Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 5 Calculus · 5.4 Tangents and normals
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Find the equation of the tangent to a curve at a given point | SL, HL | "Find the equation of the tangent to the curve at x = 2" (3 to 4 marks) |
| Find the equation of the normal at a given point | SL, HL | "Find the equation of the normal at P, in the form ax + by + d = 0" (3 to 4 marks) |
| Handle a horizontal tangent, and its vertical normal | SL, HL | "Write down the equation of the normal at the minimum point" (1 mark) |
| Work backwards from a gradient: tangent parallel or perpendicular to a line; a line that is a tangent | SL, HL | "The line y = 2x + k is a tangent to the curve. Find k" (4 to 6 marks) |
| Find where a tangent or normal meets the axes or the curve again | SL, HL | "The tangent meets the curve again at Q. Find Q" (4 to 5 marks); triangle areas |
| Find tangents and normals with technology | SL, HL | Paper 2: gradient from the GDC, then the equation (3 to 4 marks) |
Before you start
You need the three forms of a straight line from 2.1, especially the point-gradient form y − y₁ = m(x − x₁), and the rule that perpendicular gradients multiply to −1. You need 5.3 to differentiate. The formula booklet gives the straight-line forms and the derivative of xⁿ; the perpendicular rule is not in the booklet, so learn it. Everything on this page is those two old ideas put together.
1The idea in one paragraph
To write the equation of any straight line you need a point and a gradient. A tangent to a curve at a point P passes through P and has the same gradient as the curve at P, and the derivative gives you that gradient. A normal at P also passes through P, but is perpendicular to the tangent, so its gradient is the negative reciprocal. So every tangent and normal question is the same three steps: find the point, find the gradient from the derivative, put both into y − y₁ = m(x − x₁). The harder questions run those steps backwards, starting from the gradient or from a point that is not on the curve.
2The tangent at a given point
The tangent to the curve y = f(x) at the point where x = a is the straight line through (a, f(a)) with gradient f′(a):
Tangent at x = a: y − f(a) = f′(a)(x − a). Point from f, gradient from f′.
Work it for the curve y = x³ − 2x² + 1 at x = 2.
Figure 1 shows the result. Near P the tangent hugs the curve; further away the curve bends off it, which is expected. A tangent touches the curve at P but may still cross the curve somewhere else (section 6).
Two habits prevent most errors. Keep the point and the gradient on separate lines, labelled, so you never substitute into the wrong function. And if the question gives only x, find y from the curve's equation, not from the derivative: putting x = 2 into dy/dx gives 4, a gradient, not a y-coordinate.
3The normal at a given point
The normal to a curve at P is the straight line through P perpendicular to the tangent at P. Perpendicular gradients multiply to −1, so:
Normal at x = a: gradient = −1 ÷ f′(a), through (a, f(a)).
Continue the same example. The tangent gradient at (2, 1) was 4, so the normal gradient is −1/4.
The question often asks for the answer "in the form ax + by + d = 0" with integer coefficients, as here. Any non-zero multiple, such as −2x − 8y + 12 = 0, is equally correct.
A second example, with a negative power, shows the whole thing in one go. Find the tangent and normal to y = x + 4/x at x = 1.
4When the tangent is horizontal
If f′(a) = 0 the tangent is horizontal, and its equation is simply y = f(a). The normal is then vertical, and −1 ÷ 0 has no value, so the formula for its gradient breaks down. Write it as a vertical line through the point: x = a.
For y = x² − 6x + 10, dy/dx = 2x − 6 = 0 at x = 3, where y = 9 − 18 + 10 = 1. The tangent is y = 1 and the normal is x = 3, as Figure 2 shows.
This case turns up whenever the point is a turning point, so watch for it in any question that has just asked you to find a maximum or a minimum.
The reverse case is rare at this level but worth knowing: if a curve had a vertical tangent, its normal would be horizontal.
5Working backwards from a gradient
Many questions give you information about the tangent and ask you to find the point. The key move is always the same: the gradient of the tangent equals f′(x) at the point of contact, so set f′(x) equal to the gradient you know and solve for x.
A tangent parallel to a given line. Parallel lines have equal gradients. Find the point on y = x² − 3x + 2 where the tangent is parallel to y = 5x − 1, and the equation of that tangent.
Figure 3 shows the curve, the given line (grey) and the tangent (amber): parallel, with different intercepts.
A tangent perpendicular to a given line works the same way, except you first take the negative reciprocal. And a normal parallel to a given line means the tangent is perpendicular to that line.
A given line that is a tangent. "The line y = 2x + k is a tangent to y = x² − 4x + 7. Find k." Two conditions must hold at the point of contact: the gradients match, and the line and curve meet.
