How Do You Find the Points of Intersection of Two Curves Using Technology?

You graph both curves on a graphic display calculator (GDC) and use its intersection command to obtain each point. Alternatively, solve $f(x)-g(x)=0$ using technology, then substitute each solution to find the corresponding $y$-coordinate.

The Method

This is SL 2.4, Features of a graph, studied at both SL and HL. A GDC is available on every IB Mathematics: Applications and Interpretation paper.

Suppose the curves are $y=f(x)$ and $y=g(x)$. At an intersection, both curves have the same $x$- and $y$-values, so

$$f(x)=g(x).$$

Follow these steps:

1. Enter $Y_1=f(x)$ and $Y_2=g(x)$ into the graphing application.

For example, consider $y=x^2$ and $y=2x+3$. Enter both functions and use the intersection command. The GDC gives the points

$$(-1,1) \quad \text{and} \quad (3,9).$$

The equivalent equation-solver setup is

$$x^2-(2x+3)=0.$$

This gives $x=-1$ and $x=3$; substituting into either original function gives $y=1$ and $y=9$.

A common misconception is that the intersection command automatically finds every intersection. A poor graphing window may hide crossings, so inspect the graph and adjust the window or solve $f(x)-g(x)=0$ over the stated domain.

Also check for intersections where curves only touch rather than cross. These can be easy to miss visually because the graphs may appear as one curve near the contact point. Solving $f(x)-g(x)=0$ and checking the stated domain helps detect such repeated solutions. If no domain is given, use a window justified by the context and inspect the behavior beyond it.

Exam Technique

Show the equations entered or write $f(x)=g(x)$ before giving the GDC results. Examiners expect complete coordinates, not only the $x$-values, and all intersections within the specified domain.