How can I find the coordinates of crossing points between the line y=2x+1 and the curve y=x^2+x+1?
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
1 = x^2 -x+1
0 = x^2-x
Now, factor x^2-x. These 2 terms share a common factor of x.
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=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.
This question can be solved more quickly and easily by using a good graphing calculator. Texas Instruments produces a good series of graphing calculators, the latest of which is the TI-89.
When you turn on the TI-89, 6 icons will appear. The rightmost icon on the higher row ill be labeled "Graph." Go to this section. Write the two equations, y = 2x + 1 and y = x^2 + x + 1 in two different rows. Then press ENTER.
The screen will change to an x-y axis. If you wait just a little while the two graphs will appear on the screen. y = 2x +1 will be a straight line, while y = x^2 + x + 1 will be a curve. Then, find the coordinates of the point or points at which the line and the curve intersect. You can move along the graph. As you move along the graph, the x and y coordinates of each point will appear on the screen. Go to the point where the two equations intersect, and the x and y coordinates of that point will appear.
The coordinates of the crossing point have to verify the equation of the line and the equation of the curve, in the same time.
We'll have x as common factor:
From this expression, the values of x are: 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 easier one, the one of the line:
So the first pair of coordinates of crossing point: A(0,1)
So the second pair of coordinates of crossing point: B(1,3).
At the crossing points , the points the cordinates(x,y)=(a,b)
satisfies both curves. So,we have,
Eliminate b from the equations.Then,
a^2-a=0 or a*(a-1)=0
a=0 or a=1
Therefore, by substituting a=0 in the first equation , we get b=2*0+1=1
Therefore, a=0 and b=1 for a=0. Threfore, (a,b)=((0,1)
b=2a+1=2(1)+1=3. threfore, (a,b)=(1,3)
The line thus cuts the curve at (0,1) and (1,3).