TMUA: Tangents, normals and stationary points
Tangents, normals and stationary points.
- Section 1 Part 1 · MM6 Differentiation · spec MM6.3
- Papers 1 and 2
- 9 practice questions
- No calculator
What the specification covers
- Equations of tangents and normals
- Maxima and minima; increasing where f'(x) > 0
- Qualitative inflexions only
Key ideas
- Tangent at \(x=a\): gradient \(f'(a)\). Normal: gradient \(-\dfrac1{f'(a)}\).
- Stationary points: solve \(f'(x)=0\). \(f''>0\) gives a minimum, \(f''<0\) a maximum.
- \(f\) is increasing where \(f'(x)>0\) and decreasing where \(f'(x)<0\).
- '\(f'(x)>0\) for all \(x\)' for a quadratic \(f'\) means it has no real roots and a positive leading coefficient.
Common mistakes
- The normal is perpendicular to the tangent, not to the curve's axis.
- Points of inflexion are not examined, but a cubic can have stationary points that are neither maxima nor minima.
Exam tip
Factorise \(f'(x)\) to find stationary points quickly.
Worked example
Worked example
What is the distance between the two stationary points of the curve \(y=x^3-6x^2+9x+1\)?
- \(\sqrt{10}\)
- \(6\)
- \(2\sqrt{3}\)
- \(4\)
- \(2\sqrt{5}\)
Answer: E: \(2\sqrt{5}\)
\(y'=3x^2-12x+9=3(x-1)(x-3)\). The points are \((1,5)\) and \((3,1)\). Distance \(\sqrt{2^2+4^2}=2\sqrt5\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
The normal to the curve \(y=x^2\) at the point \((1,1)\) meets the curve again at the point \(P\). What is the \(x\)-coordinate of \(P\)?
- \(-1\)
- \(\frac{3}{2}\)
- \(-2\)
- \(-\frac{1}{2}\)
- \(-\frac{3}{2}\)
Show the answer and solution
Answer: E: \(-\frac{3}{2}\)
The gradient at \((1,1)\) is 2, so the normal has gradient \(-\tfrac12\): \(y=1-\tfrac12(x-1)\). Then \(x^2=\tfrac32-\tfrac x2\), \(2x^2+x-3=0\), \((2x+3)(x-1)=0\), so \(x=-\tfrac32\).
Question 2
For which values of the constant \(k\) is \(f'(x)>0\) for every real \(x\), where \(f(x)=x^3+kx^2+3x\)?
- \(k < -3\) or \(k > 3\)
- \(-3 \le k \le 3\)
- \(-3 < k < 3\)
- \(0 < k < 3\)
- \(k > 3\)
Show the answer and solution
Answer: C: \(-3 < k < 3\)
\(f'(x)=3x^2+2kx+3\), a positive quadratic, is positive for all \(x\) exactly when it has no real roots: \(4k^2-36<0\), i.e. \(-3
Question 3
What is the distance between the two stationary points of the curve \(y=x^{3} - 6 x^{2} - 3\)?
- \(32\)
- \(2 \sqrt{365}\)
- \(4 \sqrt{65}\)
- \(4 \sqrt{17}\)
- \(4\)
Show the answer and solution
Answer: C: \(4 \sqrt{65}\)
\(y'=3 x^{2} - 12 x=3 x \left(x - 4\right)\), so the stationary points are \((0,-3)\) and \((4,-35)\). Their distance is \(\sqrt{16+1024}=4 \sqrt{65}\).
Question 4
What is the distance between the two stationary points of the curve \(y=x^{3} - 9 x^{2} + 24 x + 1\)?
- \(2\)
- \(2 \sqrt{5}\)
- \(2 \sqrt{2}\)
- \(4\)
- \(2 \sqrt{362}\)
Show the answer and solution
Answer: B: \(2 \sqrt{5}\)
\(y'=3 x^{2} - 18 x + 24=3 \left(x - 4\right) \left(x - 2\right)\), so the stationary points are \((2,21)\) and \((4,17)\). Their distance is \(\sqrt{4+16}=2 \sqrt{5}\).
Question 5
What is the distance between the two stationary points of the curve \(y=x^{3} - 6 x^{2} + 1\)?
- \(4\)
- \(32\)
- \(4 \sqrt{17}\)
- \(4 \sqrt{65}\)
- \(2 \sqrt{229}\)
Show the answer and solution
Answer: D: \(4 \sqrt{65}\)
\(y'=3 x^{2} - 12 x=3 x \left(x - 4\right)\), so the stationary points are \((0,1)\) and \((4,-31)\). Their distance is \(\sqrt{16+1024}=4 \sqrt{65}\).
Question 6
What is the distance between the two stationary points of the curve \(y=x^{3} - 9 x^{2} + 15 x + 5\)?
- \(4 \sqrt{17}\)
- \(4 \sqrt{5}\)
- \(32\)
- \(4 \sqrt{65}\)
- \(4\)
Show the answer and solution
Answer: D: \(4 \sqrt{65}\)
\(y'=3 x^{2} - 18 x + 15=3 \left(x - 5\right) \left(x - 1\right)\), so the stationary points are \((1,12)\) and \((5,-20)\). Their distance is \(\sqrt{16+1024}=4 \sqrt{65}\).
Question 7
What is the distance between the two stationary points of the curve \(y=x^{3} - 9 x^{2} + 24 x + 5\)?
- \(4\)
- \(2 \sqrt{5}\)
- \(2\)
- \(2 \sqrt{530}\)
- \(2 \sqrt{2}\)
Show the answer and solution
Answer: B: \(2 \sqrt{5}\)
\(y'=3 x^{2} - 18 x + 24=3 \left(x - 4\right) \left(x - 2\right)\), so the stationary points are \((2,25)\) and \((4,21)\). Their distance is \(\sqrt{4+16}=2 \sqrt{5}\).
Question 8
What is the distance between the two stationary points of the curve \(y=x^{3} - 3 x - 3\)?
- \(2\)
- \(2 \sqrt{5}\)
- \(2 \sqrt{2}\)
- \(4\)
- \(2 \sqrt{10}\)
Show the answer and solution
Answer: B: \(2 \sqrt{5}\)
\(y'=3 x^{2} - 3=3 \left(x - 1\right) \left(x + 1\right)\), so the stationary points are \((-1,-1)\) and \((1,-5)\). Their distance is \(\sqrt{4+16}=2 \sqrt{5}\).
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