Tangent Line Activity: Secants, Slopes and Limits

Objective and materials

Students will calculate a secant slope, explain why its limit can exist when the quotient is undefined at zero, and write a tangent line through the correct point. They will also separate function height from slope and use one-sided slopes to recognize a corner.

Use a projector, paper, and a whiteboard. Student devices are optional. Students need coordinate slope, expanding a square, and the idea of a two-sided limit. The final comparison supplies the derivative of a cubic, so it does not depend on prior mastery of differentiation rules.

Open Secant to Tangent and the Tangent Line Tracer in separate tabs. Keep the teacher key off the projector until students have written their predictions.

Set the two tools to the same point

In the secant explorer, change $h$ with the slider or the second point. The slider moves in increments of $0.01$. The Halve h button rounds to this grid, so repeated halving will not reproduce an exact sequence of powers of one half. Use the specified values in the worksheet rather than assuming each button press produces an exact fraction.

The explorer keeps $|h|\ge 0.01$ and never sets $h=0$. It illustrates finite secant slopes; the algebra below establishes the limit. Reset returns the secant anchor to $a=0$, so move it back to $1$ if you reset during this lesson.

The tangent tracer's upper panel shows $(a,f(a))$ and its tangent. The lower panel shows $(a,f'(a))$ and traces the derivative only across the interval you have visited. Reset trace removes that accumulated interval; an empty part of the lower graph is not evidence that the derivative fails to exist there. Dragging need not land on an exact integer, so use the paper values for exact answers and the readouts for a visual check.

A 40-minute sequence

For each prompt, ask partners to identify the first incorrect step, then write a correction. A matching screen number is not enough: keep the coordinate calculation, limit argument, or sign explanation beside it.

Student worksheet: six mistakes to repair

Use $f(x)=x^2$ and $a=1$ for prompts 1 through 4. The anchor is $A=(1,1)$, and the second point is $B=(1+h,(1+h)^2)$. Record exact paper calculations before checking the tool.

1. Height or slope? A student says, 'The point is $(1,1)$, so the tangent slope is $1$. At $h=0.5$, the slope is $\frac{2.25-1}{1.5}=\frac{5}{6}$.' Identify the two different mistakes. Calculate the correct secant slope for $h=0.5$ and explain what the denominator measures.
2. Undefined quotient or missing limit? Complete the table below. A student substitutes $h=0$, gets $\frac{0}{0}$, and concludes that $f'(1)$ does not exist. Simplify $\frac{(1+h)^2-1}{h}$ for $h\ne0$, then use both sides of the table and the simplified expression to correct the conclusion. Explain why the tool's smallest available $h$ is not a proof.
3. Right slope, wrong line? Using the slope found in prompt 2, a student writes $y=2x$. Test that line at the anchor. Write the actual tangent in point-slope form and simplify it. State which two pieces of information determine this line.
4. Same height, same derivative? A student notices $f(-1)=f(1)=1$ and draws both lower-panel points at height $1$. Given $f'(x)=2x$, write the correct lower-panel points for $a=-1,0,1$. Explain why the top graph can be above the horizontal axis while its derivative is below it.
5. Does zero slope always mean a minimum? Switch to $g(x)=x^3-3x$, with $g'(x)=3x^2-3$. A student labels both zeros of $g'$ as minima. Find the zeros, determine the sign of $g'$ on each interval they divide, and classify each point. Give both coordinates and justify the classification from the sign change, not from comparing their heights.
6. Corner, flat point, or crossing tangent? On paper, let $r(x)=|x-1|$ and $s(x)=x^3$. A student claims, 'Both are continuous, so both have derivatives everywhere. Also, an increasing curve must have a strictly positive slope, and a tangent cannot cross its curve.' At $x=1$, calculate $\frac{r(1+h)-r(1)}{h}$ separately for negative and positive $h$. At $x=0$, calculate the difference quotient for $s$, its limit, and its tangent line. Use these two examples to correct all three claims.

The absolute-value function and $x^3$ in prompt 6 are paper examples, not function choices in the two interactives used here. Do not substitute the $x^3-3x$ preset for $x^3$.

Teacher key: calculations and reasoning

1. The height $f(1)=1$ is not a slope. A secant slope divides the change in output by the change in input, not by the second input alone. Here the rise is $2.25-1=1.25$ and the run is $1.5-1=0.5$, giving $2.5$. This is an average rate between two points, not yet the tangent slope at $1$.

2. Expand the numerator to $2h+h^2=h(2+h)$. For $h\ne0$, cancellation gives $2+h$. The original quotient remains undefined at $h=0$, but $2+h$ approaches $2$ from either side, so $f'(1)=2$. For this quadratic, the error from the tangent slope is exactly $h$, with absolute error $|h|$. Thus the screen values at $h=0.01$ and $h=-0.01$ are $2.01$ and $1.99$, not the limit itself. A finite table supports the conjecture; the algebra gives the limit.

