Understanding Gradients and Tangents Revision Notes | GCSE AQA Higher Further Maths

1. Why the Gradient of a Curve Needs a New Idea

You already know how to find the gradient of a straight line: pick two points, calculate $\frac{\text{rise}}{\text{run}}$, and you are done. The gradient is the same everywhere on that line.

A curve is different. The steepness of a curve changes from point to point — it might be climbing steeply here, levelling off there, and falling somewhere else. So asking "what is the gradient of this curve?" without specifying a point is like asking "how fast is the car going?" without saying when. The question only makes sense at a particular moment, or in this case, at a particular point.

This is where the tangent comes in. The gradient of a function at a point is defined as the gradient of the tangent to the curve at that point. That single idea is the foundation of everything in this lesson.

Think of a ball rolling along a curved ramp. At any instant, the ball is travelling in a straight-line direction — that direction is the tangent. The steepness of that straight-line direction is the gradient of the curve at that moment.

The curve above is $y = x^2$. Notice how it is steep near the edges and almost flat near the bottom. The gradient is different at every point — and to measure it precisely at any one location, you draw the tangent there and find its gradient.

2. The Key Relationship: Curve Gradient = Tangent Gradient

Here is the central fact you need to know: the gradient of a function at a point equals the gradient of the tangent drawn to the curve at that point. These two things are not just related — they are the same quantity, described two different ways.

The tangent is the tool that converts a curved, changing slope into a straight line whose gradient you can calculate. Once you have the tangent, you use the standard gradient formula on it, and that result is the gradient of the curve at that exact point.

A tangent is still a straight line, so the gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ applies to it directly. The only new step is recognising that this gradient also tells you the steepness of the curve at the point where the tangent touches it.

3. Reading the Gradient of a Tangent from a Graph

In exam questions, a tangent is often drawn for you on a graph and you are asked to find its gradient. The method is the same as for any straight line: identify two clear points on the tangent (not on the curve — on the tangent line itself), then apply the gradient formula.

Choose points that land on grid intersections where possible. This avoids reading errors and keeps your arithmetic clean. The further apart your two points are, the more accurate your gradient calculation will be.

Worked example: Finding the Gradient of a Curve at a Point Using a Tangent

A tangent is drawn to the curve $y = x^2$ at the point $(2, 4)$. The tangent line passes through the points $(0, -4)$ and $(3, 8)$. Find the gradient of the curve at $x = 2$.

1. $(x_1, y_1) = (0, -4)$ and $(x_2, y_2) = (3, 8)$ — Use points on the tangent line, not on the curve. Grid intersection points are ideal for accuracy. [M1]
2. $y_2 - y_1 = 8 - (-4) = 8 + 4 = 12$ — Subtracting a negative adds — watch the sign carefully here.
3. $x_2 - x_1 = 3 - 0 = 3$ — The horizontal distance between the two chosen points.
4. $m = \frac{12}{3} = 4$ — Rise divided by run gives the gradient of the tangent. [M1]
5. Gradient of $y = x^2$ at $x = 2$ is $4$. — The gradient of the curve at a point equals the gradient of its tangent at that point. [A1]

Answer: The gradient of the curve at $x = 2$ is $4$.

4. Positive, Negative, and Zero Gradients on a Curve

Because the gradient of a curve changes from point to point, the tangent at different locations will have different gradients — and those gradients carry meaning. A positive gradient means the curve is rising at that point. A negative gradient means it is falling. A gradient of zero means the curve is momentarily flat, which happens at a turning point.

• $m > 0$: the curve is increasing at that point (tangent slopes upward left to right)
• $m < 0$: the curve is decreasing at that point (tangent slopes downward left to right)
• $m = 0$: the curve has a turning point at that location (tangent is horizontal)

For example, on the curve $y = x^2$, the gradient is negative for $x < 0$ (the curve falls as you move right), zero at $x = 0$ (the minimum), and positive for $x > 0$ (the curve rises). The tangent at each of those points would reflect exactly that behaviour.

5. Worked Practice: Estimating Gradient from a Drawn Tangent

Exam questions often provide a curve with a tangent already drawn and ask you to estimate the gradient. The word "estimate" signals that a reading from the graph is expected — you will not get an exact algebraic answer, but your method must be clear and your chosen points must lie on the tangent.

Worked example: Estimating the Gradient of a Curve at a Given Point

The graph of $y = x^3 - 3x$ has a tangent drawn at the point $(2, 2)$. From the graph, the tangent appears to pass through $(0, -16)$ and $(2, 2)$. Estimate the gradient of the curve at $x = 2$.

1. $(x_1, y_1) = (0, -16)$ and $(x_2, y_2) = (2, 2)$ — Both points lie on the tangent line, not on the curve. Read from grid lines for accuracy. [M1]
2. $y_2 - y_1 = 2 - (-16) = 2 + 16 = 18$ — Subtracting a negative value increases the result — a common sign error to watch for.
3. $x_2 - x_1 = 2 - 0 = 2$ — Horizontal distance between the two chosen points.
4. $m = \frac{18}{2} = 9$ — This is the gradient of the tangent, and therefore the gradient of the curve at $x = 2$. [A1]
5. Gradient of $y = x^3 - 3x$ at $x = 2$ is $9$. — We can verify: $\frac{dy}{dx} = 3x^2 - 3$, so at $x = 2$: $3(4) - 3 = 9$. The tangent line is $y = 9x - 16$, confirming $(0, -16)$ and $(2, 2)$ lie on it.

Answer: The gradient of the curve at $x = 2$ is $9$.

6. Deliberate Practice

Now that the method is clear, these problems let you apply it independently. For each one, identify two points on the tangent, apply the gradient formula, and state the gradient of the curve at the given point.

7. Exam Technique and Common Mistakes

This topic appears on both calculator and non-calculator papers. The method is always the same: locate two clear points on the tangent, apply the gradient formula, and state that this gradient equals the gradient of the curve at that point. Showing both steps earns both method marks.

The most reliable way to lose marks here is to read points from the curve rather than from the tangent. The tangent is a straight line — use points that sit clearly on that line, ideally at grid intersections. If the tangent is drawn for you, do not try to re-draw it; use the one provided.

The gradient of a function at a point and the gradient of the tangent at that point are the same thing — mastering how to read and calculate that gradient from a graph is the core skill this specification requires.