A tangent follows the local direction of a curve at a chosen point. A normal passes through the same point and is perpendicular to that tangent. Differentiation supplies the tangent gradient; it does not automatically supply the whole line equation. You still need the point, the correct gradient for the requested line, and a final substitution check.
These worked examples of tangent and normal equations use polynomial curves and ordinary Cartesian axes with the same unit scale in both directions. The opening task is to find a line at a specified point; the later extension locates and classifies stationary points and interprets contextual rates of change. Every equation and sketch below is original. The sketch supports the reasoning, but all coordinates and equations needed to solve the questions also appear in text.
Keep the curve, its derivative and the line separate
For y = f(x), the expression f(x) gives the vertical coordinate. The derivative f′(x), also written dy/dx, gives the gradient as a function of x. At x = a, calculate the point P = (a, f(a)) from the original curve and calculate the tangent gradient m = f′(a) from the derivative. Those are different substitutions into different expressions.
For the polynomial powers used here, multiply the coefficient by the power and reduce the power by one. Thus the derivative of 5x³ is 15x², the derivative of −4x is −4, and a constant has derivative zero. Differentiate each term with its sign attached. Do not remove a negative coefficient while reducing its exponent.
A line through (a, b) with finite gradient m has point-gradient form y − b = m(x − a). This form is already a valid equation. Expand it only if a requested format makes that useful. When the tangent gradient is nonzero, the normal gradient is −1/m. The product of these two gradients is −1; merely changing the sign is insufficient unless the original gradient happens to be 1 or −1.
| Stage | What to write | Quick check |
|---|---|---|
| Point | b = f(a) | Does (a, b) satisfy the curve? |
| Tangent gradient | m = f′(a) | Was differentiation completed before substitution? |
| Normal gradient | −1/m, provided m ≠ 0 | Does the product with m equal −1? |
| Line | y − b = gradient × (x − a) | Does substituting (a, b) give equality? |
Subject references: OpenStax: Differentiation rules. These support the concepts; the questions, datasets, calculations and diagrams here are original.
Worked analysis 1: a positive gradient and its normal
Find the tangent and normal to y = x² + 3x − 2 at x = 2. First, y = 4 + 6 − 2 = 8, so P is (2, 8). Differentiating the entire polynomial gives dy/dx = 2x + 3. At x = 2 the tangent gradient is 7.
The tangent is y − 8 = 7(x − 2), which simplifies to y = 7x − 6. The normal gradient is −1/7, so its equation is y − 8 = −(x − 2)/7. Multiplying through by seven gives the convenient equivalent form x + 7y = 58.
Check both answers separately. The point gives 8 = 14 − 6 on the tangent and 2 + 56 = 58 on the normal. Their gradients multiply to 7 × (−1/7) = −1. A claimed normal y − 8 = −7(x − 2) would pass through the correct point, but its gradient would fail the perpendicularity check.
The original sketch shows the curve y = x² + 3x − 2, the tangent y = 7x − 6 and the normal x + 7y = 58 meeting at P(2, 8). The blue curve bends upward. The tangent rises steeply and the normal falls gently. Equal axis scales make their right-angle relationship meaningful; the written equations remain the exact evidence.
Worked analysis 2: a negative input is not a negative gradient
Find both lines for y = 2x³ − 5x + 4 at x = −1. Substituting into the curve gives y = −2 + 5 + 4 = 7. The point is therefore (−1, 7). The derivative is 6x² − 5, which becomes 6(1) − 5 = 1 at x = −1.
The tangent equation is y − 7 = 1(x + 1), or y = x + 8. The normal gradient is −1, giving y − 7 = −(x + 1), or y = −x + 6. Both return y = 7 when x = −1.
The negative coordinate did not determine the sign of the gradient. It entered a squared term in the derivative, making that contribution positive. Write brackets around a negative substitution: 6(−1)² − 5 is clearer than an expression whose exponent could be misread.
