Tangents: 7 exam questions solved step by step
What this lesson covers
- A tangent touches the curve at a point and has the same gradient as the curve there.
- The equation of a tangent needs a point and a gradient, and the gradient comes from the derivative.
7 questions, one recipe
- 1. The point: substitute x into y.
- 2. The gradient: substitute into dy/dx.
- 3. The line: y − y₁ = m(x − x₁).
- 4. Tidy to y = mx + c, then check.
Question 1 · Achieved
- y = x³ − 2x² + 1
- Find the gradient of the curve
- at the point (2, 1).
Q1: gradient at (2, 1)
- Gradient from dy/dx.
- Not from y.
- y = 8 − 8 + 1 = 1(2, 1) is on the curve
- dy/dx = 3x² − 4xdifferentiate
- m = 3(2)² − 4(2)x = 2
- = 12 − 8 = 4
- gradient 4 at (2, 1)
Question 2 · Achieved
- Find the equation of the tangent
- to y = x² + 3x − 1 at x = 1.
Q2: tangent to y = x² + 3x − 1
- Point, gradient, line.
- Check the point fits.
- y = 1 + 3 − 1 = 3point (1, 3)
- dy/dx = 2x + 3
- m = 2(1) + 3 = 5x = 1
- y − 3 = 5(x − 1)y − y₁ = m(x − x₁)
- y − 3 = 5x − 5expand
- y = 5x − 2add 3
- 5(1) − 2 = 3the point fits
Question 3 · Merit
- Find the equation of the tangent
- to y = x³ − 4x + 1 at x = −2.
- Watch the double negative.
Q3: tangent at x = −2
- Brackets round −2.
- x − (−2) = x + 2.
- y = (−2)³ − 4(−2) + 1the point
- = −8 + 8 + 1 = 1point (−2, 1)
- dy/dx = 3x² − 4
- m = 3(−2)² − 4 = 12 − 4 = 8
- y − 1 = 8(x − (−2))y − y₁ = m(x − x₁)
- y − 1 = 8(x − 2)lost the minus
- y − 1 = 8(x + 2) = 8x + 16
- y = 8x + 17add 1
- 8(−2) + 17 = 1the point fits
Question 4 · Merit
- The tangent to y = x² − 6x + 10
- has gradient 2.
- Find the equation of the tangent.
Q4: the tangent with gradient 2
- m is given: find x.
- Then the point.
- dy/dx = 2x − 6
- 2x − 6 = 2gradient 2
- 2x = 8, so x = 4
- y = 16 − 24 + 10 = 2point (4, 2)
- y − 2 = 2(x − 4)y − y₁ = m(x − x₁)
- y − 2 = 2x − 8expand
- y = 2x − 6add 2
Question 5 · Merit, in context
- A road over a hill: y = 6 − 0.1x²
- (metres, crest at x = 0).
- At x = 5 a car's headlights shine
- along the tangent. Where does the
- beam reach the ground (y = 0)?
Q5: the headlight beam
- The beam is the tangent.
- Ground level: y = 0.
- y = 6 − 0.1(5)² = 3.5point (5, 3.5)
- dy/dx = −0.2x
- m = −0.2(5) = −1x = 5
- y − 3.5 = −1(x − 5)y − y₁ = m(x − x₁)
- y = −x + 5 + 3.5expand
- y = −x + 8.5
- 0 = −x + 8.5, so x = 8.5ground: y = 0
- 8.5 m from the crest
Question 6 · Excellence
- Two tangents to y = x² + 1
- pass through the point (1, −2).
- Find both equations.
- (1, −2) is not on the curve.
Q6: a general tangent
- Touch at x = a.
- Then use (1, −2).
- 1 + 1 = 2 ≠ −2(1, −2) not on curve
- point (a, a² + 1)touches at x = a
- dy/dx = 2x, so m = 2a
- y − (a² + 1) = 2a(x − a)y − y₁ = m(x − x₁)
- y = 2ax − 2a² + a² + 1
- y = 2ax − a² + 1collect a²
- −2 = 2a(1) − a² + 1through (1, −2)
- a² − 2a − 3 = 0rearrange
Q6: two values of a, two tangents
- Each a gives a tangent.
- Check both pass (1, −2).
- y = 2ax − a² + 1
- a² − 2a − 3 = 0
- (a − 3)(a + 1) = 0factorise
- a = 3 or a = −1
- a = 3: y = 6x − 9 + 1
- = 6x − 8
- a = −1: y = −2x − 1 + 1 = −2x
- 6(1) − 8 = −2, −2(1) = −2
- y = 6x − 8 and y = −2x
Question 7 · Excellence
- Show that the line y = 2x + 1
- is a tangent to y = x² − 2x + 5,
- and find where it touches.
Q7: show it is a tangent
- Same gradient, same point.
- Prove both.
- dy/dx = 2x − 2
- 2x − 2 = 2the line's gradient
- 2x = 4, so x = 2
- curve: y = 4 − 4 + 5 = 5point (2, 5)
- line: y = 2(2) + 1 = 5same point
- x² − 2x + 5 = 2x + 1where they meet
- x² − 4x + 4 = 0
- (x − 2)² = 0only x = 2
- tangent, touching at (2, 5)
The recipe that solved all seven
- A point from y, a gradient from dy/dx, then the line.
| Step | How |
|---|---|
| Point | substitute x into y |
| Gradient | substitute x into dy/dx |
| Line | y − y₁ = m(x − x₁) |
| Tidy | y = mx + c, then check |
| Given m | set dy/dx = m to find x |
| Unknown | use a general point (a, y) |
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