# How to Find the Intersection Points of Two Graphs: Examples

TR Smith is a product designer and former teacher who uses math in her work every day.

Joined: 5 years agoFollowers: 216Articles: 455

Finding the intersection point of two or more graphs is a common problem in algebra, pre-calc, and calculus. The intersection of two graphs is an important concept in many fields. For example, the point where the graphs of supply and demand curves intersect is a key equilibrium point in finance and economics. The intersections of two quadratic functions (parabolas) may indicate where and when two falling objects are at the same height. In calculus, if you want to find the area enclosed by two curves, you need to know the point(s) where the curves cross each other.

You can use a graphing calculator to numerically estimate where two curves cross each other by graphing the equations and zooming in on the point of intersection. But in the absence of a graphing utility, you can often solve for the common points algebraically. To do this, you set the two equations f(x) and g(x) equal to each other and solve for the value(s) of x. Then you plug those x values back into one of f(x) or g(x) to find the associated y values. Here are several examples worked out.

## Example 1

Find the intersection point(s) of the graphs of a linear function and rational function

f(x) = 2 + x/2

g(x) = (x + 1)/(x - 3)

To find the x-coordinates of the intersection points, we set the two equations equal to each other to produce an equation in the variable x

2 + x/2 = (x + 1)/(x - 3)

Now we solve for x. The first step to simplify this equation is to multiply both sides by (x - 3) to clear the denominator and transform the equation into a quadratic.

(2 + x/2)(x - 3) = x + 1
2x - 6 + 0.5x^2 - 1.5x = x + 1
(1/2)x^2 + (1/2)x - 6 = x + 1
(1/2)x^2 - (1/2)x - 7 = 0
x^2 - x - 14 = 0

Now we use the quadratic formula to solve for x:

x = [1 + sqrt(1 + 56)]/2
= 4.274917

x = [1 - sqrt(1 + 56)]/2
= -3.274917

To find the corresponding y-coordinates, we plug each x-value into one of the equations. We should choose the simpler linear equation.

y = 2 + ([1 + sqrt(57)]/2)/2
= 4.137459

y = 2 + ([1 - sqrt(57)])/2
= 0.362541

Therefore, the two intersection points of these graphs are (4.274917, 4.137459) and (-3.274917, 0.362541). See image below.

Graphs of f(x) = 2 + x/2 and g(x) = (x+1)/(x-3)

## Example 2

Find the intersection point(s) of the exponential growth and decay graphs

f(x) = (1/3)(2^x) + 5/3

g(x) = (9)(1/2)^x

The equation of g(x) can be written in the equivalent forms g(x) = 9/(2^x) or g(x) = 9(2^-x). When you set the two equations equal to each other, you get

(1/3)(2^x) + 5/3 = (9)(2^-x)

Multiplying both sides by 2^x gives you

(1/3)(4^x) + (5/3)(2^x) = 9
(1/3)(4^x) + (5/3)(2^x) - 9 = 0
4^x + 5(2^x) - 27 = 0

Now you have a quadratic-type equation; instead of the terms x and x^2, you have 2^x and (2^x)^2 = 4^x. So it is a quadratic not in x, but in 2^x.

(2^x)^2 + 5(2^x) - 27 = 0

Solving for 2^x with the quadratic formula gives you two candidates for solutions:

2^x = [-5 + sqrt(25 + 108)]/2
= 3.266281

2^x = [-5 - sqrt(25 + 108)]/2
= -8.266281

The first candidate yields

2^x = 3.266281
LN(2^x) = LN(3.266281)
x*LN(2) = LN(3.266281)
x*0.693147 = 1.183652
x = 1.183652/0.693147
x = 1.707649

The second candidate can be discarded so there are no real values of x such that 2^x equals a negative number. There are, however, complex values of x that solve the equation 2^x = -8.266281.

With the x-coordinate 1.707649, we can plug it into one of the equations to find the y-coordinate.

y = (1/3)(2^1.707649) + 5/3
= 2.755427

Therefore, the intersection point of the two graphs is (1.707649, 2.755427). See image below.

Graphs of f(x) = [2^x + 5]/3 and g(x) = 9/(2^x).

## Example 3

Find the intersection points of two conic sections, the parabola

f(x) = x^2 - 3

and the circle

y^2 + x^2 = 4

There are two ways you can go about finding all the intersection points. One method is to set y = x^2 - 3, and make a substitution for y into the second equation. This gives you a 4th degree polynomial

(x^2 - 3)^2 + x^2 = 4
x^4 - 5x^2 + 5 = 0
(x^2)^2 - 5x^2 + 5 = 0

Because there are no x^3 or x terms, this quartic equation can be solved like a quadratic. Another method is to convert the circle equation into two functions:

g(x) = sqrt(4 - x^2)

h(x) = -[sqrt(4 - x^2)]

and then set each equation equal to the parabola f(x) = x^2 - 3. If you do this, you still end up with the same quartic equation x^4 - 5x^2 + 5 = 0. Solving this equation for x^2 gives us

x^2 = [5 + sqrt(25 - 20)]/2
= 3.618034

x^2 = [5 - sqrt(25 - 20)]/2
= 1.381966

The first candidate splits into two solutions

x^2 = 3.618034
x = ± sqrt(3.618034)
x = 1.902113, x = -1.902113

The second candidate also splits into two solutions

x^2 = 1.381966
x = ± sqrt(1.381966)
x = 1.175571, x = -1.175571

If we plug the four x-coordinates into the parabola equation, we get four corresponding y-coordinates

y = (1.902113)^2 - 3 = 0.618034

y = (-1.902113)^2 - 3 = 0.618034

y = (1.175571)^2 - 3 = -1.618034

y = (-1.175571)62 - 3 = -1.618034

Therefore, the four intersection points are

(1.902113, 0.618034)
(-1.902113, 0.618034)
(1.175571, -1.618034 )
(-1.175571, -1.618034)

See image below.

Graphs of the parabola y = x^2 - 3 and the circle y^2 + x^2 = 4.

0

9

24

16

3

63

2

29