Topics

Tangents and normals

Covers AS91578

The gradient at a point

$f'(x)$ is a function — it gives the gradient at every point on the curve.
$f'(a)$ is a number — the gradient at the single point where $x = a$ .
Substitute after differentiating, never before. Substituting first turns the function into a constant, whose derivative is zero.

Finding the equation of a tangent

A tangent is the straight line that touches the curve at a point and has the same gradient as the curve there.
The procedure:
- Differentiate to get $f'(x)$ .
- Substitute the $x$ -value to get the gradient $m = f'(a)$ .
- Find the $y$ -coordinate by substituting into the original function, $y_1 = f(a)$ .
- Apply the straight-line formula:

$y - y_1 = m(x - x_1)$

Rearrange into the form the question asks for.
Worked through — the tangent to $y = x^3 - 2x$ at $x = 2$ :
- $f'(x) = 3x^2 - 2$ , so $m = f'(2) = 12 - 2 = 10$
- $y_1 = f(2) = 8 - 4 = 4$ , so the point is $(2, 4)$
- $y - 4 = 10(x - 2)$ , giving $y = 10x - 16$

Finding the equation of a normal

A normal is the line perpendicular to the tangent at the same point.
Perpendicular gradients multiply to $-1$ , so:

$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

The procedure is identical to the tangent, with one extra step: flip the gradient and change its sign.
Continuing the example above: $m_{\text{normal}} = -\frac{1}{10}$ , so the normal is $y - 4 = -\frac{1}{10}(x - 2)$ , or $y = -0.1x + 4.2$ .
Special cases:
- A horizontal tangent ( $m = 0$ ) has a vertical normal, written $x = a$ — the formula $-\frac{1}{0}$ does not apply.
- A vertical tangent has a horizontal normal, $y = b$ .

When you are given the gradient instead of the point

A common exam turn: "find the point(s) where the tangent has gradient 5".
Set $f'(x) = 5$ and solve for $x$ . There may be more than one answer.
Substitute each solution back into the original function for the $y$ -coordinates.
Note the error the reports flag repeatedly: candidates "misunderstood the relationship between the gradient and the derivative, instead substituting the gradient value into the original function". The gradient goes into $f'$ , not into $f$ .

Where a tangent meets the axes or another line

These are ordinary co-ordinate geometry questions once the tangent's equation is found:
- $x$ -intercept: set $y = 0$ and solve.
- $y$ -intercept: set $x = 0$ .
- Intersection of two tangents: solve the two line equations simultaneously.
The 2025 report lists "found the coordinates of the point of intersection of two tangents to a curve" and "found the coordinates of the point where a tangent to a curve crosses the $x$ -axis" as Merit behaviours.

Tangents from an external point

If a tangent passes through a point $(p, q)$ $(p, q)$ that is not on the curve, the touch point is unknown. Call it $x = a$ $x = a$ :
- The gradient at the touch point is $f'(a)$ .
- The gradient of the line joining $(a, f(a))$ to $(p, q)$ is $\dfrac{f(a) - q}{a - p}$ .
- Set these equal and solve for $a$ .
There are often two such tangents, so expect two solutions.

Tips and Tricks

Differentiate first, substitute second. If you put $x = 2$ into $y = x^3 - 2x$ and then differentiate 4, you get 0. The reports list this confusion between $f$ and $f'$ among the most damaging Not Achieved behaviours.

The gradient of the normal is $-\frac{1}{m}$ , not $-m$ and not $\frac{1}{m}$ . Both the flip and the sign change are needed.

Finish the equation. The 2025 report is specific: "When finding equations of tangents, candidates are expected to present answers in general or gradient-intercept form, rather than leaving equations with substituted values only. Final answers must also be fully evaluated." Leaving $y - 4 = 10(x-2)$ may not earn the mark.

If the question gives you a gradient and asks for points, set $f'(x)$ equal to that gradient — and check whether the resulting equation has more than one solution. Quadratic derivatives usually give two points.

The specification says "equations of tangents may be included" in all three Level 3 Calculus standards, so this technique can appear inside a complex numbers or integration question too. It is not confined to this standard.

Worked ExampleTangent, normal, and the triangle they form

Find the equations of the tangent and the normal to $y = \dfrac{4}{x}$ at the point where $x = 2$ , and find the area of the triangle formed by the normal and the two axes.

Step 1 — Find the point on the curve

Substitute $x = 2$ into the original function:

$y = \frac{4}{2} = 2$

The point is $(2, 2)$ .

Step 2 — Differentiate

Convert to index form first, then apply the power rule:

$y = 4x^{-1} \quad \Rightarrow \quad \frac{dy}{dx} = -4x^{-2} = -\frac{4}{x^2}$

Step 3 — Find the gradient of the tangent

Now substitute $x = 2$ into the derivative:

$m_{\text{tangent}} = -\frac{4}{2^2} = -\frac{4}{4} = -1$

Step 4 — Write the tangent's equation

Using $y - y_1 = m(x - x_1)$ with the point $(2, 2)$ :

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

$y - 2 = -x + 2$

$y = -x + 4$

Step 5 — Find the gradient of the normal

Flip and change sign:

$m_{\text{normal}} = -\frac{1}{-1} = 1$

Step 6 — Write the normal's equation

Through the same point $(2, 2)$ :

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

$y = x$

Step 7 — Find where the normal meets the axes

The normal is $y = x$ .

$x$ -intercept: set $y = 0$ , giving $x = 0$
$y$ -intercept: set $x = 0$ , giving $y = 0$

Both intercepts are at the origin, so the normal passes through the origin and does not cut off a triangle at all.