A useful check: substituting y = 2x − 2 into the curve gives x² − 6x + 9 = 0, which is (x − 3)² = 0, a repeated root. A line that touches a quadratic always gives a repeated root, which is the discriminant method from 2.7 in disguise. Either method earns the marks.
Tangents from a point that is not on the curve. Find the equations of the tangents to y = x² that pass through (1, −3). This one catches people, because (1, −3) is not on the curve, so x = 1 is not the point of contact and substituting it into dy/dx gives a meaningless number. Call the unknown point of contact (a, a²) instead.
Both lines pass through (1, −3): 6 − 9 = −3 and −2 − 1 = −3. Figure 4 shows the two tangents meeting there.
6Where a tangent or normal goes next
Once you have the equation, the rest is straight-line and simultaneous-equation work.
Where it meets the curve again. A tangent only touches the curve at its point of contact, but it may cross the curve elsewhere. For y = x³ − 3x at x = 2: y = 8 − 6 = 2 and dy/dx = 3x² − 3 = 9, so the tangent is y − 2 = 9(x − 2), that is y = 9x − 16. Set the curve equal to the line:
Knowing in advance that (x − 2)² is a factor makes the cubic easy to factorise by hand on Paper 1: divide out x² − 4x + 4 and what is left is (x + 4). Figure 5 shows the tangent crossing the curve far to the left.
Where it meets the axes, and areas. For intercepts, put x = 0 and y = 0 as in 2.1. The tangent to y = x + 4/x at (1, 5) was y = −3x + 8. It meets the y-axis at (0, 8) and the x-axis where −3x + 8 = 0, at (8/3, 0). The triangle it cuts off with the axes, shaded in Figure 6, has area
½ × 8/3 × 8 = 32/3 square units.
A normal can meet the curve again too, and the method is identical: set the curve equal to the normal and solve. The point of contact is still a root, though not a double one, since the normal crosses the curve rather than touching it.
7Tangents and normals with technology
The guide asks for technology as well as analytic methods. On Paper 2 the function may be one you have not yet learned to differentiate, or one where the arithmetic is ugly. The method does not change; only the source of the numbers does.
For f(x) = 2 sin x + x² at x = 1, with the GDC in radians:
Most GDCs can also draw the tangent at a chosen x and print its equation; that is a good check, but write the point, the gradient and the point-gradient line in your working so the method marks are visible.
One rule matters more here than anywhere else: store the intermediate values and round only the final answer. If you round f′(1) to 3.08 and f(1) to 2.68 before finding the intercept, you get −0.40 rather than −0.398, and the accuracy mark goes.
8Where marks are lost
Using the normal's gradient for the tangent, or the other way round. Tangent: f′(a). Normal: −1 ÷ f′(a). Label which one you are finding.
Taking the reciprocal and forgetting the minus. Perpendicular to 4 is −1/4, not 1/4. The tangent and normal gradients must have opposite signs.
Finding y by substituting into f′. The point of contact comes from the curve's equation. f′(a) is the gradient.
Writing the normal to a horizontal tangent as y = … . If the tangent is y = 1 at (3, 1), the normal is the vertical line x = 3.
Putting an outside point's x-value into f′. If the point is not on the curve, it is not the point of contact. Use (a, f(a)) and make the line pass through the given point.
Leaving fractions in "ax + by + d = 0". When the question says a, b, d ∈ ℤ, multiply through to clear every fraction.
Early rounding on Paper 2. Store f(a) and f′(a) in the calculator and round the final equation only.
Losing a solution when working backwards. Setting f′(x) equal to a gradient can give two x-values, so two tangents. Keep both unless the question excludes one.
9Work it right
- Find the point of contact (a, f(a)) and write it down.
- Differentiate and find the gradient f′(a) on its own line.
- For a normal, write m = −1 ÷ f′(a) as a line of working. If f′(a) = 0, the normal is x = a.
- Substitute into y − y₁ = m(x − x₁), then rearrange into the form the question asks for.
- Check the equation by substituting the point of contact.
- Working backwards, set f′(x) equal to the known gradient; for a point off the curve, use a general point (a, f(a)).
- For a second intersection, set curve equal to line and use the known root to factorise.
- On Paper 2, store unrounded GDC values and give the final equation to 3 s.f.
10Try it
Marks in brackets. Q1 to Q4 are Paper 1 style, no calculator. Q5 is Paper 2 style, with a GDC.
Q1. Consider the curve y = 2x³ − 5x + 1.
(a) Find the equation of the tangent to the curve at the point where x = 1. 4 marks
(b) Find the equation of the normal to the curve at this point, giving your answer in the form ax + by + d = 0. 2 marks
Q2. Let f(x) = x² + 16/x, for x ≠ 0.
