This note shows how to use a graphing calculator (GDC) to find the key features of a graph: intercepts, turning points and asymptotes, and where two graphs meet. These skills let you solve equations that cannot be solved by algebra alone and are essential for modelling problems.
Key ideas
Once a function is entered in your GDC and drawn, the calculator can find exact-looking numerical values for most features. The skill is choosing a good window, knowing what to ask for, and reporting the answer with the right accuracy.
Key features and how to find them:
- Zeros (roots, \(x\)-intercepts). Use the "zero" or "root" feature of the graph menu. A zero of \(f\) is a solution of \(f(x) = 0\).
- \(y\)-intercept. Evaluate the function at \(x = 0\), or read it from the graph screen.
- Maximum and minimum points. Use the "maximum" or "minimum" feature. These give the coordinates of a turning point, so report both \(x\) and \(y\) if both are asked for.
- Asymptotes. A GDC may not draw them clearly. Use the shape of the graph and the algebra to decide on their equations. A vertical asymptote often comes from a denominator equal to zero.
Intersections. The \(x\)-coordinates where \(y = f(x)\) meets \(y = g(x)\) are the solutions of the equation \(f(x) = g(x)\). Draw both graphs, then use the "intersection" feature. You can also solve an equation like \(f(x) = 10\) by graphing \(y = f(x)\) and the horizontal line \(y = 10\).
Choosing a window. If you cannot see the feature you want, change the window. Think about the context first. For example, if \(x\) is a time between 0 and 10, set the \(x\)-range to match. A window that is too large can hide a turning point, and one that is too small can hide an intersection.
Counting solutions. Several intersections can exist. Check across the whole domain and find each one. The number of solutions of \(f(x) = g(x)\) is the number of times the graphs meet.
Giving answers. Unless a question says otherwise, give answers to 3 significant figures. Do not copy all the digits from the screen. In a context question, include units and say what the number means. Be sure your final answer fits the situation, for instance by rejecting a negative time.
Use technology to sketch. When a question asks you to sketch a graph with technology, copy the key features from your screen onto paper, with axes labelled and key points marked with coordinates.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Zero of \(f\) | A value of \(x\) with \(f(x) = 0\). Found with the "zero" feature. |
| Local maximum or minimum | A turning point of the graph. Found with the "maximum" or "minimum" feature. |
| Intersection of \(y = f(x)\) and \(y = g(x)\) | The solutions of \(f(x) = g(x)\). Use the "intersection" feature. |
| Solving \(f(x) = k\) | Graph \(y = f(x)\) and \(y = k\), then find the intersections. |
| Vertical asymptote | A vertical line the graph approaches. Often where a denominator is zero. |
| Horizontal asymptote | A horizontal line the graph approaches as \(x\) becomes very large or very small. |
| 3 significant figures | The usual accuracy for a final answer when none is specified. |
Worked examples
Example 1 Let \(f(x) = x^3 - 6x^2 + 4x + 8\). Use your GDC to find (a) the zeros of \(f\), (b) the \(y\)-intercept, and (c) the coordinates of the local maximum and local minimum.
Enter \(f(x)\) and draw the graph with a window such as \(-3 \leq x \leq 7\) and \(-15 \leq y \leq 15\), so that all the features are visible.
(a) Use the "zero" feature three times, once beside each crossing of the \(x\)-axis. The zeros are \(x = -0.828\), \(x = 2\) and \(x = 4.83\) (to 3 s.f.).
(b) At \(x = 0\), \(f(0) = 8\). The \(y\)-intercept is 8.
(c) Use the "maximum" feature on the left turning point: \((0.367, 8.71)\). Use the "minimum" feature on the right turning point: \((3.63, -8.71)\).
Answer: zeros at \(-0.828\), \(2\), \(4.83\); \(y\)-intercept 8; local maximum \((0.367, 8.71)\); local minimum \((3.63, -8.71)\).
Example 2 Solve \(\mathrm{e}^x = 5 - x^2\).
This equation mixes an exponential and a quadratic, so algebra cannot solve it directly. Use the graphs of both sides.
Enter \(y_1 = \mathrm{e}^x\) and \(y_2 = 5 - x^2\) and draw them. A window of \(-4 \leq x \leq 3\) and \(-2 \leq y \leq 8\) shows the curves meeting twice.
Use the "intersection" feature at each crossing. The first crossing is at \(x = -2.21\) and the second is at \(x = 1.24\) (to 3 s.f.).
Quick check at \(x = 1.24\): \(\mathrm{e}^{1.24} = 3.46\) and \(5 - 1.24^2 = 3.46\). They agree.
Answer: \(x = -2.21\) or \(x = 1.24\)
Example 3 The concentration of a medicine in the blood, in mg per litre, is modelled by \(C(t) = 20t\,\mathrm{e}^{-0.4t}\), where \(t\) is the time in hours after taking a dose, for \(t \geq 0\). (a) Find the greatest concentration and when it happens. (b) Find when the concentration is above 10 mg per litre.
Enter \(C(t)\) using \(x\) for \(t\). A suitable window is \(0 \leq x \leq 15\) and \(0 \leq y \leq 20\).
(a) Use the "maximum" feature. The maximum point is \((2.5, 18.4)\). So the greatest concentration is 18.4 mg per litre, 2.5 hours after the dose.
(b) Draw the line \(y = 10\) and use the "intersection" feature twice. The crossings are at \(t = 0.648\) and \(t = 6.36\).
The graph of \(C\) is above the line between these two values.
Answer: (a) 18.4 mg per litre at 2.5 hours; (b) the concentration is above 10 mg per litre for \(0.648 \lt t \lt 6.36\) hours.
Common errors
- Common error: giving only the \(x\)-coordinate of a maximum when the question asks for the maximum value, or giving only \(y\) when it asks when it happens.
- Common error: using a window that hides some solutions, then missing an intersection or zero. Zoom out or change the window and check.
- Common error: writing the screen value with too many digits, or rounding too early. Keep full accuracy on the calculator and round only the final answer.
- Common error: forgetting that a function might have been defined only for certain \(x\)-values, and giving a solution outside the domain.
- Common error: confusing the line \(y = 10\) with the value of \(x\). When solving \(f(x) = 10\), the answers are the \(x\)-coordinates of the intersections, not 10.
Exam tip: write down the equation or function you entered and what feature you used, for example "intersection of \(y = \mathrm{e}^x\) and \(y = 5 - x^2\)". This shows your method even if you misread a digit from the screen.
Check yourself
- Which two graphs would you draw on your GDC to solve \(x^3 - 2x = 3 - x\)?
Answer
Draw \(y = x^3 - 2x\) and \(y = 3 - x\) and find the intersection(s). Alternatively, draw \(y = x^3 - x - 3\) and find its zero. - A function has a local minimum at \((3, -4)\). State the lowest value of the function near that point and the \(x\)-value where it occurs.
Answer
The lowest value is \(-4\), and it occurs at \(x = 3\). - For the function in Example 3, find the concentration after 1 hour.
Answer
\(C(1) = 20(1)\mathrm{e}^{-0.4} = 20 \times 0.67032 = 13.4\) mg per litre (to 3 s.f.). - Explain why the equation \(f(x) = -5\) has no solutions if the minimum value of \(f\) is 2.
Answer
The graph of \(f\) never goes below \(y = 2\), so it never reaches the horizontal line \(y = -5\). The graphs do not intersect, so there are no solutions.