A tangent is the straight line that touches a curve at one point and has the same gradient as the curve there. To find its equation you need two things: the gradient from dy/dx, and the y-coordinate from the original curve.
This skill sits at the start of tangents, normals and rates. It relies on the rules from differentiation techniques.
What is the method, step by step?
- Find the point. Substitute the given x into the curve equation to get y.
- Differentiate. Write dy/dx for the curve.
- Find the gradient. Substitute the same x into dy/dx. This number is m.
- Write the line. Use y − y₁ = m(x − x₁) with the point from step 1.
- Tidy up into the form the question asks for, and check that the point fits your final equation.
The last check takes five seconds. Put x back in and confirm you get the same y.
Worked example
Find the equation of the tangent to y = x³ − 4x + 1 at the point where x = 2.
Step 1, the point: y = 2³ − 4(2) + 1 = 8 − 8 + 1 = 1. The point is (2, 1).
Step 2, differentiate: dy/dx = 3x² − 4.
Step 3, the gradient: at x = 2, m = 3(4) − 4 = 8.
Step 4, the line: y − 1 = 8(x − 2), so y = 8x − 16 + 1 = 8x − 15.
Step 5, check: when x = 2, 8(2) − 15 = 1. ✓
The tangent is y = 8x − 15.
The mistake to watch for
A common slip is to use the wrong substitution: putting x = 2 into the curve and treating the answer as the gradient.
Mistaken working: y = 1 at x = 2, so m = 1 and the tangent is y − 1 = 1(x − 2), giving y = x − 1.
The value 1 is the height of the curve, not its steepness. The steepness comes only from dy/dx.
The correction is a two-column habit. Write “point” and “gradient” as two labels, and fill the first from the curve and the second from the derivative. Then you can see which number belongs where.
Check yourself
Try these, then open each answer.
1. Find the tangent to y = x² − 3x + 5 at x = 4.
Show answer
y = 16 − 12 + 5 = 9, so the point is (4, 9). dy/dx = 2x − 3, so m = 2(4) − 3 = 5.
y − 9 = 5(x − 4), so y = 5x − 11. Check: 5(4) − 11 = 9. ✓
2. Find the tangent to y = 2x³ at x = −1.
Show answer
y = 2(−1)³ = −2, so the point is (−1, −2). dy/dx = 6x², so m = 6.
y + 2 = 6(x + 1), so y = 6x + 4. Check: 6(−1) + 4 = −2. ✓
3. Find the tangent to y = √x at x = 9, giving your answer in the form ax + by + c = 0.
Show answer
y = √9 = 3, so the point is (9, 3). Write y = x^(1/2), so dy/dx = 1/(2√x) and m = 1/6.
y − 3 = (1/6)(x − 9). Multiply by 6: 6y − 18 = x − 9, so x − 6y + 9 = 0. Check: 9 − 18 + 9 = 0. ✓
Where this leads next
The same gradient gives the perpendicular line, so continue with forming a normal using the correct reciprocal sign. When you want mixed questions, use the tangents, normals and rates practice set. The non-calculator working trainer helps with the fraction arithmetic in gradients, and the quadratic structure explorer lets you see a curve and its tangent side by side.
Some students can do every step on a clean example yet lose the thread when the curve has a root or a fraction. That is a pattern our teachers can spot quickly in online one-to-one Additional Mathematics tuition.