The equation of the tangent line to y = f(x) at x = a is y − f(a) = f'(a)(x − a).
Example ①
Find the equation of the tangent line to y = x² at x = 3.
Answer:f'(x) = 2x, f'(3) = 6, f(3) = 9.
y − 9 = 6(x − 3) → y = 6x − 9
Find the equation of the tangent line to y = x² at x = 3.
Answer:f'(x) = 2x, f'(3) = 6, f(3) = 9.
y − 9 = 6(x − 3) → y = 6x − 9
Example ②
Find the equation of the line tangent to y = x³ − x at the point (2, 6).
Answer:f'(x) = 3x²−1, f'(2) = 11, f(2) = 6.
y − 6 = 11(x − 2) → y = 11x − 16
Find the equation of the line tangent to y = x³ − x at the point (2, 6).
Answer:f'(x) = 3x²−1, f'(2) = 11, f(2) = 6.
y − 6 = 11(x − 2) → y = 11x − 16
Example ③ (Tangent from an External Point)
Find the equations of the tangent lines drawn from the point (1, −1) (which is not on the curve) to the curve y = x².
Answer:Let the point of tangency be (a, a²). The slope is = 2a. Tangent line: y − a² = 2a(x − a) → y = 2ax − a².
Since it passes through (1, −1), −1 = 2a − a² → a² − 2a − 1 = 0 → a = 1 ± √2.
Tangent lines: y = (2+2√2)x − (3+2√2) and y = (2−2√2)x − (3−2√2)
Find the equations of the tangent lines drawn from the point (1, −1) (which is not on the curve) to the curve y = x².
Answer:Let the point of tangency be (a, a²). The slope is = 2a. Tangent line: y − a² = 2a(x − a) → y = 2ax − a².
Since it passes through (1, −1), −1 = 2a − a² → a² − 2a − 1 = 0 → a = 1 ± √2.
Tangent lines: y = (2+2√2)x − (3+2√2) and y = (2−2√2)x − (3−2√2)
Key Points
- Tangent line: y − f(a) = f'(a)(x − a)
- Tangent from an external point: represent the point of tangency with a variable and use the condition that the line passes through the point
Practice Problems
- Find the equation of the tangent line to y = x³ at x = −1.
- Find the coordinates of the point on y = x² − 2x where the tangent has slope 4, and find the equation of the tangent line.
- Find the equations of the tangent lines drawn from the point (0, −1) to y = x².
Answers and Explanations
- Answer:y = 3x + 2
Explanation: f'(x)=3x², f'(−1)=3, f(−1)=−1. y+1=3(x+1)→y=3x+2. - Answer:Point (3, 3), tangent line y = 4x − 9
Explanation: f'(x)=2x−2=4→x=3, f(3)=9−6=3. y−3=4(x−3). - Answer:y = 2x − 1 and y = −2x − 1
Explanation: Point of tangency (a,a²). Since y=2ax−a² passes through (0,−1): −1=−a²→a=±1.