f(x) = x^2 + x + 1

g(x) = 2x + 1

The intercepting point is where f(x) = g(x)

Then,

x^2 + x + 1 = 2x + 1

Now grouo similar terms:

==> x^2 - x = 0

Now factor x:

==> x( x-1) = 0

==> x1= 0 ==> f(x1) = f(0) = 1

==> x2= 1 ==> f(x2) = f(1) = 3

**Then we have two intercepting points at:**

**A(0, 1) and B(1, 3) **

f(x) = x^2+x+1 and f(x) = 2x+1.

To find the inerceting points.

Solution:

Let us denote f(x) by y on y axis. Then at the intercepting point the y coordinate is same on both graph.

So we can write x^2+x+1 = 2x+1.

x^2+x+1-2x-1 = 0

x^2-x = 0

x(x-1) = 0

x = 0 or x=1 are x coordinatesof the point of intersection.

When x = 0 , the 2nd equatio y = 2x+1 gives y = 2*0+1 =1.

When x=1, y =2x+1 gives: y = 2*1+1 = 3.

So the interception coordinates at 2 points are: (0 ,1) and (1, 3).

To determine the intercepting points between the graphs of the given fucntions, we'll have to solve the system formed by the equations of the functions:

y = 2x + 1 (1)

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

The coordinates of the intercepting points are the solutions of the system, so they have to verify both equations.

We'll put (1)=(2):

2x + 1 = x^2+x+1

We'll eliminate like terms:

x^2-x=0

We'll factorize:

x*(x-1)=0

We'll put each factor as 0:

x=0 and x-1=0, x=1

Now, we'll substitute the x values in one of the 2 equation, and we'll do it in the linear equation,because it's more easy to calculate:

y=2x+1

x=0

y=2*0+1, y=1

x=1

y=2*1+1=3

**So the intercepting points are: A(0,1) and B(1,3).**