We have to find the common points of y=1+2x and the parabola y=x^2+x+1
substitute y = 2x + 1 in y=x^2+x+1
=> 2x + 1 = x^2 + x + 1
=> x^2 - x = 0
=> x(x - 1) = 0
=> x = 0 and x = 1
For x = 0 , y = 1 and for x = 1, y = 3
The common points are (0,1) and (1, 3)
The common point that lies on the line and parabola in the same time is the intercepting point of the line and parabola.
So, the y coordinate of the point verify the equation of the line and the equatin of the parabola, in the same time.
We'll move all term to one side and we'll combine like terms:
We'll factorize by x:
We'll put each factor as zero:
We'll add 1 both sides:
Now, we'll substitute the value of x in the equation of the line, because it is much more easier to compute y.
So the first pair of coordinates of crossing point: A(0,1)
So the second pair of coordinates of crossing point: B(1,3).
So, the common points are: A(0,1) and B(1,3).