Tangent Line & Normal Line Mastery Worksheet | Calculus

📐 The 4-Step Process:

Find derivative f'(x)
Evaluate f'(a) for slope m
Find point (a, f(a))
Use point-slope form

y - f(a) = m(x - a)

Tangent Line:

Slope = f'(a)

Normal Line:

Slope = -1/f'(a)

(Perpendicular to tangent)

Problem 1: Basic Polynomial Tangent Beginner

f(x) = x² - 4x + 1

Find the equation of the tangent line at x = 3. Also find the normal line at this point.

1 Find derivative:

f'(x) = 2x - 4

2 Slope at x = 3:

m_tangent = f'(3) = 2(3) - 4 = 2

m_normal = -1/m_tangent = -1/2

3 Point on curve:

y₁ = f(3) = 3² - 4(3) + 1 = 9 - 12 + 1 = -2

Point: (3, -2)

4 Tangent line equation:

y - (-2) = 2(x - 3)

y + 2 = 2x - 6

y = 2x - 8

5 Normal line equation:

y - (-2) = (-1/2)(x - 3)

y + 2 = (-1/2)x + 3/2

y = (-1/2)x - 1/2

💡 Verification: Tangent and normal slopes multiply to -1: 2 × (-1/2) = -1 ✓

Problem 2: Rational Function Tangent Intermediate

f(x) = (x + 1)/(x - 2)

Find the equation of the tangent line at x = 3. Where does this tangent line intersect the x-axis?

1 Find derivative (quotient rule):

f'(x) = [(1)(x-2) - (x+1)(1)]/(x-2)²

f'(x) = [x - 2 - x - 1]/(x-2)² = -3/(x-2)²

2 Slope at x = 3:

m = f'(3) = -3/(3-2)² = -3/1 = -3

3 Point on curve:

y₁ = f(3) = (3+1)/(3-2) = 4/1 = 4

Point: (3, 4)

4 Tangent line equation:

y - 4 = -3(x - 3)

y - 4 = -3x + 9

y = -3x + 13

5 Find x-intercept (y=0):

0 = -3x + 13

3x = 13

x = 13/3 ≈ 4.333

X-intercept: (13/3, 0)

💡 Geometric Insight: The tangent line touches the curve at (3,4) and crosses the x-axis at (13/3,0).

Problem 3: Implicit Tangent Line Intermediate

x² + y² = 25

Find the equation of the tangent line to the circle at the point (3, 4).

1 Differentiate implicitly:

d/dx(x²) + d/dx(y²) = d/dx(25)

2x + 2y·dy/dx = 0

2 Solve for dy/dx:

2y·dy/dx = -2x

dy/dx = -x/y

3 Slope at (3, 4):

m = -3/4

4 Tangent line equation:

y - 4 = (-3/4)(x - 3)

y - 4 = (-3/4)x + 9/4

y = (-3/4)x + 9/4 + 16/4

y = (-3/4)x + 25/4

5 Alternative verification:

Multiply by 4: 4y = -3x + 25

3x + 4y = 25

💡 Geometric Insight: For circle x² + y² = r², tangent at (x₁,y₁) is: x₁x + y₁y = r²
Check: 3·3 + 4·4 = 9 + 16 = 25 ✓

Problem 4: Horizontal Tangents (Critical Points) Advanced

f(x) = x³ - 6x² + 9x + 2

Find all points where the tangent line is horizontal. Determine if these are local maxima, minima, or neither.

1 Find derivative:

f'(x) = 3x² - 12x + 9

2 Horizontal tangent when f'(x) = 0:

3x² - 12x + 9 = 0

x² - 4x + 3 = 0

(x - 1)(x - 3) = 0

x = 1 or x = 3

3 Corresponding points:

f(1) = 1 - 6 + 9 + 2 = 6 ⇒ Point: (1, 6)

f(3) = 27 - 54 + 27 + 2 = 2 ⇒ Point: (3, 2)

4 Second derivative test:

f''(x) = 6x - 12

f''(1) = 6(1) - 12 = -6 < 0 ⇒ Local maximum at (1, 6)

f''(3) = 6(3) - 12 = 6 > 0 ⇒ Local minimum at (3, 2)

5 Tangent equations:

At (1, 6): y = 6 (horizontal line)

At (3, 2): y = 2 (horizontal line)

💡 Interpretation: The cubic has a local max at (1,6) and local min at (3,2). Both have horizontal tangents (slope = 0).

📊 Tangent Line Quick Reference

Function Type	Tangent Strategy	Common Pitfalls
Explicit: y = f(x)	1. Find f'(x) 2. Evaluate f'(a) 3. Use point-slope	Forgetting to find f(a) for the point
Implicit: F(x,y) = 0	1. Differentiate implicitly 2. Solve for dy/dx 3. Use point-slope	Forgetting chain rule on y terms
Parametric: x(t), y(t)	dy/dx = (dy/dt)/(dx/dt) Evaluate at t₀	Dividing by dx/dt when it's zero
Normal Line	Same point, slope = -1/m where m = tangent slope	Forgetting negative reciprocal

💪 Additional Practice Problems

Problem 5: f(x) = √x at x = 4
Answer: y = (1/4)x + 1

Problem 6: f(x) = sin(x) at x = π/4
Answer: y = (√2/2)x + (√2/2)(1 - π/4)

Problem 7: f(x) = ln(x) at x = 1
Answer: y = x - 1

Problem 8: x² + xy + y² = 7 at (1, 2)
Answer: y = (-4/5)x + 14/5