Grade by Grade
D. Characteristics of Functions · Rates of Change

The Difference Quotient: Tangent Slopes Algebraically

D1.6 D1.8About 70 minutes
Learning goals
  • I can write and simplify the difference quotient (f(a + h) − f(a)) ÷ h for polynomial functions.
  • I can find the exact slope of a tangent by letting h approach 0 after simplifying.
  • I can find tangent slopes at a general point x = a and spot patterns.
  • I can use exact tangent slopes to find velocities, tangent lines and points with horizontal tangents.

Words to know

difference quotient
The expression (f(a + h) − f(a)) ÷ h, the slope of the secant from x = a to x = a + h.
let h approach 0
Make h smaller and smaller. After simplifying, you can find the value the expression approaches by putting h = 0.
exact value
A value found algebraically, without rounding, such as a tangent slope of exactly 5.
slope function
A formula that gives the slope of the tangent at any x, such as 2a for y = x². In MCV4U this is called the derivative.

From estimates to exact answers

So far you have estimated tangent slopes with tables: 4.1, 4.01, 4.001… "so probably 4". That works, but it never proves the answer is exactly 4. In this lesson you will use algebra to get the exact slope, for polynomial functions like the ones in Polynomial Functions and Equations.

The tool is the difference quotient:

(f(a + h) − f(a)) ÷ h

It is just the slope of the secant from x = a to x = a + h, written with letters instead of numbers.

xyaa + hhf(a + h) − f(a)(a, f(a))(a + h, f(a + h))difference quotientf(a + h) − f(a)h= slope of the secantAs h → 0: slope ofthe tangent at x = a.
The difference quotient is rise over run for the secant from (a, f(a)) to (a + h, f(a + h)). The run is h; the rise is f(a + h) − f(a).

The method

  1. Find f(a) and f(a + h). Expand f(a + h) carefully.
  2. Subtract: f(a + h) − f(a). Every term without h should cancel.
  3. Divide by h. Every remaining term has a factor of h, so you can cancel it (h ≠ 0).
  4. Now let h approach 0: every term that still contains h disappears. What is left is the exact tangent slope.

The order matters. You cannot put h = 0 at the start (that gives 0 ÷ 0). You simplify first, and only then let h approach 0.

Worked example

Example 1: f(x) = x² at x = 3

Find the exact slope of the tangent to y = x² at x = 3.

  1. f(3) = 9 and f(3 + h) = (3 + h)² = 9 + 6h + h².
  2. f(3 + h) − f(3) = 6h + h².
  3. Divide by h: (6h + h²) ÷ h = 6 + h.
  4. As h approaches 0, 6 + h approaches 6.

The tangent slope is exactly 6. (Compare: the table in the last lessons gave 6.1, 6.01, … which is 6 + h for h = 0.1, 0.01.)

Worked example

Example 2: A quadratic and its tangent line

For f(x) = 2x² − 3x + 1, find the slope of the tangent at x = 2 and the equation of the tangent.

  1. f(2) = 8 − 6 + 1 = 3.
  2. f(2 + h) = 2(4 + 4h + h²) − 3(2 + h) + 1 = 8 + 8h + 2h² − 6 − 3h + 1 = 3 + 5h + 2h².
  3. f(2 + h) − f(2) = 5h + 2h². Divide by h: 5 + 2h.
  4. As h → 0, the slope approaches 5.
  5. Tangent: y − 3 = 5(x − 2), so y = 5x − 7.

Slope 5; tangent line y = 5x − 7.

Worked example

Example 3: A cubic

Find the slope of the tangent to f(x) = x³ at x = 1.

  1. Expand: (1 + h)³ = 1 + 3h + 3h² + h³. (Use (1 + h)² = 1 + 2h + h², then multiply by (1 + h).)
  2. f(1 + h) − f(1) = 3h + 3h² + h³.
  3. Divide by h: 3 + 3h + h².
  4. As h → 0: 3.

Slope 3, so the tangent is y − 1 = 3(x − 1), or y = 3x − 2. This is the tangent you met in the lesson Instantaneous Rate of Change: The Slope of a Tangent, now proved exactly.

Worked example

Example 4: The slope at any point

Find the slope of the tangent to f(x) = x² at a general point x = a. Use it to list the slopes at x = −2, −1, 0, 1 and 2.

  1. f(a + h) − f(a) = (a² + 2ah + h²) − a² = 2ah + h².
  2. Divide by h: 2a + h. As h → 0: 2a.
  3. Slopes: at −2: −4; at −1: −2; at 0: 0; at 1: 2; at 2: 4.

The tangent slope of y = x² at x = a is 2a. One calculation gives every tangent slope at once.

−2−112246xyslope −4slope −2slope 0slope 2slope 4y = x²tangent slope at x = a:2aNegative on the left,zero at the vertex,positive on the right.
Short pieces of the tangent lines to y = x² at x = −2, −1, 0, 1 and 2. The slopes are 2a: −4, −2, 0, 2 and 4.
Worked example

Example 5: Velocity of the ball

For h(t) = −4.9t² + 20t + 1.5, find the velocity at any time t = a. Use it to find the velocity at t = 3 and the time when the ball reaches its maximum height.

