Tangents and normals: from gradient to finished equation

y = x³ − 3x + 1 at (2, 3)

• The tangent has the curve’s gradient.
• The normal is at right angles.
• Gradients multiply to −1.

The tangent at x = 2

• A point, then a gradient.
• Finish as y = mx + c.

1. x = 2: y = 8 − 6 + 1 = 3the point
2. dy/dx = 3x² − 3differentiate
3. x = 2: m = 3(4) − 3 = 9the gradient
4. y − 3 = 9(x − 2)point-gradient form
5. y − 3 = 9x − 18expand
6. y = 9x − 15add 3

The normal at x = 2

• Negative reciprocal: −1/9.
• Either form is finished.

1. tangent m = 9
2. normal m = −1/9−1 ÷ 9
3. y − 3 = −(1/9)(x − 2)through (2, 3)
4. y = −x/9 + 2/9 + 3expand
5. y = −x/9 + 29/93 = 27/9
6. 9y = −x + 29× 9
7. x + 9y − 29 = 0general form
8. 9 × (−1/9) = −1at right angles

The 2025 error: unfinished equations

• Values substituted is not finished.
• y = mx + c, or general form.
• Evaluate every number fully.

A point with a given gradient

• Set dy/dx equal to 9.
• Two points: find both y.

1. dy/dx = 3x² − 3
2. 3x² − 3 = 9gradient 9
3. 3x² = 12, so x² = 4
4. x = ±2
5. x = 2: y = 8 − 6 + 1 = 3
6. x = −2: y = −8 + 6 + 1 = −1
7. (2, 3) and (−2, −1)

Same gradient: parallel tangents

• Same gradient, 9.
• Parallel lines never meet.

1. at (−2, −1): y + 1 = 9(x + 2)
2. y + 1 = 9x + 18expand
3. y = 9x + 17
4. at (2, 3): y = 9x − 15as before
5. m = 9 for both: parallelnever meet

Where a tangent meets an axis

• y = e²ˣ at x = 1.
• Then set y = 0.

1. x = 1: y = e²the point (1, e²)
2. dy/dx = 2e²ˣchain rule
3. x = 1: m = 2e²the gradient
4. y − e² = 2e²(x − 1)point-gradient form
5. y = 2e²x − 2e² + e²expand
6. y = 2e²x − e²
7. y = 0: 2e²x = e²meets the x-axis
8. x = 1/2divide by 2e²

Where two tangents meet

• Find each tangent first.
• Set the equations equal.

1. x = 0: y = 1, m = 3(0) − 3 = −3
2. y − 1 = −3(x − 0)point-gradient form
3. y = −3x + 1
4. x = 2: y = 9x − 15from before
5. −3x + 1 = 9x − 15set them equal
6. 16 = 12x, so x = 4/3add 3x + 15
7. y = −3(4/3) + 1 = −3substitute back
8. they meet at (4/3, −3)

A normal to y = tan x

• The same steps with trig.
• Normal gradient −1/2.

1. x = π/4: y = tan(π/4) = 1
2. dy/dx = sec² xderivative of tan x
3. cos²(π/4) = 1/2
4. m = sec²(π/4) = 21 ÷ (1/2)
5. normal m = −1/2−1 ÷ 2
6. y − 1 = −(1/2)(x − π/4)point-gradient form
7. y = −x/2 + π/8 + 1expand

The answers: questions 1 and 2

1. 1. x = e: y = e ln e = ethe point
2. dy/dx = ln x + 1 = 2product rule
3. y − e = 2(x − e), y = 2x − e
4. y = 0: x = e/2meets the x-axis
5. 2. x = 2: y = √9 = 3the point
6. dy/dx = 2/√(4x + 1) = 2/3chain rule
7. normal m = −3/2−1 ÷ (2/3)
8. y − 3 = −(3/2)(x − 2)
9. y = −3x/2 + 6expand

The answers: question 3

1. 3. dy/dx = 3x² − 12
2. 3x² − 12 = 15gradient 15
3. 3x² = 27, x² = 9, x = ±3
4. x = 3: y = 27 − 36 = −9
5. x = −3: y = −27 + 36 = 9
6. (3, −9) and (−3, 9)

The answers: question 4

1. 4. dy/dx = 2x
2. x = 1: point (1, 1), m = 2
3. y − 1 = 2(x − 1), y = 2x − 1
4. x = 3: point (3, 9), m = 6
5. y − 9 = 6(x − 3), y = 6x − 9
6. 2x − 1 = 6x − 9set them equal
7. 4x = 8, so x = 2
8. y = 2(2) − 1 = 3substitute back
9. they meet at (2, 3)