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Cambridge IGCSE0606

Quadratic functions

Mathematics Additional 0606 Chapter Notes

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Quadratic functions
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Quadratic functions notes

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In this chapter

  1. 1Completing the square
  2. 2Maximum and minimum by differentiation
  3. 3Sketching the graph and finding the range
  4. 4The discriminant and the roots
  5. 5A line and a curve
  6. 6Solving quadratic equations
  7. 7Quadratic inequalities

1. Completing the square

Write ax² + bx + c in the form a(x + p)² + q. The minimum point or maximum point of the curve is then (-p, q).

Take a out of the x² and x terms first. Half the x coefficient goes in the bracket. Then subtract what the bracket adds.

x² + bx + c = (x + b/2)² + c - b²/4

ax² + bx + c = a(x + b/2a)² + c - b²/4a

Examiner insight

“Those candidates who chose to complete the square often made sign errors.” “When a form for an answer is given in the question candidates should ensure that their final answer fits this.”

Worked example 12 marks

The point A is the minimum point on the curve y = x² - 6x + 10. Using the method of completing the square, find the coordinates of A.

  1. 1

    y = (x - 3)² + 1 [1]

  2. 2

    minimum point A = (3, 1) [1]

2. Maximum and minimum by differentiation

The gradient is zero at the maximum or minimum point. Differentiate, set the gradient equal to 0 and solve for x.

Then put that x back into the function to find the maximum value or minimum value. The question asks for the value of the function, not just for x.

f(x) = ax² + bx + c ⇒ f'(x) = 2ax + b = 0 ⇒ x = -b/2a

Examiner insight

“Successful candidates used differentiation or completing the square to find a minimum value.” “Occasionally candidates omitted to find the greatest value of the function, but this was rare.”

Worked example 13 marks

Find the minimum value of f(x) = 3x² - 12x + 5 by differentiation.

  1. 1

    f'(x) = 6x - 12 [1]

  2. 2

    6x - 12 = 0, so x = 2 [1]

  3. 3

    f(2) = 3(4) - 12(2) + 5 = -7 [1]

3. Sketching the graph and finding the range

A quadratic graph is a parabola. If a > 0 it has a minimum point. If a < 0 it has a maximum point. The turning point is (-p,q) when the function is written as a(x + p)² + q.

For all real x, the range starts or ends at the turning value. For a given domain, work out the value of the function at each end of the domain. Include the turning value only if the turning point lies in the domain. The range runs from the lowest of these values to the highest.

Write the range with correct notation: f(x) ≥ q or y ≥ q. An inequality in x is a domain, not a range.

Examiner insight

“Far too many responses were of the type x > 6, rather than the acceptable f ( x ) > 6 or y > 6.” “Some candidates made sign errors, notation errors or assumed the range was the set of real numbers.”

Worked example 14 marks

The function f is defined by f(x) = 2x² - 6x + 4 for all real values of x.

(a) Write f(x) in the form a(x + b)² + c where a, b and c are constants.

(b) Hence find the range of f.

  1. 1

    (a) a = 2 [1]

  2. 2

    2x² - 6x = 2[(x - 3/2)² - 9/4], so b = -3/2 [1]

  3. 3

    c = 4 - 9/2 = -1/2, so f(x) = 2(x - 3/2)² - 1/2 [1]

  4. 4

    (b) f(x) ≥ -1/2 [1]

4. The discriminant and the roots

For ax² + bx + c = 0, the discriminant is b² - 4ac. Its value tells you how many real roots there are.

Greater than zero: two real roots. Equal to zero: two equal roots. Less than zero: no real roots. To show there are no real roots, work out the discriminant and state the conclusion in words.

The same conditions decide how a quadratic graph meets the x-axis: it cuts it twice, touches it once, or misses it.

b² - 4ac > 0 (two real roots)

b² - 4ac = 0 (two equal roots)

b² - 4ac < 0 (no real roots)

Examiner insight

“Candidates should be aware that, as in the question, the comment should relate to real solutions not just solutions.” “Candidates needed to show, rather than just state, that 4x2 – 8x + 13 has no real roots, usually by showing that the discriminant is negative.”

Worked example 13 marks

Show that there is no real value of k for which the equation (k - 2)x² + (2k + 3)x + 3 = 0 has two equal roots.

  1. 1

    b² - 4ac = (2k + 3)² - 4(k - 2)(3) [1]

  2. 2

    = 4k² + 12k + 9 - 12k + 24 = 4k² + 33 [1]

  3. 3

    4k² + 33 = 0 is not possible for real k, so there are no equal roots [1]

5. A line and a curve

To find where a line meets a curve, put the two equations equal. Rearrange to a quadratic in x equal to zero. Then use the discriminant of that quadratic.

Intersect: b² - 4ac > 0 for two points. Tangent: b² - 4ac = 0. Does not intersect: b² - 4ac < 0.

For a tangent the discriminant is set equal to zero. Solving that equation gives values of the unknown, not an inequality.

