# Prove that the graphs of the functions f(x)=2x+1 and g(x)=x^2+x+1 have a point of intersection.

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To prove that the two graphs of the functions f(x)=2x+1 and g(x)=x^2+x+1 have a point of intersection, let equate f(x) and g(x) and see if there is a valid solution.

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

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

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

=> x^2 - x = 0

=> x(x - 1) = 0

=> x can be 0 or 1.

For x = 0, f(x) = 1 and g(x) = 1

For x = 1, f(x) = 3 and g(x) = 3

**Therefore the point of intersection is (0, 1) and (1, 3).**

To prove that the graphs of the functions f(x)=2x+1 and g(x)=x^2+x+1 have a point of intersection.

f(x) = y = 2x+1...(1)

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

At the point of intersection, the ordinates y of graph is common to both graph. So we solve for the x coordinate values.

=> x^2+x+1 = 2x+1.

=> x^2+x+1-2x-1 = 2x+1-2x-1.

=> x^2-x = 0.

=> x(x-1) = 0.

=> x = 0, or x-1 = 0.

=> x= 0, or x= 1.

When x = 0, from (1) , we get : y = 2x+1 = 2*0+1 = 1.

When x = 1, from (1), we get : y = 2x+1 , y = 2*1+1 = 3.

So the two points of intersection are (0,1) and (1,3).

The intercepting point that is located on the line described by f(x) and parabola described by g(x), in the same time, is the intercepting point of the line and parabola.

So, the y coordinate of the point verify the equation of f(x) and the equation of g(x), in the same time.

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

We'll move all term to one side and we'll combine like terms:

x^2-x=0

We'll factorize by x:

x*(x-1)=0

We'll put each factor as zero:

x=0

x-1=0

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.

y=2x+1

x=0

y=2*0+1, y=1

So the first pair of coordinates of intercepting point: M(0,1)

x=1

y=2*1+1=3

So the second pair of coordinates of intercepting point: N(1,3).

**So, the graphs of the functions f and g are intercepting and their intercepting points are: M(0,1) and N(1,3). **