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Topic 2

Quadratic functions

Syllabus 2.1–2.5 · Papers 1 and 2

🎯What you need to be able to do

  • Find the maximum or minimum value of \( ax^{2} + bx + c \) by completing the square, or by differentiation.
  • Use that value to sketch the graph, and to find the range of the function on a given domain.
  • Use the discriminant to decide whether a quadratic has two, one or no real roots.
  • Use the same idea to decide whether a line cuts a curve, touches it, or misses it.
  • Solve quadratic equations by factorising, the formula and completing the square.
  • Solve quadratic inequalities, and write the solution set in the correct form.

📚The mathematics

Completing the square

Every quadratic can be written as \( a(x + p)^{2} + q \). That form tells you at a glance:

vertex \( (-p,\ q) \)
line of symmetry \( x = -p \)
least value \( q \) if \( a > 0 \); greatest value \( q \) if \( a < 0 \)

When \( a \ne 1 \), take \(a\) out of the \( x^{2} \) and \(x\) terms only, complete the square inside the bracket, then multiply back:

\[ 3x^{2} - 12x + 5 = 3\left(x^{2} - 4x\right) + 5 = 3\left[(x - 2)^{2} - 4\right] + 5 = 3(x - 2)^{2} - 7 \]

The step that goes wrong is multiplying the \( -4 \) by 3. Expand your answer back out — it takes ten seconds and catches it every time.

The syllabus also allows differentiation: \( \dfrac{d}{dx}\left(3x^{2} - 12x + 5\right) = 6x - 12 = 0 \) at \( x = 2 \), where \( y = -7 \). Use whichever the question asks for; if it says “by completing the square”, calculus gets no credit.

Range on a domain

To find the range for a restricted domain, sketch the parabola only over that domain and read off the lowest and highest points. Check three places: both ends of the domain, and the vertex if it lies inside the domain.

The parabola y = 3x squared minus 12x plus 5 drawn solid for 0 to 5 and dashed outside. The vertex (2, minus 7) and the end points (0, 5) and (5, 20) are marked, and a bracket on the y-axis shows the range from minus 7 to 20.
For \( 0 \le x \le 5 \) the lowest point is the vertex, not an end point, so the range is \( -7 \le f(x) \le 20 \).

The discriminant

For \( ax^{2} + bx + c = 0 \) the roots are \( x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \) (the formula is in the List of formulas). The part under the square root, \( b^{2} - 4ac \), decides how many real roots there are:

\( b^{2} - 4ac > 0 \): two distinct real roots
\( b^{2} - 4ac = 0 \): two equal roots
\( b^{2} - 4ac < 0 \): no real roots

Lines and curves. To find where \( y = mx + c \) meets a quadratic curve, set them equal and rearrange to \( (\ldots)x^{2} + (\ldots)x + (\ldots) = 0 \). That equation has two, one or no roots exactly when the line cuts the curve twice, is a tangent, or does not meet it. “Intersects” on its own means \( \ge 0 \) (touching counts); “at two distinct points” means \( > 0 \).

The curve y = 3x squared minus 12x plus 5 with three lines through (0, 2): y = minus 6x plus 2 touching it at (1, minus 4), y = minus 18x plus 2 touching it at (minus 1, 20), and the horizontal line y = 2 cutting it twice.
Three lines \( y = kx + 2 \) and the worked example’s curve: \( k = -6 \) and \( k = -18 \) give a discriminant of zero (tangents); \( k = 0 \) gives a positive one (two points).

Quadratic inequalities

Find the critical values (the roots), sketch the parabola, then read off where it is above or below the axis. For \( a > 0 \):

  • \( (\ldots) < 0 \) — between the roots: one interval, \( \alpha < x < \beta \).
  • \( (\ldots) > 0 \) — outside the roots: two pieces, \( x < \alpha \) or \( x > \beta \).

The syllabus is explicit about the form: \( -3 < x < 4 \) and \( x < 1 \) or \( x > 6 \). Never write the second as \( 6 < x < 1 \), which describes no numbers at all. On Paper 2 a correct solution set earns the marks without working, but on Paper 1 the working is where most of them are.

✏️Worked example

\( f(x) = 3x^{2} - 12x + 5 \). (a) Write \( f(x) \) in the form \( a(x + p)^{2} + q \), and state the coordinates of the minimum point. [3] (b) Find the range of \(f\) for the domain \( 0 \le x \le 5 \). [2] (c) Find the values of \(k\) for which the line \( y = kx + 2 \) is a tangent to \( y = f(x) \). [3] (d) Hence write down the values of \(k\) for which the line meets the curve at two distinct points. [1]

(a) As above, \( f(x) = 3(x - 2)^{2} - 7 \). The minimum point is \( (2, -7) \).

