The Logic of the Tangent#
A tangent to a curve is a straight line that “touches” the curve at exactly one point and has the same slope as the curve at that point.
In Grade 10/11 you found the gradient of a straight line using $m = \frac{y_2 - y_1}{x_2 - x_1}$. But a curve doesn’t have a constant gradient — its slope changes at every point. The derivative gives you the gradient at any specific point.
1. The Method#
To find the equation of the tangent to $f(x)$ at the point where $x = a$:
Step 1: Find $f'(x)$#
Differentiate the function using the rules of differentiation.
Step 2: Calculate $m_{tan} = f'(a)$#
Substitute $x = a$ into the derivative to get the gradient of the tangent at that specific point.
Step 3: Find the point of contact#
If you don’t already have the $y$-coordinate, calculate $f(a)$ to get the point $(a; f(a))$.
Step 4: Use the point-gradient formula#
$$ y - y_1 = m(x - x_1) $$Substitute $m_{tan}$ and the point $(a; f(a))$, then simplify.
2. Worked Examples#
Example 1: Basic Tangent#
Find the equation of the tangent to $f(x) = x^2 - 3x + 2$ at $x = 4$.
Step 1: $f'(x) = 2x - 3$
Step 2: $m_{tan} = f'(4) = 2(4) - 3 = 5$
Step 3: $f(4) = (4)^2 - 3(4) + 2 = 16 - 12 + 2 = 6$ Point: $(4; 6)$
Step 4:
$$ y - 6 = 5(x - 4) $$$$ y = 5x - 20 + 6 $$$$ y = 5x - 14 $$Example 2: Tangent to a Cubic#
Find the equation of the tangent to $g(x) = x^3 - 6x^2 + 9x$ at $x = 1$.
Step 1: $g'(x) = 3x^2 - 12x + 9$
Step 2: $m_{tan} = g'(1) = 3(1) - 12(1) + 9 = 3 - 12 + 9 = 0$
Step 3: $g(1) = 1 - 6 + 9 = 4$ Point: $(1; 4)$
Step 4:
$$ y - 4 = 0(x - 1) $$$$ y = 4 $$The tangent is a horizontal line because $x = 1$ is a turning point (stationary point) of the cubic!
Example 3: Finding the Point Given the Gradient#
The tangent to $f(x) = -x^2 + 4x + 5$ has a gradient of $-2$. Find the point of tangency and the equation of the tangent.
Step 1: $f'(x) = -2x + 4$
Step 2: Set $f'(x) = -2$:
$$ -2x + 4 = -2 $$$$ -2x = -6 $$$$ x = 3 $$Step 3: $f(3) = -(3)^2 + 4(3) + 5 = -9 + 12 + 5 = 8$ Point: $(3; 8)$
Step 4:
$$ y - 8 = -2(x - 3) $$$$ y = -2x + 6 + 8 $$$$ y = -2x + 14 $$Example 4: Tangent Parallel to a Given Line#
Find the equation of the tangent to $f(x) = x^3 - 3x + 2$ that is parallel to the line $y = 9x + 1$.
Parallel lines have the same gradient: $m = 9$.
Step 1: $f'(x) = 3x^2 - 3$
Step 2: Set $f'(x) = 9$:
$$ 3x^2 - 3 = 9 $$$$ 3x^2 = 12 $$$$ x^2 = 4 $$$$ x = 2 \text{ or } x = -2 $$Two tangent points!
For $x = 2$: $f(2) = 8 - 6 + 2 = 4$. Point: $(2; 4)$
$$ y - 4 = 9(x - 2) \Rightarrow y = 9x - 14 $$For $x = -2$: $f(-2) = -8 + 6 + 2 = 0$. Point: $(-2; 0)$
$$ y - 0 = 9(x + 2) \Rightarrow y = 9x + 18 $$3. The Normal to a Curve#
The normal is the line perpendicular to the tangent at the same point. If the tangent gradient is $m_{tan}$, then the normal gradient is:
$$ m_{normal} = -\frac{1}{m_{tan}} $$Example#
Find the equation of the normal to $f(x) = x^2$ at $x = 3$.
$f'(x) = 2x$, so $m_{tan} = f'(3) = 6$.
$m_{normal} = -\frac{1}{6}$
$f(3) = 9$. Point: $(3; 9)$.
$$ y - 9 = -\frac{1}{6}(x - 3) $$$$ y = -\frac{1}{6}x + \frac{1}{2} + 9 $$$$ y = -\frac{1}{6}x + \frac{19}{2} $$4. Finding Where a Line is Tangent to a Curve#
Sometimes the question gives you the tangent equation and asks you to find the point of contact.
Example: The line $y = 2x + k$ is a tangent to $f(x) = x^2 + 1$. Find $k$.
Step 1: The gradient of the tangent is $2$, so $f'(x) = 2x = 2$, giving $x = 1$.
Step 2: Point of contact: $f(1) = 1 + 1 = 2$. Point: $(1; 2)$.
Step 3: Substitute into the tangent equation: $2 = 2(1) + k$, so $k = 0$.
The tangent is $y = 2x$.
🚨 Common Mistakes#
- Using $f'(x)$ instead of $f(x)$ for the $y$-coordinate: After finding $x$, you must substitute into the ORIGINAL function $f(x)$ to get $y$, not into $f'(x)$. Substituting into $f'(x)$ just gives you the gradient again.
- Forgetting the point-gradient formula: Many students find the gradient correctly but then don’t know how to write the line equation. Use $y - y_1 = m(x - x_1)$ every time.
- Multiple tangent points: When a gradient value leads to $x^2 = k$, there are TWO solutions and thus TWO tangent lines. Don’t stop at one.
- Horizontal tangent: If $m_{tan} = 0$, the tangent is simply $y = f(a)$ (a horizontal line). Students sometimes write $y = 0$ instead.
- Normal vs Tangent: If the question asks for the normal, remember to use the negative reciprocal of the gradient, not the same gradient.
💡 Pro Tip: Tangent at a Turning Point#
At a turning point, $f'(x) = 0$, so the tangent is always a horizontal line $y = q$ (where $q$ is the $y$-value of the turning point). This is the simplest tangent equation you’ll ever write — don’t overthink it!
⏮️ Power Rule | 🏠 Back to Calculus | ⏭️ Cubic Functions