3. The proposed $y=2x$ gives $y=2$ at $x=1$, so it misses $(1,1)$. The correct tangent is $y-1=2(x-1)$, or $y=2x-1$. A point and a slope determine the line. Its slope is $2$, and substituting $x=1$ gives the required height $1$.

4. The lower points are $(-1,-2)$, $(0,0)$, and $(1,2)$. Their heights are slopes, not the upper graph's outputs. At $a=-1$, the upper point has positive height $1$ but the function is decreasing as $x$ increases nearby, so the tangent slope is negative. At $0$, the upper and lower points both happen to be $(0,0)$, but they represent different quantities.

5. Solving $3x^2-3=0$ gives $x=-1$ and $x=1$. Factoring $g'(x)=3(x-1)(x+1)$ gives a positive derivative on $(-\infty,-1)$, negative on $(-1,1)$, and positive on $(1,\infty)$. At $(-1,2)$, increasing changes to decreasing, so there is a local maximum. At $(1,-2)$, decreasing changes to increasing, so there is a local minimum. Zero slope alone would not establish either classification.

6. For $r$, the quotient is $\frac{|h|}{h}=-1$ when $h<0$ and $1$ when $h>0$. The one-sided limits disagree, so $r'(1)$ does not exist even though $r$ is continuous at $1$. For $s$, the quotient at $0$ is $\frac{h^3}{h}=h^2$ for $h\ne0$, whose limit is $0$. Thus $s'(0)=0$ and the tangent is $y=0$. The function $x^3$ is strictly increasing, yet has zero derivative at $0$: strictly increasing does not imply strictly positive derivative at every point. Finally, $x^3$ is negative to the left of $0$ and positive to the right, so its graph crosses its tangent there. A tangent is defined by the local slope, not by a rule forbidding crossings.

Exit check and response to errors

Without the tool, use $q(x)=-(x-3)^2+4$ and $q'(x)=-2(x-3)$. Find the tangent line at $x=2$. State where the derivative is positive, zero, and negative. Explain whether the horizontal tangent at $x=3$ is a local maximum or a local minimum.

Key: $q(2)=3$ and $q'(2)=2$, so $y-3=2(x-2)$, giving $y=2x-1$. The derivative is positive for $x<3$, zero at $3$, and negative for $x>3$. The point $(3,4)$ is a local maximum because the derivative changes from positive to negative.

• If height and slope are still mixed up, label $(a,f(a))$ and $(a,f'(a))$ separately, then revisit prompt 4.
• If the limit is rejected because of division by zero, revisit the cancellation condition $h\ne0$ and distinguish the original expression from its limit.
• If the equation has the right slope but misses the point, substitute the anchor into the proposed line before simplifying anything.
• If every zero slope is called an extremum, compare prompt 5 with the $x^3$ counterexample in prompt 6.

Optional preset checks: steep is not undefined

For a follow-up demonstration, choose $e^{\frac{x}{2}}$ in the tangent tracer. Its derivative $\frac{1}{2}e^{\frac{x}{2}}$ stays positive: at $x=-3$ it is approximately $0.112$, and at $x=3$ approximately $2.241$. This preset has no horizontal tangent.

The $\sqrt{x}$ preset starts at $a=2.5$, where the output is approximately $1.581$ and the slope approximately $0.316$. Moving toward the leftmost reachable point makes the tangent steep, but the tracer does not reach $x=0$. Every displayed point in that reachable positive interval has finite slope $\frac{1}{2\sqrt{x}}$. Do not use the screen's edge as evidence for what happens at the endpoint or as an example of a corner.

For the cubic preset, the initial anchor is $a=0$, with slope $-3$. Predict the two zero-slope locations before moving to $a=-1$ and $a=1$. Keep the derivative's sign changes, not the relative heights of the two points, as the reason for the classifications.

No-tech version and next practice

Use the same six prompts and timing without a device. Plot $y=x^2$ from $x=-2$ to $x=2$ on graph paper. Mark $(1,1)$ and the six second points in the worksheet table. Draw several secants, then draw $y=2x-1$ through the anchor and compare slopes. Small separations are difficult to measure on paper, so use the calculated table rather than a ruler to establish the decimal slopes.

On a second set of axes, plot $(-1,-2)$, $(0,0)$, and $(1,2)$ for the derivative. Sketch $x^3-3x$ with its two stationary points, then use the supplied derivative to complete the sign chart. Draw $|x-1|$ and $x^3$ for the final comparison. The optional exponential and square-root demonstrations can be omitted; the paper worksheet still covers all six core prompts.

Students who need another explanation can read the limit definition of the derivative. For equation-building practice, use the Tangent and Normal Line Finder after solving the point-slope equation on paper. Its custom mode uses a numerical slope estimate, so use the exact algebra in this lesson as the justification. Students ready to practice differentiation rules can continue to the Derivative Practice Checker.