Calculus / Unit 3: The Derivative
3.6 Equation of a Tangent Line
Notes
To write the tangent line at $x = a$: find the point $(a, f(a))$, find the slope $f'(a)$, then use point-slope form $y - f(a) = f'(a)(x - a)$.
- Example: $f(x) = x^2$ at $x = 3$. The point is $(3, 9)$ and the slope is $6$. So $y - 9 = 6(x - 3)$, or $y = 6x - 9$.
Practice
Find the tangent line to $f(x) = x^3$ at $x = 1$.
The point is $(1, 1)$ and $f'(1) = 3$. Then $y - 1 = 3(x - 1)$.
Find the tangent line to $\displaystyle f(x) = \frac{1}{x}$ at $x = 2$ (use $\displaystyle f'(x) = -\frac{1}{x^2}$).
The point is $\displaystyle (2, \frac{1}{2})$ and the slope is $\displaystyle -\frac{1}{4}$. Then $\displaystyle y = \frac{1}{2} - \frac{1}{4}(x - 2) = -\frac{1}{4}x + 1$.
A function has $f(3) = 10$ and $f'(3) = -2$. What is the tangent line at $x = 3$?
Use point-slope form with the point $(3, 10)$ and slope $-2$.
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