Worked analysis 3: a horizontal tangent needs a vertical normal
For y = 3x² − 12x + 5 at x = 2, the point is (2, −7). The derivative 6x − 12 gives zero there. Consequently the tangent is the horizontal line y = −7.
Do not calculate −1/0. A vertical line has no finite gradient, and the vertical line through this point is x = 2. That is the normal. Every point on it has x-coordinate 2, while its y-coordinate can vary.
A zero gradient does not mean that no tangent exists. It means the tangent is horizontal. Similarly, the normal cannot be forced into y = mx + c with an ordinary finite m. Recognising this exception is more important than memorising the negative-reciprocal rule without its condition.
Worked analysis 4: preserve a negative tangent gradient
Find the two lines to y = x³ − 4x² + 2x + 6 at x = 1. The original curve gives y = 1 − 4 + 2 + 6 = 5. Differentiating yields 3x² − 8x + 2, so m = 3 − 8 + 2 = −3.
The tangent is y − 5 = −3(x − 1), hence y = −3x + 8. Its normal has gradient +1/3, giving y − 5 = (x − 1)/3, or 3y = x + 14. Substitution of (1, 5) verifies both equations, and −3 × 1/3 = −1 confirms perpendicularity.
Leaving the normal in point-gradient form avoids unnecessary fractions in the intercept. If you do expand, multiply every term on both sides consistently. A line with the right gradient but the wrong intercept is parallel to the required line, not the required answer.
Repair the premature-substitution error
A learner starts worked analysis 1 by writing “At x = 2, y = 8; therefore dy/dx = 0 because 8 is constant.” The calculation has replaced the whole curve by one coordinate before asking how the curve changes nearby. The derivative of the unrelated constant function y = 8 is indeed zero, but that is not the derivative of the original polynomial.
The repair is to keep y = x² + 3x − 2 intact until its derivative 2x + 3 has been found. Substitute x = 2 into that derivative only afterwards. It is harmless to calculate the point first on a separate line; the mistake is treating its fixed y-coordinate as the original function.
Three solved practice tasks
Practice 1: two equations from a quadratic
Find the tangent and normal to y = 2x² − x + 3 at x = 1. The point is (1, 4); dy/dx = 4x − 1, so m = 3. The tangent is y = 3x + 1. The normal is x + 3y = 13. Each contains (1, 4), and their gradients 3 and −1/3 multiply to −1.
Practice 2: use brackets at a negative coordinate
For y = x³ + x at x = −2, the point is (−2, −10). The derivative 3x² + 1 gives 13. Thus the tangent is y + 10 = 13(x + 2) and the normal is y + 10 = −(x + 2)/13. At x = −2, both right-hand sides vanish.
Practice 3: locate the horizontal tangent
Find the point where y = x² − 6x + 11 has a horizontal tangent, then write both lines. Solve 2x − 6 = 0 to get x = 3. Substitution gives y = 2. The tangent is y = 2 and the normal is x = 3. The equation dy/dx = 0 supplied the location; the original curve supplied its height.
Stationary points and contextual rates of change
The tangent work above gives gradients at specified points. To locate a stationary point of a differentiable curve, solve f′(x) = 0 and then use f(x) to find its height. Classify it by the sign of f′ on the two sides: positive to negative gives a local maximum; negative to positive gives a local minimum. Zero gradient alone does not establish a turning point.
Worked case: find and classify every stationary point
For y = x³ − 3x² − 9x + 5, find the stationary points. Differentiate: y′ = 3x² − 6x − 9 = 3(x − 3)(x + 1). Thus x = −1 or x = 3. Substituting into the original curve gives y(−1) = −1 − 3 + 9 + 5 = 10 and y(3) = 27 − 27 − 27 + 5 = −22.
| Interval | Sign of x − 3 | Sign of x + 1 | Sign of y′ | Behaviour |
|---|---|---|---|---|
| x < −1 | − | − | + | Increasing |
| −1 < x < 3 | − | + | − | Decreasing |
| x > 3 | + | + | + | Increasing |
Therefore (−1,10) is a local maximum and (3,−22) is a local minimum. An independent check uses y″ = 6x − 6: y″(−1) = −12 < 0 and y″(3) = 12 > 0. These are local turning points, not a global maximum and minimum over all real x; a cubic grows without bound in one direction and decreases without bound in the other.
