Grade by Grade
A. Rate of Change · Derivatives and Differentiation Rules

Derivatives of Polynomials: Tangents, Normals and Points with a Given Slope

A3.3About 70 minutes
Learning goals
  • I can find the instantaneous rate of change of a polynomial at a point.
  • I can write the equations of tangent and normal lines.
  • I can find the points where a polynomial has a given rate of change by solving f′(x) = k.
  • I can find tangents parallel or perpendicular to a given line.

Words to know

tangent line
The line that touches a curve at a point and has the same slope as the curve there.
normal line
The line through a point of a curve that is perpendicular to the tangent at that point.
negative reciprocal
For a slope m ≠ 0, the number −1/m. Perpendicular lines have slopes that are negative reciprocals.
horizontal tangent
A tangent with slope 0, found by solving f′(x) = 0.

Putting the rules to work

You can now differentiate any polynomial in a line or two. In this lesson you will use those derivatives to answer two kinds of questions:

  1. Given a point, find the rate: what is the slope of the tangent at x = a, and what are the equations of the tangent and normal lines there?
  2. Given a rate, find the points: where on the curve is the slope equal to a given number, such as 0 (horizontal tangents) or 36?

The second kind is new: you set f′(x) equal to a number and solve an equation, often a quadratic or a factorable cubic from MHF4U's Polynomial Functions and Equations.

Tangent and normal lines

  1. Find f′(x).
  2. Slope: m = f′(a). Point: (a, f(a)). Use f for the point and f′ for the slope.
  3. Tangent: y − f(a) = m(x − a).
  4. Normal (the line perpendicular to the tangent at the same point): slope −1/m, so y − f(a) = (−1/m)(x − a). If m = 0, the normal is the vertical line x = a.
xy1234512345(3, 2)curve: y = x² − 4x + 5tangent: slope 2y = 2x − 4normal: slope −1/2y = −x/2 + 7/2The slopes multiply to −1,so the lines are perpendicular.
The tangent and the normal to y = x² − 4x + 5 at (3, 2), drawn on a grid with equal scales so the right angle looks right. The normal is perpendicular to the tangent at the point of tangency.
Worked example

Example 1: Tangent and normal

Find the equations of the tangent and the normal to y = x² − 4x + 5 at x = 3.

  1. dy/dx = 2x − 4. Slope at x = 3: 2(3) − 4 = 2.
  2. Point: y = 9 − 12 + 5 = 2, so (3, 2).
  3. Tangent: y − 2 = 2(x − 3), so y = 2x − 4.
  4. Normal slope: −1/2. Normal: y − 2 = −(1/2)(x − 3), so y = −x/2 + 7/2 (or x + 2y − 7 = 0).

Tangent y = 2x − 4; normal x + 2y − 7 = 0.

Worked example

Example 2: Where is the slope 36?

Find the points on the graph of f(x) = x³ + 3x² − 9x + 1 where the tangent slope is 36, and the equations of the tangents there.

  1. f′(x) = 3x² + 6x − 9.
  2. Set 3x² + 6x − 9 = 36: 3x² + 6x − 45 = 0, so x² + 2x − 15 = 0.
  3. Factor: (x + 5)(x − 3) = 0, so x = −5 or x = 3.
  4. f(3) = 27 + 27 − 27 + 1 = 28 and f(−5) = −125 + 75 + 45 + 1 = −4.
  5. Tangent at (3, 28): y − 28 = 36(x − 3), so y = 36x − 80. Tangent at (−5, −4): y + 4 = 36(x + 5), so y = 36x + 176.

Points (3, 28) and (−5, −4); tangents y = 36x − 80 and y = 36x + 176.

xy-5-3-113-202040(3, 28)(−5, −4)f(x) = x³ + 3x² − 9x + 1
The two points on f(x) = x³ + 3x² − 9x + 1 where the tangent slope is 36. The tangents (dashed) are parallel because they have the same slope. The vertical scale is compressed, so the slopes look less steep than 36.
Worked example

Example 3: Horizontal tangents

Find all points on y = x⁴ − 8x² + 3 where the tangent is horizontal.

  1. dy/dx = 4x³ − 16x = 4x(x² − 4) = 4x(x − 2)(x + 2).
  2. dy/dx = 0 when x = 0, 2 or −2.
  3. y(0) = 3; y(2) = 16 − 32 + 3 = −13; y(−2) = −13 too (the function is even).

