the result of the square expanding:
(x-2)^2 = x^2-4x+4
We'll re-write (3):
(x-2) = (x-2)^2
We'll subtract (x-2) both sides:
(x-2)^2 - (x-2) = 0
We'll factorize by (x-2):
(x-2)[(x-2)-1] = 0
We'll set each factor as 0:
x-2 = 0
We'll add 2 both sides;
x = 2
From (1) => y = x-2 => y = 2-2 = 0
x-3 = 0
x = 3
From (1) => y = x-2 => y = 3-2 = 1
The coordinates of the intercepting points are: (2 , 0) and (3 , 1).
To find the intercepting ponits of x^2-4x+4 and x-2.
Let y = x^2-4x+4 and y = x-2 be the curve and the line.
Since the y coordinates at the intersection is same for both curve and the line,
x^2-4x+4 = x-2
(x-2)^2 = (x-2)
(x-2)^2 - (x-2 ) = 0
(x-2)(x-2-1) = 0
(x-2) (x-3) = 0
So x= 2 or x=3 are the x coordinates of the intersections.
At x =2, y = (x-2) = 2-2 = 0
At x=3, y = x-2 = 3-2 = 1.
So (2,0) and (3, 1) are coordinates of intersection of x^2-4x+4 and x-2.