(b) The vertex \( x = 2 \) is inside the domain, so the least value is \(-7\). At the ends, \( f(0) = 5 \) and \( f(5) = 75 - 60 + 5 = 20 \). So \( -7 \le f(x) \le 20 \).

(c) Set the line equal to the curve:

\[ 3x^{2} - 12x + 5 = kx + 2 \;\Longrightarrow\; 3x^{2} - (12 + k)x + 3 = 0 \]

A tangent means equal roots, so \( b^{2} - 4ac = 0 \): \( (12 + k)^{2} - 36 = 0 \), so \( 12 + k = \pm 6 \), giving \( k = -6 \) or \( k = -18 \).

(d) The discriminant \( (12 + k)^{2} - 36 \) is a positive quadratic in \(k\) with roots \(-18\) and \(-6\), so it is positive outside them: \( k < -18 \) or \( k > -6 \).

Check it. With \( k = -6 \) the equation is \( 3x^{2} - 6x + 3 = 3(x - 1)^{2} = 0 \): one repeated root, \( x = 1 \), so the line touches at \( (1, -4) \). With \( k = 0 \) (which is \( > -6 \)) the line \( y = 2 \) should cross twice: \( 3x^{2} - 12x + 3 = 0 \) has discriminant \( 144 - 36 = 108 > 0 \). Both agree with the diagram.
The wrong range. Taking only the end values in (b) gives \( 5 \le f(x) \le 20 \), which misses everything below 5. Whenever the vertex is inside the domain, it is one end of the range.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Find the values of \(k\) for which the equation \( kx^{2} + 4x + k - 3 = 0 \) has two equal roots. [3]
Equal roots: \( 16 - 4k(k - 3) = 0 \Rightarrow k^{2} - 3k - 4 = 0 \Rightarrow (k - 4)(k + 1) = 0 \), so \( k = 4 \) or \( k = -1 \). Both are allowed: neither makes the \( x^{2} \) coefficient zero (if \( k = 0 \) the equation would not be a quadratic at all).
2. (No calculator.) Find the set of values of \(k\) for which the line \( y = 2x + k \) meets the curve \( y = x^{2} - 4x + 11 \) at two distinct points. [3]
\( x^{2} - 4x + 11 = 2x + k \Rightarrow x^{2} - 6x + (11 - k) = 0 \). Two distinct points: \( 36 - 4(11 - k) > 0 \Rightarrow 4k - 8 > 0 \Rightarrow k > 2 \).
3. (No calculator.) Solve \( 2x^{2} - 7x - 15 < 0 \). [3]
\( (2x + 3)(x - 5) < 0 \), critical values \( -\tfrac{3}{2} \) and 5. The parabola opens upwards, so it is negative between them: \( -\tfrac{3}{2} < x < 5 \).
4. (No calculator.) Find the values of \(x\) for which \( x(x + 4) \ge 2x + 15 \). [3]
Collect on one side first: \( x^{2} + 2x - 15 \ge 0 \), so \( (x + 5)(x - 3) \ge 0 \). Outside the roots: \( x \le -5 \) or \( x \ge 3 \). Dividing by \(x\) at the start would lose solutions — never divide an inequality by something whose sign you do not know.
5. (No calculator.) \( f(x) = 4 + 6x - x^{2} \). (a) Write \( f(x) \) in the form \( a - (x - b)^{2} \). (b) Find the range of \(f\) for \( -1 \le x \le 4 \). [4]
(a) \( -(x^{2} - 6x) + 4 = -\left[(x - 3)^{2} - 9\right] + 4 = 13 - (x - 3)^{2} \). (b) The maximum 13 is at \( x = 3 \), inside the domain. Ends: \( f(-1) = 4 - 6 - 1 = -3 \), \( f(4) = 4 + 24 - 16 = 12 \). The least is \(-3\), so \( -3 \le f(x) \le 13 \).
6. (No calculator.) Find the values of \(m\) for which the line \( y = mx - 1 \) does not meet the curve \( y = x^{2} + 3 \). [3]
\( x^{2} + 3 = mx - 1 \Rightarrow x^{2} - mx + 4 = 0 \). No intersection: \( m^{2} - 16 < 0 \), so \( (m - 4)(m + 4) < 0 \) and \( -4 < m < 4 \).
7. (Calculator.) Solve \( 3x^{2} - 5x - 4 = 0 \), giving your answers correct to 3 significant figures. [2]
\( x = \dfrac{5 \pm \sqrt{25 + 48}}{6} = \dfrac{5 \pm \sqrt{73}}{6} \), so \( x = 2.26 \) or \( x = -0.591 \). Keep the full calculator value until the last step; rounding \( \sqrt{73} \) to 8.5 first gives \(-0.583\), which is wrong at 3 s.f.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — put a slider on \(k\) in \( y = kx + 2 \) and watch the line become a tangent at the two critical values
  • Khan Academy — completing the square and quadratic inequalities