Function, gradient and local change

A function value and a rate answer different questions

For the function f(x) = x², the value f(2) = 4 gives the height of the point (2,4) on its graph. It does not give the gradient there. To describe change, compare an output change with its corresponding input change.

This lesson develops polynomial tangents. It supports the graphical relationship between a function and its derivative. Other differentiation methods are developed in separate lessons. Independent questions and later checks remain separate from this lesson’s public models.

You need to expand a quadratic bracket, simplify fractions and use the equation of a straight line. If those steps are unfamiliar, pause at the algebra shown below and work through it before the tangent calculation. This support does not change your selected course or establish earlier-course mastery.

Begin with a secant gradient

Take the points with inputs a and a + h, where h ≠ 0. The gradient of the secant through them is

For f(x) = x², expansion gives

At a = 2, the secant gradient is 4 + h:

The gradients approach four from both sides. The table suggests a limit; a finite list of values does not prove it. The exact simplification 2a + h explains the limit for every real a.

Do not put h = 0 into the original quotient. It would be 0/0, which is undefined. Instead, take the limit through nonzero values of h.

The derivative is the limiting rate

The derivative at a, when this limit exists, is

For f(x) = x², the simplified quotient is 2a + h for every nonzero h, so f'(a) = 2a. Thus the derivative function is f'(x) = 2x on the whole real line.

The tangent gradient at input two is f'(2) = 4. Here it happens to equal the function value. That coincidence can conceal a misconception: at input three, f(3) = 9 but f'(3) = 6. Height and gradient are different quantities even when their numbers sometimes agree.

Check your understanding 1 Separate the two quantities

For f(x) = x² at x = -1, give the point on the curve and the tangent gradient. Is the curve rising or falling as x increases through that point?

Point, gradient and local direction

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The point is (-1,1) because f(-1) = 1. The tangent gradient is f'(-1) = -2, so the curve is locally falling as x increases. A positive height does not imply a positive gradient. Compare all three parts of your response yourself.

Read derivative information from a graph

For the parabola y = x², the tangent slopes are negative when x < 0, zero at x = 0 and positive when x > 0. This agrees with the derivative graph y = 2x: it lies below, on and above the horizontal axis in those respective regions.

The derivative graph’s height at an input is the original graph’s slope there. It is not the original graph shifted or rescaled vertically. For example, the original parabola is never below the horizontal axis, while its derivative is negative on the negative half-line.

Try the model. You can change your answer or look at the worked model. This activity has no score.

Compare your derivative sketch with the public relationship f'(x)=2x. The sketch check does not verify a symbolic proof. You can also draw on paper and compare with the public answer.

Show solution

x: -2 to 2; derivative f'(x): -4 to 4. Coordinates are mathematical values; the viewport is a display window.

Points and lines in the drawing

Object	Values	Given or drawn
derivative	(-2, -4) → (2, 4); start included, end included	Drawn

Use the negative, zero and positive derivative values to explain the parabola’s falling, stationary and rising regions. You can enter coordinates with the keyboard or draw on paper and compare with the worked graph.

At the vertex, f'(0) = 0. The sign change from negative to positive shows a local minimum. A zero derivative alone is insufficient to establish a maximum or minimum: g(x) = x³ has derivative zero at zero but keeps increasing through it. Further stationary-point analysis belongs to the wider topic.

A sketch, a numerical approximation or an interactive exploration can help you notice this relationship. It does not prove a symbolic derivative or supply independent assessment evidence.

Find the tangent using both height and gradient

Let f(x) = x²-3x + 1. Find the tangent at x = 2.

First find the point: f(2) = 4-6 + 1 = -1, so the point is (2,-1).

Differentiate the polynomial: f'(x) = 2x-3. This follows by taking the difference quotient term by term; the constant contributes zero and the linear term contributes -3. Therefore f'(2) = 1.

A line of gradient one through (2,-1) has equation

Check the point: at x = 2, the line gives y = -1. Check the gradient: the coefficient of x is one. These checks test different requirements. A line through the point with the wrong slope is not the requested tangent.

When a question asks for working, show how the gradient and point were obtained. A correct final line alone does not establish your derivative calculation or your reasoning.

The original domain decides whether a tangent exists

Consider

For every allowed input, q(x) = x + 1. Its derivative is one on that original domain. At x = 2, the curve point is (2,3) and its tangent is y = x + 1.

At x = 1, however, the original function has no point. The simplified line would pass through (1,2), but that point is missing from the graph of q. There is no tangent to this function at that input. Simplifying the formula does not silently extend its domain.

A sketch must show the excluded point with an open marker and retain the restriction in text. Sampling nearby points or obtaining a derivative value from the simplified expression cannot supply the missing function value.

Check your understanding 2 A missing input

For r(x) = (x²-4)/(x + 2), find the tangent at x = 1. Then explain why the same procedure does not give a tangent to r at x = -2.

Domain, point, gradient, tangent and excluded-input explanation

Show answer

The original domain excludes -2. On that domain, r(x) = x-2 and r'(x) = 1. At x = 1, the point is (1,-1) and the tangent is y + 1 = x-1, or y = x-2. At x = -2, the original function is undefined, so there is no point there at which to draw its tangent. The line’s extension is a different domain choice.

Transfer to a derivative with zero slope

Check your understanding 3 A horizontal tangent

For p(x) = x³-3x + 2, find the tangent at x = 1. Explain why answering only “the gradient is zero” is incomplete.

Derivative, point, line and explanation

Show answer

p'(x) = 3x²-3, so p'(1) = 0. The point has height p(1) = 1-3 + 2 = 0. A horizontal line through (1,0) is y = 0. A zero gradient specifies the line’s direction but not its height: many horizontal lines have gradient zero. The point is needed to select this tangent.

Your derivative, substitution, line and explanation are separate parts to review. Comparing your prose with this answer is self-review, not automatic verification of an argument.

Before you move on

Cover an earlier tangent solution. Starting from the original function, name its domain, obtain its derivative, find the point and gradient, and form the line. Test it at the point. If you use a graph to anticipate the answer, still give the symbolic argument the question requests.

Plain-text inputs in topic checks use forms such as x^2-3*x+1. Follow the field’s instructions about whether to enter a whole equation or just its right-hand expression. Unsupported syntax needs a different representation or review; it is not evidence of a mathematical misconception.