Home / IGCSE Add Maths 0606 / Quadratic functions / Sketching a quadratic and finding its range
Cambridge IGCSE · Topic 2.2
IGCSE Add Maths 0606 / Topic 2 / IGCSE

Find the vertex, and the sketch follows.

Once a quadratic is written as \(a(x - p)^{2} + q\), its graph is almost drawn: the sign of \(a\) gives the shape, \((p, q)\) is the vertex and \(x = p\) is the line of symmetry. Use that to sketch the curve, to write its range in correct notation over any domain, and to work back from a sketch to the equation.

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The key idea

For \(f(x) = a(x - p)^{2} + q\): the vertex is \((p, q)\) and the line of symmetry is \(x = p\). If \(a > 0\) the curve is U-shaped with minimum \(q\), so over all real \(x\) the range is \(f(x) \ge q\); if \(a < 0\) it is an upside-down U with maximum \(q\) and range \(f(x) \le q\). On a restricted domain, check whether \(x = p\) lies inside it, then work out the values at the ends.

x y O x = 1 −1 3 −3 (1, −4) y = x² − 2x − 3   = (x − 1)² − 4 a = 1 > 0: U-shape minimum point (1, −4) symmetry: x = 1 y-intercept: −3 roots: x = −1, x = 3 range for all real x: f(x) ≥ −4
Fig. 1: a sketch of y = x² - 2x - 3 = (x - 1)² - 4. The curve is symmetrical about x = 1, so the roots -1 and 3 sit 2 units either side of the vertex
Section 01

Five features make a sketch.

A sketch is not a plotted graph: it shows the correct shape and labels the key points. In the lab, turn on the sketch view and change \(a\), \(b\) and \(c\): watch the vertex, the intercepts and the line of symmetry move together. Complete the square first, then read off the features below.

FeatureHow to find it for \(y = a(x - p)^{2} + q\)
Shape\(a > 0\): U-shape (minimum); \(a < 0\): upside-down U (maximum)
Vertex\((p, q)\): the minimum or maximum point
Line of symmetry\(x = p\)
\(y\)-interceptput \(x = 0\); for \(ax^{2} + bx + c\) it is simply \(c\)
\(x\)-interceptssolve \(a(x - p)^{2} + q = 0\); there are none if the vertex is above a U or below an upside-down U
Worked example A minimum curve
Sketch \(y = x^{2} - 2x - 3\), showing the coordinates of the vertex and of the points where the curve meets the axes.
  1. Complete the square: \(x^{2} - 2x - 3 = (x - 1)^{2} - 1 - 3 = (x - 1)^{2} - 4\).
  2. \(a = 1 > 0\), so a U-shape with minimum point \((1, -4)\) and line of symmetry \(x = 1\).
  3. \(y\)-intercept: \(x = 0\) gives \(y = -3\).
  4. \(x\)-intercepts: \((x - 1)^{2} = 4\), so \(x - 1 = \pm 2\), giving \(x = 3\) or \(x = -1\).
  5. Draw a smooth U through these points, symmetrical about \(x = 1\) (Fig. 1).
Answer: U-shape, minimum \((1, -4)\), crosses the axes at \((-1, 0)\), \((3, 0)\) and \((0, -3)\)
Worked example A maximum curve with surd roots
Sketch \(y = 5 + 4x - 2x^{2}\), showing the maximum point and the intercepts.
  1. \(5 + 4x - 2x^{2} = -2(x^{2} - 2x) + 5 = -2\left[(x - 1)^{2} - 1\right] + 5 = 7 - 2(x - 1)^{2}\).
  2. \(a = -2 < 0\), so an upside-down U with maximum point \((1, 7)\).
  3. \(y\)-intercept: \(5\).
  4. \(x\)-intercepts: \(2(x - 1)^{2} = 7\), so \((x - 1)^{2} = \dfrac{7}{2}\) and \(x = 1 \pm \sqrt{\dfrac{7}{2}} = 1 \pm \dfrac{\sqrt{14}}{2}\).
  5. On a non-calculator paper leave these exact; one root is negative (just below \(-0.8\)) and one is a little under 3.
Answer: Maximum \((1, 7)\), \(y\)-intercept \(5\), roots \(x = 1 \pm \frac{\sqrt{14}}{2}\)
Your turn
Sketch \(y = x^{2} + 4x + 7\). Does it meet the \(x\)-axis?
Examiner trap

