Once you know the derivative at a point, the tangent line is ordinary algebra: a point and a slope give a line by Point-Slope Form.
Key idea
The tangent line to $y = f(x)$ at $x = a$ passes through $(a, f(a))$ with slope $f'(a)$: $$y = f(a) + f'(a)(x - a).$$ Near $a$, this line is the best straight-line description of the curve.
Why it works
Point-slope form says a line through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$. The tangent line touches the curve at $(a, f(a))$ and has the curve's slope there, $f'(a)$. Putting those in gives the formula. The slope must be a single number, because a line has one slope everywhere.
Worked example: From given values
$f(2) = 5$ and $f'(2) = -3$. Write the tangent line at $x = 2$.
- Point $(2, 5)$, slope $-3$.
- $y - 5 = -3(x - 2)$.
- $y = -3x + 6 + 5 = -3x + 11$.
Answer: $y = -3x + 11$
Worked example: From a formula
Find the tangent line to $f(x) = x^2$ at $x = 3$.
- Point: $f(3) = 9$.
- Slope: $\frac{(3 + h)^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h \to 6$, so $f'(3) = 6$.
- $y - 9 = 6(x - 3)$, so $y = 6x - 9$.
Answer: $y = 6x - 9$
Worked example: A cubic, and a surprise
Find the tangent line to $f(x) = x^3$ at $x = -1$. Where else does it meet the curve?
- Point: $f(-1) = -1$.
- Slope: $$\frac{(-1 + h)^3 - (-1)}{h} = \frac{3h - 3h^2 + h^3}{h} = 3 - 3h + h^2 \to 3$$.
- Line: $y = -1 + 3(x + 1) = 3x + 2$.
- Set $x^3 = 3x + 2$: $x^3 - 3x - 2 = (x + 1)^2(x - 2) = 0$, so the line meets the curve again at $x = 2$.
Answer: $y = 3x + 2$. It touches the curve at $x = -1$ and crosses it at $(2, 8)$.
The surprise is worth keeping: "tangent" is a local idea. A tangent line hugs the curve near the point of tangency; far away it may cross the curve. That local closeness is what makes tangent lines so useful later, for linear approximation and for Newton's Method.
Common mistake
Using the derivative formula as the slope, for example writing $y - 9 = 2x(x - 3)$ for the parabola above. That is a parabola, not a line. The slope of the tangent line is the number $f'(3) = 6$. Check: your answer should have the form $y = mx + b$, and it should pass through $(a, f(a))$.
If $f'(a) = 0$, the tangent line is horizontal: $y = f(a)$. Those points will matter a lot when we look for maximum and minimum values.