(a) Find f′(x). 2 marks
(b) Show that the tangent to the graph of f at x = 2 is horizontal, and write down its equation. 3 marks
(c) Write down the equation of the normal to the graph of f at x = 2. 1 mark
Q3. The line y = −2x + k is a tangent to the curve y = x² + 4x − 1. Find the value of k. 5 marks
Q4. The tangent to the curve y = x³ at the point P(1, 1) meets the curve again at the point Q. Find the coordinates of Q. 6 marks
Q5. Let f(x) = x²e−0.5x, for x ∈ ℝ. The point P on the graph of f has x-coordinate 1.
(a) Find the gradient of the graph of f at P. 2 marks
(b) Find the equation of the tangent to the graph of f at P. 2 marks
(c) The tangent at P meets the x-axis at A, and the normal at P meets the x-axis at B. Find the distance AB. 4 marks
11In one breath
A tangent passes through the point of contact with the curve's gradient there, and a normal passes through the same point at right angles, so both come from one point and one gradient put into y − y₁ = m(x − x₁). Get the point from the curve's equation and the gradient from the derivative, and for the normal take the negative reciprocal, −1 ÷ f′(a); if the tangent is horizontal, the normal is the vertical line x = a. Working backwards, set f′(x) equal to the gradient you know (the same as a parallel line, the negative reciprocal of a perpendicular one); for a point off the curve, call the contact point (a, f(a)) and make the tangent pass through the given point. Afterwards, intercepts come from x = 0 and y = 0, and a second meeting point from setting curve equal to line, where the point of contact is a double root. On Paper 2 take f(a) and f′(a) from the GDC, keep them unrounded, and round only the final equation.
Answers
Q1. (a) At x = 1, y = 2 − 5 + 1 = −2. dy/dx = 6x² − 5, so the gradient is 6 − 5 = 1. Tangent: y + 2 = 1(x − 1), so y = x − 3. A1 for y = −2, A1 for dy/dx, A1 for gradient 1, A1 for the equation in any correct form.
(b) Normal gradient = −1 ÷ 1 = −1. y + 2 = −(x − 1), so y = −x − 1, that is x + y + 1 = 0. M1 for the negative reciprocal of their gradient used with (1, −2), A1 for the equation in the required form.
Q2. (a) f(x) = x² + 16x⁻¹, so f′(x) = 2x − 16x⁻² (or 2x − 16/x²). A1 for 2x, A1 for −16x⁻².
(b) f′(2) = 4 − 16/4 = 4 − 4 = 0, so the gradient of the tangent is 0 and the tangent is horizontal. f(2) = 4 + 8 = 12, so the tangent is y = 12. M1 for substituting x = 2 into their f′, A1 for showing f′(2) = 0 with a conclusion, A1 for y = 12.
(c) x = 2. A1. "y = 2" or a gradient of −1/0 scores 0.
Q3. dy/dx = 2x + 4. For the tangent, 2x + 4 = −2, so x = −3 and y = 9 − 12 − 1 = −4. The line passes through (−3, −4), so −4 = 6 + k and k = −10. A1 for dy/dx, M1 for setting dy/dx = −2, A1 for x = −3, A1 for y = −4, A1 for k = −10. Alternative: x² + 4x − 1 = −2x + k gives x² + 6x − (1 + k) = 0; M1 for setting the discriminant to 0, A1 for 36 + 4(1 + k) = 0, A1 for k = −10.
Q4. dy/dx = 3x², so the gradient at P is 3. Tangent: y − 1 = 3(x − 1), so y = 3x − 2. Setting x³ = 3x − 2 gives x³ − 3x + 2 = 0. Since x = 1 is a double root, x³ − 3x + 2 = (x − 1)²(x + 2), so x = −2 and y = (−2)³ = −8. Q is (−2, −8). A1 for gradient 3, A1 for the tangent, M1 for equating the curve and the tangent, M1 for factorising using (x − 1), A1 for x = −2, A1 for Q(−2, −8).
Q5. (a) Using the GDC, f′(1) = 0.910 (0.909795…). A2 for 0.910; accept the exact value 1.5e−0.5 if found by hand.
(b) f(1) = e−0.5 = 0.606530… . y − 0.60653 = 0.909796(x − 1), so y = 0.910x − 0.303. M1 for the point-gradient form with their values, A1 for the equation to 3 s.f.
(c) A: 0 = 0.909796(x − 1) + 0.60653, so x = 1 − 0.66667 = 0.333 (exactly 1/3). Normal gradient = −1 ÷ 0.909796 = −1.09915; B: 0 = −1.09915(x − 1) + 0.60653, so x = 1.55182. AB = 1.55182 − 0.33333 = 1.22 (3 s.f.). A1 for A, M1 for the normal gradient as the negative reciprocal, A1 for B, A1 for AB = 1.22. Rounding A and B too early can change the third figure of AB and lose the final A1.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 5.4 Tangents and normals. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.