Step 8 — Re-read and use the tangent instead

The normal forms no triangle, so the intended triangle must come from the tangent $y = -x + 4$ :

$x$ -intercept: $0 = -x + 4$ , so $x = 4$ — the point $(4, 0)$
$y$ -intercept: $y = 4$ — the point $(0, 4)$

Step 9 — Find the area

The triangle has its right angle at the origin, with legs along the axes:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$

The lesson: always check whether a line actually cuts off a region before computing an area. A line through the origin never does.

Tangents and normals

Covers AS91578

The gradient at a point

$f'(x)$ is a function — it gives the gradient at every point on the curve.
$f'(a)$ is a number — the gradient at the single point where $x = a$ .
Substitute after differentiating, never before. Substituting first turns the function into a constant, whose derivative is zero.

Finding the equation of a tangent

A tangent is the straight line that touches the curve at a point and has the same gradient as the curve there.
The procedure:
- Differentiate to get $f'(x)$ .
- Substitute the $x$ -value to get the gradient $m = f'(a)$ .
- Find the $y$ -coordinate by substituting into the original function, $y_1 = f(a)$ .
- Apply the straight-line formula:

$y - y_1 = m(x - x_1)$

Rearrange into the form the question asks for.
Worked through — the tangent to $y = x^3 - 2x$ at $x = 2$ :
- $f'(x) = 3x^2 - 2$ , so $m = f'(2) = 12 - 2 = 10$
- $y_1 = f(2) = 8 - 4 = 4$ , so the point is $(2, 4)$
- $y - 4 = 10(x - 2)$ , giving $y = 10x - 16$

Finding the equation of a normal

A normal is the line perpendicular to the tangent at the same point.
Perpendicular gradients multiply to $-1$ , so:

$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

The procedure is identical to the tangent, with one extra step: flip the gradient and change its sign.
Continuing the example above: $m_{\text{normal}} = -\frac{1}{10}$ , so the normal is $y - 4 = -\frac{1}{10}(x - 2)$ , or $y = -0.1x + 4.2$ .
Special cases:
- A horizontal tangent ( $m = 0$ ) has a vertical normal, written $x = a$ — the formula $-\frac{1}{0}$ does not apply.
- A vertical tangent has a horizontal normal, $y = b$ .

When you are given the gradient instead of the point

A common exam turn: "find the point(s) where the tangent has gradient 5".
Set $f'(x) = 5$ and solve for $x$ . There may be more than one answer.
Substitute each solution back into the original function for the $y$ -coordinates.
Note the error the reports flag repeatedly: candidates "misunderstood the relationship between the gradient and the derivative, instead substituting the gradient value into the original function". The gradient goes into $f'$ , not into $f$ .

Where a tangent meets the axes or another line

These are ordinary co-ordinate geometry questions once the tangent's equation is found:
- $x$ -intercept: set $y = 0$ and solve.
- $y$ -intercept: set $x = 0$ .
- Intersection of two tangents: solve the two line equations simultaneously.
The 2025 report lists "found the coordinates of the point of intersection of two tangents to a curve" and "found the coordinates of the point where a tangent to a curve crosses the $x$ -axis" as Merit behaviours.

Tangents from an external point

If a tangent passes through a point $(p, q)$ $(p, q)$ that is not on the curve, the touch point is unknown. Call it $x = a$ $x = a$ :
- The gradient at the touch point is $f'(a)$ .
- The gradient of the line joining $(a, f(a))$ to $(p, q)$ is $\dfrac{f(a) - q}{a - p}$ .
- Set these equal and solve for $a$ .
There are often two such tangents, so expect two solutions.

Tips and Tricks

The gradient of the normal is $-\frac{1}{m}$ , not $-m$ and not $\frac{1}{m}$ . Both the flip and the sign change are needed.

Worked ExampleTangent, normal, and the triangle they form

Find the equations of the tangent and the normal to $y = \dfrac{4}{x}$ at the point where $x = 2$ , and find the area of the triangle formed by the normal and the two axes.

Step 1 — Find the point on the curve

Substitute $x = 2$ into the original function:

$y = \frac{4}{2} = 2$

The point is $(2, 2)$ .

Step 2 — Differentiate

Convert to index form first, then apply the power rule:

$y = 4x^{-1} \quad \Rightarrow \quad \frac{dy}{dx} = -4x^{-2} = -\frac{4}{x^2}$

Step 3 — Find the gradient of the tangent

Now substitute $x = 2$ into the derivative:

$m_{\text{tangent}} = -\frac{4}{2^2} = -\frac{4}{4} = -1$

Step 4 — Write the tangent's equation

Using $y - y_1 = m(x - x_1)$ with the point $(2, 2)$ :

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

$y - 2 = -x + 2$

$y = -x + 4$

Step 5 — Find the gradient of the normal

Flip and change sign:

$m_{\text{normal}} = -\frac{1}{-1} = 1$

Step 6 — Write the normal's equation

Through the same point $(2, 2)$ :

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

$y = x$

Step 7 — Find where the normal meets the axes

The normal is $y = x$ .

$x$ -intercept: set $y = 0$ , giving $x = 0$
$y$ -intercept: set $x = 0$ , giving $y = 0$

Both intercepts are at the origin, so the normal passes through the origin and does not cut off a triangle at all.

Step 8 — Re-read and use the tangent instead

The normal forms no triangle, so the intended triangle must come from the tangent $y = -x + 4$ :

$x$ -intercept: $0 = -x + 4$ , so $x = 4$ — the point $(4, 0)$
$y$ -intercept: $y = 4$ — the point $(0, 4)$

Step 9 — Find the area

The triangle has its right angle at the origin, with legs along the axes:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$

The lesson: always check whether a line actually cuts off a region before computing an area. A line through the origin never does.