Examiner insight

“Some candidates, having found the critical values for k, went on to form an inequality, showing a basic misunderstanding of the problem.” “The most successful method was to first eliminate y from the two given equations and then to set the discriminant of this quadratic in x to zero to lead to a quadratic in k.”

Worked example 14 marks

The line y = x + k and the curve y = x² - 3x + 5 are given. Find the value of k for which the line is a tangent to the curve, and the values of k for which the line does not intersect the curve.

  1. 1

    x² - 3x + 5 = x + k, so x² - 4x + 5 - k = 0 [1]

  2. 2

    b² - 4ac = 16 - 4(5 -k) = 4k - 4 [1]

  3. 3

    tangent: 4k - 4 = 0, so k = 1 [1]

  4. 4

    no intersection: 4k - 4 < 0, so k < 1 [1]

6. Solving quadratic equations

Solve for real roots by factorisation, by the quadratic formula or by completing the square. The formula is given in the List of formulas.

Write the formula down before you substitute into it. Keep surds exact unless the question asks for a decimal. On the calculator paper, correct answers are acceptable without working.

x = (-b ± √(b² - 4ac))/2a

Examiner insight

“As has been reported in previous years, it is advisable to quote the quadratic formula before using it to reduce the risk of incorrect substitution.” “Candidates are recommended to use either a factorisation method or the quadratic formula to solve quadratic equations and then use their calculator as a check.”

Worked example 13 marks

Solve 3x² - 5x - 1 = 0. Give your answers in exact form.

  1. 1

    x = (5 ± √((-5)² - 4(3)(-1)))/(2(3)) [1]

  2. 2

    = (5 ± √37)/6 [1]

  3. 3

    x = (5 + √37)/6 or x = (5 - √37)/6 [1]

7. Quadratic inequalities

Rearrange to a quadratic compared with zero. Solve the matching equation to find the critical values. Then sketch the curve to decide which region is wanted.

If the curve is a U-shape (a > 0): less than zero is the region between the critical values. Greater than zero is the outside region, written as two separate ranges.

Write the solution in the correct form, for example -3 < x < 4 or x < 1 or x > 6.

Examiner insight

“Forming the required inequality involved identifying the outside region and this was best recognised by the use of a diagram.” “Candidates seemed to be determined to express the required region as a single inequality which was either the inside region and hence incorrect or a mathematically incorrect statement.”

Worked example 16 marks

(a) Write 2x² - 6x + 5 in the form a(x + b)² + c, where a, b and c are constants.

(b) Hence or otherwise solve the inequality 2x² - 6x + 5 > 13.

  1. 1

    (a) a = 2 [1]

  2. 2

    b = -3/2 [1]

  3. 3

    c = 1/2, so 2(x - 3/2)² + 1/2 [1]

  4. 4

    (b) 2(x - 3/2)² + 1/2 > 13, so (x - 3/2)² > 25/4 [1]

  5. 5

    x - 3/2 = ±5/2, so the critical values are x = 4 and x = -1 [1]

  6. 6

    x < -1 or x > 4 [1]

Recap

  • Completing the square: ax² + bx + c = a(x + p)² + q, with turning point (-p, q).
  • By differentiation: set the gradient to 0, solve for x, then find the value of the function.
  • Range: start from the turning value and check both ends of a given domain. Write it as f(x) or y, never x.
  • Discriminant b² - 4ac: positive for two real roots, zero for two equal roots, negative for no real roots.
  • Line and curve: equate, rearrange, use the discriminant. Tangent means equal to zero.
  • Solve by factorising, formula or completing the square. Keep surds exact.
  • Inequalities: find critical values, sketch, then read off between or outside.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. Write x² + 8x + 3 in the form (x + a)² + b.2 marks
  2. Find the minimum value of y = x² - 10x + 7 by differentiation.3 marks
  3. Do not use a calculator. The function f is defined by f(x) = 5 - 2x - x² for all real x. Find the maximum value of f(x) and write down the range of f.3 marks
  4. The function f is defined by f(x) = x² - 4x + 1 for 0 ≤ x ≤ 5. Find the range of f.3 marks
  5. Do not use a calculator. Show that the equation 3x² - 4x + 2 = 0 has no real roots.2 marks
  6. Do not use a calculator. Solve 2x² + 3x - 5 = 0.3 marks
  7. The line y = 2x + k is a tangent to the curve y = x² + 3. Find the value of k.4 marks
  8. Do not use a calculator. Solve x² - 2x - 8 < 0.3 marks
  9. Do not use a calculator. The line y = x + 1 and the curve y = x² + kx + k have at least one point in common. Find the set of possible values of k.5 marks
  10. A factory makes x tonnes of jam each week, where 0 ≤ x ≤ 5. Its weekly profit is P thousand dollars, where P = -2x² + 12x - 10. A negative value of P is a loss. (a) Write P in the form a(x + b)² + c and state the greatest weekly profit. (b) Find the range of P. (c) Find the set of values of x for which the factory makes a loss.8 marks

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