D. Characteristics of Functions · Rates of Change

Secants Approaching a Tangent

D1.7 D1.8About 70 minutes

Learning goals

I can find the slopes of secants PQ as Q moves closer to a fixed point P, from both sides.
I can use those slopes to estimate the slope of the tangent at P for quadratic, exponential and sinusoidal functions.
I can write the equation of the tangent line once I know its slope.
I can connect secant slopes to average rates and the tangent slope to the instantaneous rate.

Words to know

fixed point P
The point where we want the tangent slope. It stays put while Q moves.

approach
Get closer and closer to a value without necessarily reaching it.

tangent line equation
y − f(a) = m(x − a), where m is the slope of the tangent at (a, f(a)).

linear approximation
Using the tangent line to estimate values of a function close to the point of tangency.

Watching a line swing

Pin one end of a ruler at a point P on a curve. Lay the ruler across the curve so it also crosses at a second point Q. That is a secant. Now slide Q along the curve toward P. The ruler swings. As Q gets very close to P, the ruler settles into one position: the tangent at P.

In the last lesson you shrank intervals using numbers. Today you will see the same idea on the graph: secants approaching a tangent.

The method

Fix the point P(a, f(a)).
Choose points Q(x, f(x)) on the curve closer and closer to P, from the right (x > a) and from the left (x < a).
Calculate each secant slope: m(PQ) = (f(x) − f(a)) ÷ (x − a).
See what value the slopes approach from both sides. That value is the slope of the tangent at P.

Each secant slope is an average rate of change. The tangent slope they approach is the instantaneous rate of change at P.

Worked example

Example 1: y = x² at P(2, 4)

Find the slopes of secants PQ for Q with x = 3, 2.5, 2.1, 2.01 and x = 1, 1.5, 1.9, 1.99. Estimate the slope of the tangent at P and write its equation.

Right side: Q(3, 9): (9 − 4) ÷ (3 − 2) = 5. Q(2.5, 6.25): 2.25 ÷ 0.5 = 4.5. Q(2.1, 4.41): 0.41 ÷ 0.1 = 4.1. Q(2.01, 4.0401): 0.0401 ÷ 0.01 = 4.01.
Left side: Q(1, 1): (1 − 4) ÷ (1 − 2) = 3. Q(1.5, 2.25): (−1.75) ÷ (−0.5) = 3.5. Q(1.9, 3.61): (−0.39) ÷ (−0.1) = 3.9. Q(1.99, 3.9601): 3.99.
From both sides the slopes approach 4.
Tangent: y − 4 = 4(x − 2), so y = 4x − 4.

Tangent slope 4; tangent line y = 4x − 4. The instantaneous rate of change of y = x² at x = 2 is 4.

Secants from P(2, 4) to Q₁, Q₂ and Q₃ on y = x². Their slopes (6, 5, 4.5) decrease toward the tangent slope, 4, as Q slides toward P.

Each dot shows the slope of one secant PQ, plotted against the x-coordinate of Q. The dots close in on 4 from both sides. There is no secant when Q = P, so that spot is an open circle.

Worked example

Example 2: An exponential, y = 2ˣ at P(0, 1)

Estimate the slope of the tangent to y = 2ˣ at P(0, 1) using secants. Write the tangent's equation.

Slope of PQ = (2ˣ − 1) ÷ (x − 0) = (2ˣ − 1) ÷ x.
Right: x = 1: 1. x = 0.1: 0.7177. x = 0.01: 0.6956. x = 0.001: 0.6934.
Left: x = −1: 0.5. x = −0.1: 0.6697. x = −0.01: 0.6908. x = −0.001: 0.6929.
The slopes squeeze in on about 0.693 from both sides.

Tangent slope ≈ 0.693; tangent line y ≈ 0.693x + 1. Notice the slopes from the right are all a bit bigger and the slopes from the left a bit smaller; the true value is trapped between.

Worked example

Example 3: A sinusoidal function, y = sin x at P(0, 0)

Estimate the slope of the tangent to y = sin x at the origin (x in radians).

Slope of PQ = (sin x − 0) ÷ (x − 0) = sin x ÷ x.
x = 1: 0.8415. x = 0.1: 0.99833. x = 0.01: 0.99998.
Because sin(−x) = −sin x, the left side gives exactly the same values (for example x = −0.1 gives 0.99833).
The slopes approach 1.

Tangent slope 1, so the tangent at the origin is y = x. This is why, for small angles in radians, sin x ≈ x. (This only works in radians, one more reason calculus uses them.)

With graphing technology

In a free graphing tool such as Desmos or GeoGebra: graph f, make a point P on it, and make a second point Q(a + h, f(a + h)) using a slider for h. Draw the line through P and Q and display its slope. Drag the slider toward 0 from both sides and watch the slope settle. Most tools can also draw the tangent and report its slope directly, which is a good check.

Worked example

Example 4: Using the tangent to estimate

Use the tangent y = 4x − 4 to y = x² at (2, 4) to estimate 2.05². Compare with the true value.

