AP Calculus AB and BC

Tangent Line Tracer: A 40-Minute Classroom Lesson

This 40-minute lesson runs the tangent line tracer at /interactives/tangent-line-tracer as predict-then-reveal: drag the point to read the slope sign, find where it hits zero, and see that a smooth curve keeps one finite slope even at its steepest reachable edge, unlike a true corner.

Objective and materials

By the end of the period, students can state, for any point they choose on a smooth curve, whether the tangent line's slope there is positive, negative, or zero, and can explain why a single steep but finite slope reading is not the same thing as a sharp corner where no tangent line exists at all.

A shared screen or projector open to /interactives/tangent-line-tracer, with the $x^2$ preset selected (it loads first, with the point sitting at $x = 1$ )
One notecard or half sheet per student for written predictions
A whiteboard, or one shared document, where the class prediction gets recorded before every reveal
Optional: one device per pair so pairs can drag the point and try the keyboard controls themselves after the exit ticket

Nothing needs to be built ahead of time. The tracer already opens on $f(x) = x^2$ with the point sitting at $x = 1$ , where the $a$ readout reads $1.000$ and the $f'(a)$ readout reads $2.000$ , so the only setup is loading the page and leaving the point untouched until Move 1.

The hook, 4 minutes

Draw a smooth hill on the board, rising then falling, no axes needed. Ask a student to describe, in one word, how steep the curve feels right at the very top of the hill compared to a point partway up the rise. Someone will say flat, or zero.

Then ask the room a single question: could two completely different smooth curves share that exact same flat-at-the-top shape, and still disagree everywhere else about how steep they are? Take a show of hands and move straight into Move 1 without resolving it. The tracer resolves it in the next few minutes.

Guided exploration: four teacher moves, 26 minutes

Each move follows the same shape: state exactly what is about to change, collect a written prediction from every student before the change happens, then drag and compare. Do not move the point ahead of the prediction. The prediction is the lesson; the drag is only the check.

Move 1, moving the point, 0:04 to 0:09. The tracer is still sitting at its default: $f(x) = x^2$ with the point at $x = 1$ , where the top panel shows a tangent line and the bottom panel shows a single dot on the $f'$ graph at $x = 1$ . Before touching anything, ask the class to predict, in words, what will happen to both panels as the point drags to the left: will the tangent line's tilt change, and will the bottom panel gain a growing curve or stay a single dot? Then drag the point from $x = 1$ to $x = -1$ , narrating as you go. The tangent line visibly rotates to match the curve's local tilt, and the bottom panel grows a curve connecting every x-value the point has crossed, not just the two endpoints, since the caption beneath that panel says the derivative graph is traced only where you have dragged.
Move 2, sign of the slope, 0:09 to 0:15. With the point now at $x = -1$ , the $a$ readout reads $-1.000$ and the $f'(a)$ readout reads $-2.000$ . Ask the class to predict, before you drag, whether the slope at $x = 1$ will read as a bigger negative number, a smaller negative number, or a positive number. Drag back to $x = 1$ : $f'(a)$ reads $2.000$ . State the rule out loud: reading left to right, wherever the curve is falling the slope readout is negative and the bottom dot sits below the zero line, and wherever the curve is climbing the slope readout is positive and the dot sits above it. Ask a student to point at the exact spot on the top panel where the readout must flip sign, before dragging there to check.
Move 3, where the slope is zero, 0:15 to 0:22. Drag toward the spot just named. At $x = 0.000$ the $f'(a)$ readout reads $0.000$ and the bottom dot sits exactly on the zero line drawn across the lower panel; this is the vertex of the parabola. Switch presets to $x^3 - 3x$ , which opens with the point at $x = 0$ and a starting slope reading of $-3.000$ . Ask the class to predict how many x-values on this new curve will give a slope reading of exactly $0.000$ , given that the curve rises, then falls, then rises again. Drag to $x = -1$ (slope reads $0.000$ , a local high point) and to $x = 1$ (slope reads $0.000$ , a local low point). Then switch to $e^{x/2}$ and ask the class to predict whether dragging across its whole reachable range will ever produce a $0.000$ reading. Drag from the far left to the far right of that range: the $f'(a)$ readout stays positive the entire time, from about $0.112$ at the left edge to about $2.241$ at the right edge, and never once touches zero, because this curve is always climbing.
Move 4, a cusp or corner, 0:22 to 0:30. Switch to $\sqrt{x}$ , which opens with the point at $x = 2.500$ . Ask the class to predict what the $f'(a)$ readout will do as the point drags toward the steep left edge of the curve: will the number climb toward a very large value, or will it eventually read as undefined? Focus the point (click it once, or tab to it) and press the Home key to jump straight to the leftmost $x$ the point can reach. The $a$ readout lands at about $0.012$ , not at $0.000$ , and the $f'(a)$ readout reads about $4.472$ , a large but perfectly ordinary number. Tell the class plainly: the domain here stops just short of $x = 0$ by design, which is itself the teaching point, so no matter how far left the point travels, it will never show a genuinely undefined slope. Then draw a sharp V-shaped corner on the board, something like $f(x) = |x - 1|$ , and ask a student to try drawing one single tangent line at the corner point itself. They cannot: a line hugging the left branch has one tilt, a line hugging the right branch has a completely different tilt, and no single line matches both. That is what a genuine corner looks like, and it is a different situation from the very steep, but still perfectly finite, slope the tracer shows at its own edges.