Worked case: instantaneous rate has its own unit
A fictional filling model gives water volume V = 2t² + 3t + 10 litres for 0 ≤ t ≤ 5 minutes. Find the instantaneous filling rate at t = 2 and compare it with the average rate between t = 1 and t = 3. This is a supplied mathematical model, not measured apparatus data.
dV/dt = 4t + 3 litres per minute, so at t = 2 the instantaneous rate is 11 L/min. The volume at that instant is V(2) = 24 L, a different quantity. For the average, V(1) = 15 L and V(3) = 37 L, giving (37 − 15)/(3 − 1) = 11 L/min.
The equality here comes from a quadratic volume and an interval symmetric about t = 2. It is not a general rule that average and instantaneous rates agree. From t = 2 to t = 3, the average is (37 − 24)/1 = 13 L/min. A direct derivative check uses [V(2 + h) − V(2)]/h = 11 + 2h for h ≠ 0, which approaches 11 as h approaches zero.
Independent derivative tasks
- Find and classify the stationary point of y = 2x² − 8x + 7. Give both coordinates and a sign or second-derivative check.
- A fictional particle position is s = t³ − 6t² + 9t metres for 0 ≤ t ≤ 4 seconds. Find velocity and acceleration at t = 2. At which times is it instantaneously at rest? Does it change direction at both?
Explained answers
- y′ = 4x − 8 = 0 at x = 2; y(2) = 8 − 16 + 7 = −1. The answer is (2,−1), a minimum, since y″ = 4 > 0, or because y′ changes from negative to positive. Completing the square, y = 2(x − 2)² − 1, independently confirms the minimum.
- v = ds/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3) m/s; a = dv/dt = 6t − 12 m/s². At t = 2, v = −3 m/s and a = 0 m/s²; speed is 3 m/s, so zero acceleration does not mean rest. Rest occurs at t = 1 s and 3 s. Velocity is positive before 1, negative between 1 and 3, and positive after 3 within the domain, so direction changes at both times.
Misconception check: a horizontal tangent need not turn
For y = x³, y′ = 3x² is zero at x = 0 but positive on both sides. The stationary point (0,0) is neither a local maximum nor a local minimum. Its second derivative is also zero there, so the second-derivative test is inconclusive; the derivative sign test settles the question. Do not label every solution of y′ = 0 a maximum or minimum.
Frequently asked questions
Should I substitute the x-coordinate before differentiating?
Find the point whenever convenient, but differentiate the original function before evaluating its gradient. A single substituted coordinate no longer describes how the curve changes.
Is dy/dx the equation of the tangent?
No. It describes the gradient of the curve. Evaluate it at the required point, then combine that gradient with the point in a line equation.
Does a normal have the negative tangent gradient?
Generally it has the negative reciprocal, −1/m, when m is nonzero. Changing only the sign usually does not make the lines perpendicular.
What is the normal when the tangent is horizontal?
It is the vertical line through the point, written x = a. A vertical line has no finite gradient, so division by zero is unnecessary.
Can a tangent cross the curve?
Yes. Tangency describes local direction at a point; it does not promise that the line meets the curve only once or stays on one side everywhere.
Must the final equation be written as y = mx + c?
Not unless the question requires that format. Point-gradient form and equivalent rearrangements are valid. A vertical normal must instead be written as x = a.
How can I check both answers quickly?
Check that both lines contain the chosen point. Then check that the tangent gradient equals the derivative there and that finite tangent and normal gradients multiply to −1.
Continue with targeted practice
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