A tangent question is a calculus question followed by a straight-line question. First find the gradient of the curve at the required point. Then use that gradient and point to construct the tangent equation. A normal uses the perpendicular gradient.
This separation is important. Many errors happen because students differentiate correctly but then use the wrong point, or find the correct gradient but place it into an incorrect line equation.
1. The derivative supplies the tangent gradient
If y=f(x), then f′(a) is the tangent gradient at x=a, provided the derivative exists there.
The point on the curve is (a,f(a)). You usually need both pieces before writing the tangent.
2. Point-gradient form
A line of gradient m through (x₁,y₁) can be written
y−y₁=m(x−x₁).
This form is often safer than trying to guess the intercept immediately.
3. Worked tangent example
Find the tangent to y=x²+3x−1 at x=2.
First find the point: y=4+6−1=9, so P=(2,9).
Differentiate: dy/dx=2x+3. At x=2, m=7.
Therefore the tangent is
y−9=7(x−2), or y=7x−5.
4. The normal is perpendicular to the tangent
For finite nonzero tangent gradient m, the normal gradient is −1/m.
In the previous example, the normal gradient is −1/7, so the normal is
y−9=−(1/7)(x−2).
5. Special case: horizontal tangent
If f′(a)=0, the tangent is horizontal: y=f(a). Its normal is vertical: x=a.
The negative-reciprocal formula should not be forced through division by zero.
6. Worked horizontal-tangent example
For y=(x−3)²+5, dy/dx=2(x−3). At x=3, the derivative is zero and the point is (3,5).
Tangent: y=5. Normal: x=3.
7. Tangent at an unknown x-coordinate
Suppose y=x² and the tangent gradient is 6. Since dy/dx=2x, solve 2x=6 to obtain x=3. The point is (3,9), so the tangent is y−9=6(x−3).
Here the derivative equation locates the contact point rather than merely evaluating a supplied point.
8. Tangent parallel to a given line
If a tangent is parallel to y=4x−7, its gradient must be 4. Set f′(x)=4, solve for the possible x-values, then find each corresponding point on the curve.
There may be more than one point where a curve has the same gradient.
9. Tangent perpendicular to a given line
If the given nonvertical line has gradient m, a perpendicular tangent has gradient −1/m. Set the derivative equal to that required tangent gradient.
Handle horizontal and vertical special cases separately.
10. Trigonometric tangent example
For y=sin x at x=π/3, the point is (π/3,√3/2). The derivative is cos x, so the tangent gradient is 1/2.
Tangent:
y−√3/2=(1/2)(x−π/3).
Exact values keep the geometry transparent.
11. Exponential tangent example
For y=e^x at x=0, the point is (0,1) and dy/dx=e^x gives gradient 1. Thus the tangent is y=x+1.
The exponential function’s value and gradient happen to be equal here because d(e^x)/dx=e^x.
12. Tangent equation as a local approximation
Near the contact point, a smooth curve and its tangent have very similar values. This is why a tangent line can act as a local linear approximation.
The approximation becomes less reliable farther from the contact point because the curve’s gradient may change.
13. Check the tangent equation
- Does the contact point satisfy the curve?
- Does the contact point satisfy the tangent line?
- Does the line gradient equal f′(a)?
- For a normal, is the perpendicular relationship correct?
14. Common mistakes
- Using f(a) as the gradient instead of f′(a).
- Using x=a but forgetting to calculate y=f(a).
- Using −m instead of −1/m for a normal.
- Applying negative reciprocal to a horizontal tangent without recognising the vertical normal.
- Rounding an exact point or gradient before constructing the line.
15. Independent practice
- Find the tangent to y=x³ at x=2.
- Find the normal to y=x² at x=1.
- Find points on y=x²−4x where the tangent is horizontal.
- Find the point on y=x² where the tangent gradient is −6.
- Find the tangent to y=cos x at x=0.
- For y=e^x, find the tangent at x=ln2.
16. Answers
- Point (2,8), gradient 12: y−8=12(x−2).
- Point (1,1), tangent gradient 2, normal gradient −1/2: y−1=−(1/2)(x−1).
- dy/dx=2x−4=0 gives x=2; point (2,−4).
- 2x=−6 gives x=−3; point (−3,9).
- Point (0,1), derivative −sin0=0; tangent y=1.
- Point (ln2,2), gradient 2: y−2=2(x−ln2).
17. What mastery looks like
A student has tangent-and-normal control when they can move cleanly through the chain curve → derivative → gradient → point → line, handle special cases, and verify both the calculus and coordinate geometry independently.
Continue with What Differentiation Means, How to Check a Derivative, or return to the Additional Mathematics Hub.