Preset	x	f(x)	f'(x)
$x^2$	-1	1.000	-2.000
$x^2$	0	0.000	0.000
$x^2$	1	1.000	2.000
$x^3 - 3x$	0	0.000	-3.000
$x^3 - 3x$	-1	2.000	0.000
$x^3 - 3x$	1	-2.000	0.000
$e^{x/2}$	-3	0.223	0.112
$e^{x/2}$	3	4.482	2.241
$\sqrt{x}$	2.500	1.581	0.316
$\sqrt{x}$	0.012	0.112	4.472

That table is the answer key for Moves 1 through 4, using the exact numbers the tracer displays at each x-value above. Every entry is what the readouts show, so it doubles as a projector-free reference if a laptop fails partway through.

Check for understanding, 4 minutes

Move away from the screen for this part; it tests whether the rule transfers, not whether students can read a readout. Pose the question without the tracer: $f(x) = -(x - 3)^2 + 4$ is a downward-opening parabola with its peak at $x = 3$ . Without touching the tracer, state whether the slope is positive or negative for $x$ less than $3$ , positive or negative for $x$ greater than $3$ , and exactly what the slope equals at $x = 3$ .

What a correct answer sounds like

Since the parabola opens downward and peaks at $x = 3$ , the curve climbs for every $x$ less than $3$ , so the slope there is positive, and falls for every $x$ greater than $3$ , so the slope there is negative. The peak itself is where the tangent runs flat, so the slope equals $0$ at $x = 3$ . Students who answer by citing the shape rather than by guessing have met the objective; students who cannot explain why should stay for Move 3's rule during the exit ticket review.

Give students thirty seconds of silent thinking, thirty seconds to compare with a neighbor, then take a thumbs up or down on all three parts at once. Cold-call one thumbs-down student to hear the reasoning of a neighbor who answered correctly before moving on.

Exit ticket, 6 minutes

For a smooth curve that is climbing the whole way across an interval, is the slope reading positive, negative, or zero across that whole interval?
Name one preset in the tangent line tracer whose slope reading never once equals $0.000$ , no matter where the point is dragged, and explain why.
True or false, and correct it if false: any point where the curve looks very steep must be a corner where no tangent line exists.
For $f(x) = x^3 - 3x$ , the tracer shows a slope reading of $0.000$ at both $x = -1$ and $x = 1$ . Without dragging, state which of those two x-values is a local high point and which is a local low point.

Exit ticket answer key

1) Positive across the whole interval, since a climbing curve has a tangent tilted upward everywhere on it. 2) $e^{x/2}$ , because one half of $e^{x/2}$ is always a positive number for every real $x$ ; it is a curve that keeps climbing forever and never levels off. 3) False. A very steep reading is still one finite number describing one direction. A genuine corner has two different directions meeting at a point, and no tangent line fits either branch. 4) $x = -1$ is the local high point, since $f(-1) = 2$ and the curve falls away on both sides of it; $x = 1$ is the local low point, since $f(1) = -2$ and the curve rises away on both sides of it.

No-tech variant with graph paper

Run the identical four moves with nothing but paper when a screen is not available. Print or draw one set of axes per student showing $y = x^2$ from $x = -2$ to $x = 2$ , with gridlines every 0.5 units on both axes, and mark the curve at $x = -2, -1.5, -1$ , up through 2 so students can connect the dots with a smooth curve rather than draw the parabola freehand. Hand out a second, pre-drawn graph of $f(x) = |x|$ , a plain V shape with its point at the origin, to use in Move 4.

Move 1, moving the point by hand. Students mark a point at $x = 1$ on their curve, where $y = 1$ , and, using a straightedge, sketch the single line that just grazes the curve there without crossing it, eyeballing the tangent. Before they draw, have them predict in one word whether that tangent will tilt up or down.
Move 2, sign of the slope by hand. Repeat at $x = -1$ , where $y = 1$ as well. Predict first whether this second tangent will tilt the same way as the one at $x = 1$ or the opposite way, then draw it. Students should see the two tangents tilt in opposite directions, one up and one down, matching the sign flip from Move 2 on screen.
Move 3, where the slope is zero by hand. Have students find the one x-value between -1 and 1 where a hand-drawn tangent line would be perfectly horizontal, mark it, and sketch that flat tangent. On $y = x^2$ this is $x = 0$ , the vertex, matching what the zero line shows in the bottom panel on screen.
Move 4, a cusp or corner by hand. Students take the pre-drawn $f(x) = |x|$ graph and attempt the same straightedge tangent line exactly at the point of the V. Most will draw two different, equally reasonable-looking lines, one following the left branch's tilt and one following the right. Discuss as a class why neither line is correct on its own, and why this is a different situation from a very steep tangent on a smooth curve like $y = x^2$ far from its vertex.

