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Tangent line equations to a function graph

A tangent line equation is the equation of a straight line that touches a function graph at a given point. On this page: the general formula, problems on finding the tangent line by abscissa x₀, by ordinate y₀, and problems on tangent and parallel lines, with detailed solutions.

Tangent line equation at a point with abscissa x₀

General form of the tangent line equation:

y = f (x 0) + f' (x 0) \cdot (x - x 0)

Problem 3a: f(x) = x³ − 3x², x₀ = −1

Solution:

f (- 1) = (- 1) 3 - 3 \cdot (- 1) 2 = - 1 - 3 = - 4

f' (x) = 3 x 2 - 6 x

f' (- 1) = 3 \cdot (- 1) 2 - 6 \cdot (- 1) = 3 + 6 = 9

y = - 4 + 9 (x + 1) = 9 x + 5

Answer: $y = 9 x + 5$

Problem 3b: f(x) = 2x² − 5x + 1, x₀ = 2

Solution:

f (2) = 2 \cdot 4 - 5 \cdot 2 + 1 = 8 - 10 + 1 = - 1

f' (x) = 4 x - 5

f' (2) = 8 - 5 = 3

y = - 1 + 3 (x - 2) = 3 x - 7

Answer: $y = 3 x - 7$

Problem 3c: f(x) = −x² + x − 1, x₀ = −2

Solution:

f (- 2) = - 4 - 2 - 1 = - 7

f' (x) = - 2 x + 1

f' (- 2) = 4 + 1 = 5

y = - 7 + 5 (x + 2) = 5 x + 3

Answer: $y = 5 x + 3$

Problem 3d: f(x) = 3x² − 4x − 2, x₀ = −1

Solution:

f (- 1) = 3 + 4 - 2 = 5

f' (x) = 6 x - 4

f' (- 1) = - 6 - 4 = - 10

y = 5 - 10 (x + 1) = - 10 x - 5

Answer: $y = - 10 x - 5$

Problem 3e: f(x) = x² − x³, x₀ = −1

Solution:

f (- 1) = 1 - (- 1) = 2

f' (x) = 2 x - 3 x 2

f' (- 1) = - 2 - 3 = - 5

y = 2 - 5 (x + 1) = - 5 x - 3

Answer: $y = - 5 x - 3$

Problem 3f: f(x) = x³ + x, x₀ = 2

Solution:

f (2) = 8 + 2 = 10

f' (x) = 3 x 2 + 1

f' (2) = 12 + 1 = 13

y = 10 + 13 (x - 2) = 13 x - 16

Answer: $y = 13 x - 16$

Tangent line equation at a point with ordinate y₀

Problem 4a: f(x) = 2x − x², y₀ = −3

Solution:

2 x - x 2 = - 3

x 2 - 2 x - 3 = 0

x 1 = - 1, x 2 = 3

For x₀ = −1:

f' (x) = 2 - 2 x

f' (- 1) = 4

y = - 3 + 4 (x + 1) = 4 x + 1

For x₀ = 3:

f' (3) = - 4

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

Answer: $y = 4 x + 1$ or $y = - 4 x + 9$

Problem 4b: f(x) = 5x^(2/5) + 27, y₀ = 32

Solution:

5 x \frac{2}{5} + 27 = 32

x \frac{2}{5} = 1

x = 1

f' (x) = 2 x \frac{- 3}{5}

f' (1) = 2

y = 32 + 2 (x - 1) = 2 x + 30

Answer: $y = 2 x + 30$

Problem 4c: f(x) = x² − 4x, y₀ = −3

Solution:

x 2 - 4 x = - 3

x 2 - 4 x + 3 = 0

x 1 = 1, x 2 = 3

For x₀ = 1:

f' (x) = 2 x - 4

f' (1) = - 2

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

For x₀ = 3:

f' (3) = 2

y = - 3 + 2 (x - 3) = 2 x - 9

Answer: $y = - 2 x - 1$ or $y = 2 x - 9$

Problem 4d: f(x) = 11 − 3x^(4/3), y₀ = 8

Solution:

11 - 3 x \frac{4}{3} = 8

x \frac{4}{3} = 1

x = 1

f' (x) = - 4 x \frac{1}{3}

f' (1) = - 4

y = 8 - 4 (x - 1) = - 4 x + 12

Answer: $y = - 4 x + 12$

Tangent line and parallel lines

Problem 5a: f(x) = x² − 3x − 4, x₀ = 1

Solution:

f (1) = 1 - 3 - 4 = - 6

f' (x) = 2 x - 3

f' (1) = - 1

y = - 6 - 1 (x - 1) = - x - 5

A parallel line has the same slope k = -1:

y = - x + C, C \neq - 5

Answer: $y = - x - 5$ , parallel line: $y = - x + 2$

Problem 5b: f(x) = x³ − 3x², x₀ = −1

Solution:

f (- 1) = - 1 - 3 = - 4

f' (x) = 3 x 2 - 6 x

f' (- 1) = 3 + 6 = 9

y = - 4 + 9 (x + 1) = 9 x + 5

A parallel line has the same slope k = 9:

y = 9 x + C, C \neq 5

Answer: $y = 9 x + 5$ , parallel line: $y = 9 x - 2$

Problem 5c: f(x) = x² − 2x − 3, x₀ = 2

Solution:

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

f' (x) = 2 x - 2

f' (2) = 2

y = - 3 + 2 (x - 2) = 2 x - 7

A parallel line has the same slope k = 2:

y = 2 x + C, C \neq - 7

Answer: $y = 2 x - 7$ , parallel line: $y = 2 x + 1$

Problem 5d: f(x) = −x³ + x + 1, x₀ = −2

Solution:

f (- 2) = 8 + 2 + 1 = 11

f' (x) = - 3 x 2 + 1

f' (- 2) = - 12 + 1 = - 11

y = 11 - 11 (x + 2) = - 11 x - 11

A parallel line has the same slope k = -11:

y = - 11 x + C, C \neq - 11

Answer: $y = - 11 x - 11$ , parallel line: $y = - 11 x + 5$

Tangent line equations

Tangent line equations to a function graph.