You can use a graphing calculator. There are also various places on-line that will graph both for you.

In order to find the crossing points, you will find the point at which the y's are equal. You can set both equations equal to each other and solve for x. Once...

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You can use a graphing calculator. There are also various places on-line that will graph both for you.

In order to find the crossing points, you will find the point at which the y's are equal. You can set both equations equal to each other and solve for x. Once you have that, you can plug that value back in and find x.

**y = 2x+1 and y = x^2+x+1**

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

You need to get this equal to 0, since it is a quadratic equation.

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

**-2x -2x**

**--------------------**

**1 = x^2 -x+1**

**-1 -1**

**---------------------**

**0 = x^2-x**

Now, factor x^2-x. These 2 terms share a common factor of x.

**0=x^2-x**

**0=x(x-1)**

Set each factor equal to 0.

**x=0 and x-1=0**

**So, x=1 and x=0.**

Plug those 2 values in for either equation.

**y=2x+1**

**y=2(1) + 1**

**y = 3**

**So that point will be (1,3)**

**y = 2(0) + 1**

**y = 1**

**So that point will be (0, 1)**

Remember that the reason that you have 2 crossing points is that the equation with an exponent of 2, will for a parabola. It will be curved, because of that curve, it will cross the line 2 times.