The check for understanding and the exit ticket run exactly as written above; neither one needs a screen. Collect the graph paper itself as the artifact of the lesson: three hand-drawn tangent lines and one attempted corner tangent are a complete record of whether the objective was met.

Common misconceptions

Assuming a curve that looks steep on screen must be broken or undefined there. On the tracer, dragging the point as far as it can go on $\sqrt{x}$ produces a large slope reading like $4.472$ , not an undefined one; the point simply cannot reach the one x-value, $x = 0$ , where the slope would actually stop being a single number.
Treating a genuine corner and a merely steep point as the same problem. A corner, like the point of $f(x) = |x - 1|$ , has two different one-sided tilts meeting with no single line that fits both; a steep point on a smooth curve, like $\sqrt{x}$ near its left edge, still has exactly one well-defined tilt, just a large one. A genuine corner is a different shape from anything a smooth-curve tracer draws, so sketch one by hand on the board instead.
Believing every smooth curve must level off somewhere. $e^{x/2}$ never does; drag it across its whole reachable range and the $f'(a)$ readout stays positive the entire time. A curve can keep climbing forever without ever handing back a slope reading of $0.000$ .
Reading the bottom panel as a pre-drawn graph of $f'$ rather than a trace of only the visited x-values. A student who switches presets and expects to immediately see the whole new derivative curve will be confused when the panel starts nearly empty; explain up front that the curve below only exists where the point has actually been dragged, and the Reset trace button clears it back to a single dot.
Confusing which readout answers which question. The $a$ readout is the input, the location on the curve; $f(a)$ is the height of the curve there; $f'(a)$ is the tangent's slope there. Have students point to each of the three readouts by name at least once before Move 1 begins.

Worked examples

Worked example

Confirming f'(1) for f(x) = x^2 from the limit definition

The tracer's $f'(a)$ readout shows $2.000$ when the point sits at $x = 1$ on $f(x) = x^2$ . Confirm this value using the limit definition of the derivative rather than the shortcut power rule.

Write the difference quotient at x = 1: $f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{(1+h)^2 - 1^2}{h}$ .
Expand the numerator: $(1+h)^2 - 1 = 1 + 2h + h^2 - 1 = 2h + h^2$ .
Divide by h and simplify, valid since h approaches but never equals 0: $\frac{2h + h^2}{h} = 2 + h$ .
Take the limit as h approaches 0: $\lim_{h \to 0} (2 + h) = 2$ . This matches the tracer's f'(a) readout of 2.000 exactly at x = 1.

$f'(1) = 2$ , matching the tracer's readout of $2.000$ .

Worked example

Classifying the two zero-slope points on x^3 - 3x

The tracer shows an $f'(a)$ readout of $0.000$ at both $x = -1$ and $x = 1$ on $f(x) = x^3 - 3x$ . Determine which point is a local maximum and which is a local minimum without relying on the tracer.

Find the derivative and set it to zero: $f'(x) = 3x^2 - 3 = 0$ gives $x^2 = 1$ , so $x = -1$ or $x = 1$ , matching the two x-values the tracer flags.
Check the sign of the slope just to the left and right of x = -1. At x = -2, $f'(-2) = 3(4) - 3 = 9$ , positive; at x = 0, $f'(0) = 3(0) - 3 = -3$ , negative. The slope goes from positive to negative through x = -1, so this point is a local maximum.
Check the sign of the slope just to the left and right of x = 1. At x = 0, the slope is -3, negative; at x = 2, $f'(2) = 3(4) - 3 = 9$ , positive. The slope goes from negative to positive through x = 1, so this point is a local minimum.
Confirm with the function values themselves: $f(-1) = -1 + 3 = 2$ and $f(1) = 1 - 3 = -2$ . The higher output sits at the local maximum, x = -1, and the lower output sits at the local minimum, x = 1, matching the classification found from the slope's sign change.

$x = -1$ is a local maximum, $f(-1) = 2$ ; $x = 1$ is a local minimum, $f(1) = -2$ .

Frequently asked questions

What grade level or unit is this tangent line lesson for?

AP Calculus AB or BC, timed for a standard 40-minute period around CED Unit 2, right after students meet the derivative as the slope of a tangent line and before or alongside their first formal differentiation rules. The same predict-then-drag structure also works later in the course as a review of what a derivative graph actually means.

What if I only have a projector and no student devices?

The lesson is written for exactly that case. One shared screen driven by the teacher covers all four moves; students write predictions on notecards rather than on their own device. Student devices are listed as optional, for pairs to drag the point and try the keyboard controls themselves after the exit ticket.

How do I grade or check this without collecting a stack of papers?

Collect only the exit ticket, four short prompts with a one-line answer key included above, checkable in under a minute per student. The written predictions during the guided moves are formative and are meant to be looked at in the room, thumbs up or down, rather than graded afterward.

Does this work if I cannot get to a computer at all that day?

Yes. The no-tech variant above runs the same four moves on graph paper using $f(x) = x^2$ , since its tangent lines at $x = -1$ , $x = 0$ , and $x = 1$ are clean and easy to sketch by hand. The check for understanding and the exit ticket are already screen-free and need no change.