Near x = 2, the curve and its tangent are almost the same (local linearity).
Tangent at x = 2.05: 4(2.05) − 4 = 8.2 − 4 = 4.2.
True value: 2.05² = 4.2025.

Estimate 4.2, true value 4.2025: off by only 0.0025. This is called a linear approximation; it is used constantly in science and engineering.

Just for fun

In this lesson, Q spends the whole time walking toward P and is never allowed to arrive. Mathematicians call this a limit. Q calls it unfair.

Hands-on activity

Swing the ruler

You need: graph paper, a ruler, a pencil, a calculator (or a free graphing site on a computer).

Draw y = x² for 0 ≤ x ≤ 4 on graph paper, as large as the page allows. Mark P(2, 4).
Mark Q at x = 4, 3, 2.5 and 2.2. For each, draw the secant PQ (extend it across the page), and calculate its slope from the coordinates.
Hold your pencil tip on P and rest the ruler against it. Slowly rotate the ruler as you imagine Q moving closer to P. Draw the line where it just touches the curve at P.
Measure the slope of your drawn tangent using two points far apart on it. Compare with 4.
Repeat on the left side with Q at x = 0, 1 and 1.5.

What to look for: the secant slopes on the right (6, 5, 4.5, 4.2) decrease toward 4; on the left (2, 3, 3.5) they increase toward 4. Your drawn tangent should have slope close to 4 and pass close to (1, 0) and (3, 8).

Your turn

For y = x² and P(2, 4), find the slope of PQ when Q has x = 2.001.

For y = x² and P(1, 1), find the slopes of PQ for x = 1.1 and x = 1.01. Estimate the tangent slope and write the tangent equation.

For y = x³ and P(1, 1), find the slopes of PQ for x = 1.1, 1.01 and 0.99. Estimate the tangent slope.

Write the equation of the tangent to y = x³ at (1, 1).

For y = 3ˣ and P(0, 1), find the slope of PQ when x = 0.001.

For y = cos x (radians) and P(0, 1), find the slope of PQ when x = 0.1. What is the tangent at P?

Why can't we just put Q at P to get the tangent slope directly?

More practice: warm-up

Find the slope of PQ for P(2, 4) and Q(3, 9).

Find the slope of PQ for P(1, 1) and Q(1.5, 2.25).

The tangent to a curve at (2, 4) has slope 4. Write its equation.

Secant slopes from the left are 3.9, 3.99, 3.999 and from the right are 4.1, 4.01, 4.001. What is the tangent slope?

As Q moves closer to P along a curve, what do the secants PQ approach?

More practice: on your own

For f(x) = x² − 3x and P(4, 4), find the secant slopes for x = 4.1 and 4.01. Estimate the tangent slope and write the tangent equation.

For f(x) = √x and P(4, 2), find the secant slopes for x = 4.01 and x = 3.99. Estimate the tangent slope and write the tangent equation.

Use your tangent from the previous question to estimate √4.1. Compare with a calculator.

A substance decays as f(x) = 10(0.5)ˣ grams after x days. Use the secant from P(1, 5) to Q at x = 1.001 to estimate the rate of decay at x = 1.

Explain in your own words how secant slopes are related to average and instantaneous rates of change.

Challenge

For y = x², let P = (a, a²) and Q = (a + h, (a + h)²). Show that the slope of PQ is 2a + h. Then find the tangent slope and the equation of the tangent at x = a.

Find the point on y = x² where the tangent is parallel to y = 6x + 1, and write that tangent's equation.

The tangent to y = 2ˣ at x = 0 has slope about 0.693. Show that the secant slope from x = 3 to x = 3 + h is 8 times the secant slope from x = 0 to x = h, and use this to find the tangent slope at x = 3.

Check your understanding

Quiz

On y = x², what is the slope of the secant from P(2, 4) to Q(2.1, 4.41)?

As Q moves along a curve toward a fixed point P, the slopes of the secants PQ approach:

Secants of y = 2ˣ through P(0, 1) have slopes 0.7177, 0.6956, 0.6934 from the right and 0.6697, 0.6908, 0.6929 from the left. What is the best estimate of the tangent slope?

The tangent to y = x² at (2, 4) has slope 4. What is its equation?

With x in radians, what is the slope of the tangent to y = sin x at the origin?

Which of these is NOT a valid way to estimate a tangent slope?

For parents and teachers

Big ideas (D1.7, D1.8): secants through a fixed point P and a moving point Q have slopes (average rates) that approach the tangent slope (instantaneous rate) as Q approaches P. Once the slope is known, the tangent's equation follows from y − f(a) = m(x − a).

How to help: Ask your teen to show you the table of secant slopes and explain why the numbers from the left and the right "squeeze" the answer. If you have a computer, try a free graphing tool together with a slider for Q.

Common difficulties:

Sign errors on the left side: both f(x) − f(a) and x − a are negative, so the slope is positive for y = x² near (2, 4).
Forgetting to check both sides.
Using degree mode for sin x: the slope at 0 would come out about 0.01745 instead of 1.
Writing the tangent equation with the wrong point (it must use the point of tangency).