(0, 3), (2, −13) and (−2, −13). (Using the sign of dy/dx from the first lesson, (0, 3) is a local max and the other two are local mins.)

Worked example

Example 4: A tangent perpendicular to a given line

Find the point on y = x² + 3x where the tangent is perpendicular to the line x − 5y + 2 = 0.

  1. Rewrite the line: y = x/5 + 2/5, so its slope is 1/5.
  2. A perpendicular slope is the negative reciprocal: −5.
  3. dy/dx = 2x + 3 = −5, so x = −4.
  4. y = 16 − 12 = 4.

The point is (−4, 4), where the tangent is y = −5x − 16.

Just for fun

A tangent line and a normal line meet at a party. The normal says, "I'm just here to be perpendicular." The tangent says, "Me too, I only touched on the topic."

Hands-on activity

Tangent lines with a string and a grid

You need: graph paper, a pencil, a ruler, a piece of thread or thin string and a square corner (the corner of a book or a sheet of paper).

  1. Plot y = x² − 4x + 5 carefully for x from 0 to 5 using whole-number and half-number x-values. Use the same scale on both axes.
  2. Using Example 1, draw the tangent line y = 2x − 4 with the ruler. Check that it touches the curve only at (3, 2) and stays below the curve elsewhere.
  3. Place the square corner of a book at (3, 2) with one edge on the tangent. Draw along the other edge: that is the normal.
  4. Check that your normal passes through (5, 1) and (1, 3), as y = −x/2 + 7/2 predicts.
  5. Stretch the string from (3, 2) along the normal. Where does it cross the y-axis? (It should be 3.5.)

What to look for: a drawn normal that passes within about one small square of (5, 1) and (1, 3). The normal is perpendicular only because the scales are equal; with unequal scales a right angle will not look like one.

Your turn

1

Find the slope of the tangent to y = 2x³ − x at x = −1.

2

Find the equation of the tangent to y = x³ − 2x² at x = 2.

3

Find the equation of the normal to y = x² at (1, 1).

4

Find the points on y = x³ − 12x + 1 where the tangent is horizontal.

5

At what point on y = 3x² − 5x is the tangent slope 7?

6

Find the point on y = x² − x where the tangent is parallel to y = 3x + 8.

7

The tangent at some point on a curve is horizontal. What is the normal there?

More practice: warm-up

1

Slope of the tangent to y = x² + 1 at x = 4?

2

A tangent has slope 3. What is the slope of the normal?

3

A tangent has slope −2/5. Slope of the normal?

4

Write the line through (2, 5) with slope 4.

5

For f(x) = x³, find f(2) and f′(2).

More practice: on your own

1

Find the equations of the tangent and normal to y = x³ − 3x + 2 at x = 1.

2

Find the points on y = x³ − 3x² − 9x + 4 where the tangent slope is −9.

3

Find the equation of the normal to y = 2x² − 3x + 1 at x = 1.

4

Find the point on y = x² + 4x where the tangent is perpendicular to y = −x/2 + 9.

5

Show that the tangent to y = x³ − x at x = 1 passes through the point (0, −2).

Challenge

1

Find the equations of the two tangents to y = x² + 2 that pass through the origin.

2

The normal to y = x² at (1, 1) meets the parabola again. Find the other point.

3

Find k so that the line y = 4x + k is tangent to y = x³ − 8x.

Check your understanding

Quiz

1.

To write the tangent to y = f(x) at x = a, you need:

2.

The tangent to a curve at a point has slope 4. What is the slope of the normal there?

3.

At which x-values does y = x³ − 27x have horizontal tangents?

4.

What is the tangent to y = x² + 1 at x = 2?

5.

At what point on y = x² − 6x is the tangent parallel to y = 2x − 1?

6.

For f(x) = x³ + 3x² − 9x + 1, how many points have tangent slope 36?

For parents and teachers

Big idea (A3.3): derivatives of polynomials answer two kinds of question: "what is the rate at this point?" (substitute into f′) and "where is the rate equal to this?" (solve f′(x) = value).

How to help: When your teen writes a tangent line, ask "Which number came from f and which from f′?" Mixing them up is the most common error. For normal lines, ask them to say "flip and change the sign".

Common difficulties:

  • Using f′(a) as the y-coordinate of the point.
  • Forgetting the second (or third) solution when solving f′(x) = k.
  • Writing the normal slope as −m instead of −1/m.
  • Not rewriting a line in the form y = mx + b before reading its slope.