  1. The letter h is already used for height, so call the small step k: h(a + k) = −4.9(a² + 2ak + k²) + 20(a + k) + 1.5.
  2. h(a + k) − h(a) = −9.8ak − 4.9k² + 20k.
  3. Divide by k: −9.8a − 4.9k + 20. As k → 0: v(a) = 20 − 9.8a.
  4. At t = 3: v = 20 − 29.4 = −9.4 m/s (falling).
  5. Maximum height when the velocity is 0: 20 − 9.8a = 0, so a = 20 ÷ 9.8 ≈ 2.04 s.

v(t) = 20 − 9.8t; at t = 3 s the ball falls at 9.4 m/s; the top is reached at t ≈ 2.04 s. (In MCR3U you would have found this vertex with t = −b ÷ (2a) = 20 ÷ 9.8: the same answer.)

Looking ahead to MCV4U

The slope function you found in Example 4 (2a for y = x²) is called the derivative. In Calculus and Vectors (MCV4U), the unit Limits and Rates of Change writes this process as a limit, lim (h → 0) of the difference quotient, and the next units find shortcuts so you rarely need to expand by hand.

Just for fun

Calculus students cancel h so often that h has started taking it personally. Don't worry: it always comes back in the next question.

Hands-on activity

Slope patterns on paper

You need: paper, a pencil and graph paper.

  1. Use the difference quotient to find the tangent slope of y = x³ at a general point x = a. (Expand (a + h)³ = a³ + 3a²h + 3ah² + h³.)
  2. Use your formula to fill in a table of slopes at x = −2, −1, 0, 1, 2.
  3. On graph paper, plot the points (x, slope). What shape do they make?
  4. Do the same for y = x² (slope 2a). Compare the two "slope graphs".

What to look for: for y = x³ the slope is 3a², giving 12, 3, 0, 3, 12. These points lie on a parabola. For y = x² the slope points (−4, −2, 0, 2, 4) lie on a straight line. The slope graph of a degree-n polynomial is a polynomial of degree n − 1. You will use slope graphs again in the lesson Reading Rates of Change from Graphs.

Your turn

1

Use the difference quotient to find the tangent slope of f(x) = x² at x = 5.

2

Find the tangent slope of f(x) = 3x² at x = 1.

3

Find the tangent slope of f(x) = x² + 4x at x = −1.

4

Find the tangent slope of f(x) = 5 − x² at x = 2, and the tangent equation.

5

Find the tangent slope of f(x) = x³ at x = 2, and the tangent equation.

6

Use v(t) = 20 − 9.8t to find the ball's velocity at t = 1. Does it agree with the earlier tangent slope of 10.2?

7

Find the difference quotient of f(x) = 2x + 5 at any a. Explain the result.

More practice: warm-up

1

Expand (3 + h)².

2

Expand (1 + h)³.

3

Simplify (6h + h²) ÷ h, for h ≠ 0.

4

What does 6 + h approach as h approaches 0?

5

Using "slope = 2a" for y = x², what is the tangent slope at x = 4?

More practice: on your own

1

Find the tangent slope of f(x) = x² − 6x + 2 at x = 3. What does your answer tell you about the point (3, f(3))?

2

Find the tangent slope and the tangent equation for f(x) = 4x − x³ at x = 1.

3

Use "slope = 2a" for y = x² to find the tangent slopes at x = −3, 0 and 2.5.

4

A company's profit is P(x) = −2x² + 40x − 50 (thousands of dollars) when it sells x thousand items. Find the instantaneous rate of change of profit at x = 5 and interpret it.

5

Where is the tangent to y = x² − 4x horizontal?

Challenge

1

Show that the tangent slope of f(x) = ax² + bx + c at x = p is 2ap + b. Use this to explain the vertex formula x = −b ÷ (2a) from Grade 11.

2

The tangent slope of y = x³ at x = a is 3a². Find all points where the tangent slope is 12, and the tangent at each.

3

Find the tangent slope of y = x⁴ at x = 1. Using your results for x², x³ and x⁴ at x = 1, predict the tangent slope of y = x¹⁰ at x = 1.

Check your understanding

Quiz

1.

What does the difference quotient (f(a + h) − f(a)) ÷ h represent?

2.

For f(x) = x², what does (f(3 + h) − f(3)) ÷ h simplify to?

3.

What is the exact tangent slope of f(x) = 2x² − 3x + 1 at x = 2?

4.

What is the tangent slope of y = x³ at x = 2?

5.

A ball has velocity v(t) = 20 − 9.8t m/s. What is its velocity at t = 3 s?

6.

Why do we simplify and cancel h before letting h approach 0?

For parents and teachers

Big ideas (D1.6, D1.8): the difference quotient (f(a + h) − f(a)) ÷ h is the slope of a secant. For polynomials, expanding, cancelling h and then letting h approach 0 gives the exact tangent slope. Doing it at a general point x = a gives a slope formula, which is the idea of the derivative in MCV4U.

How to help: The algebra is the hard part. Ask your teen to say each step out loud: "expand, subtract, divide by h, let h go to 0". Check expansions together: (a + h)² has three terms, (a + h)³ has four.

Common difficulties:

  • Expanding (3 + h)² as 9 + h² (forgetting the middle term 6h).
  • Forgetting to distribute a coefficient or a minus sign: −(4 + 4h + h²) = −4 − 4h − h².
  • Letting h → 0 before cancelling, which gives 0 ÷ 0.
  • Mixing up the letter h for height and h for the small step; use another letter (such as k) when needed.