Finding a tangent: understand the method, then check your working

Separate the curve from the line

Consider this original practice question: find the equation of the tangent to y = x² at x = 2. The curve is a parabola. The tangent is a straight line that passes through the point on the curve and has the same gradient there.

Before calculating, identify the two things needed to describe that line: a point and a gradient. This is a calculus example for students studying differentiation; check that it belongs to your course and current school topic. It is not an official HSC examination question.

Use the original function to find the point

Substitute x = 2 into y = x². The result is y = 4, so the point of contact is (2, 4). Keep this point visible beside your working.

A common mistake is to substitute into the derivative and treat that output as the y-coordinate. The original function tells you the height of the curve. Its derivative tells you how steep the curve is. In this particular example the two outputs happen to be equal at x = 2, so checking what each calculation means matters more than noticing the repeated number.

Differentiate, then evaluate the gradient

Differentiate y = x² to get dy/dx = 2x. This expression gives the gradient at different positions along the curve; it is not yet a single numerical gradient.

At x = 2, the gradient is m = 2 × 2 = 4. State both steps in your working: first the derivative, then its value at the requested x-coordinate. If you skip straight to a number, it is harder to spot whether an error came from differentiation or substitution.

Combine the point and gradient

Use point-gradient form: y − y₁ = m(x − x₁). With the point (2, 4) and gradient 4, this becomes y − 4 = 4(x − 2). Expanding gives y = 4x − 4.

The derivative 2x is not the tangent equation. It describes the gradient of the curve as x changes. The tangent equation describes a particular straight line. Keeping those roles separate is useful whenever a question moves between a function and its derivative.

Check the point and the gradient independently

Put x = 2 into your line: y = 4(2) − 4 = 4. It passes through the required point. Its coefficient of x is 4, matching the derivative at that point. Both checks agree.

You can also compare the line with the curve: x² − (4x − 4) = (x − 2)². This is zero at x = 2 and positive elsewhere, consistent with the tangent lying below this parabola. That extra check is specific to this example, not a rule that all tangent lines sit below all curves.

When reviewing an error, label the step that failed: function value, derivative, substitution or line equation. Practise a new question aimed at that step instead of copying the entire solution again.