Watch the sign of \(p\). \((x + 2)^{2} + 3\) is \((x - (-2))^{2} + 3\), so the vertex is at \(x = -2\), not \(x = 2\). The \(q\) value keeps its sign: \((x - 1)^{2} - 4\) has vertex \((1, -4)\).

Section 02

The range, and the domain that limits it.

The range is the set of output values, so write it in terms of \(f(x)\) (or \(y\)), never \(x\). Over all real \(x\), a quadratic's range runs from its vertex value to infinity in one direction. On a restricted domain you need more care.

Worked example Vertex inside the domain
Find the range of \(f(x) = x^{2} - 2x - 3\) for \(0 \le x \le 4\).
  1. \(f(x) = (x - 1)^{2} - 4\): vertex at \(x = 1\), which is inside \(0 \le x \le 4\), so the least value is \(-4\).
  2. Ends: \(f(0) = -3\) and \(f(4) = 9 - 4 = 5\).
  3. The greatest value is the larger end value, \(5\). Both ends are included, so both inequalities are \(\le\).
Answer: \(-4 \le f(x) \le 5\)
Worked example Vertex outside the domain
Find the range of \(g(x) = x^{2} - 6x + 11\) for \(x > 4\).
  1. \(g(x) = (x - 3)^{2} + 2\): the vertex is at \(x = 3\), which is NOT in the domain.
  2. For \(x > 3\) the curve is rising, so on \(x > 4\) the values start just above \(g(4) = 1 + 2 = 3\) and increase without limit.
  3. \(x = 4\) is not in the domain, so \(3\) is not reached: strict inequality.
Answer: \(g(x) > 3\) (not \(g(x) \ge 2\): the minimum value 2 is never reached here)
Your turn
Find the range of \(h(x) = 7 + 4x - x^{2}\) for \(-1 \le x < 3\).
Examiner trap

Do not just substitute the two ends. For \(0 \le x \le 4\) above, \(f(0) = -3\) and \(f(4) = 5\) would give \(-3 \le f(x) \le 5\), missing the minimum \(-4\) at \(x = 1\). Always check where the vertex is first.

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

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What are the coordinates of the vertex of \(y = (x + 3)^{2} - 5\)?

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Which describes the graph of \(y = 4 - (x - 1)^{2}\)?

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Find the range of \(f(x) = (x - 2)^{2} + 1\) for \(x \in \mathbb{R}\).

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Find the range of \(f(x) = x^{2}\) for \(-2 \le x \le 3\).

Section 03

From a sketch back to the equation.

If a sketch gives you the vertex \((p, q)\), start from \(y = a(x - p)^{2} + q\) and use one more point on the curve to find \(a\). If it gives the two roots \(\alpha\) and \(\beta\), start from \(y = a(x - \alpha)(x - \beta)\); the vertex lies on the line halfway between the roots.

Worked example Vertex and one point
A quadratic curve has minimum point \((2, -8)\) and crosses the \(y\)-axis at \((0, 4)\). Find its equation in the form \(y = ax^{2} + bx + c\).
  1. Vertex form: \(y = a(x - 2)^{2} - 8\).
  2. Through \((0, 4)\): \(4 = a(0 - 2)^{2} - 8 = 4a - 8\), so \(a = 3\).
  3. Expand: \(y = 3(x^{2} - 4x + 4) - 8 = 3x^{2} - 12x + 4\).
Answer: \(y = 3x^{2} - 12x + 4\)
Worked example Roots and a maximum value
A quadratic curve meets the \(x\)-axis at \(x = -1\) and \(x = 5\) and has maximum value 18. Find its equation.
  1. The line of symmetry is halfway between the roots: \(x = \dfrac{-1 + 5}{2} = 2\). So the maximum point is \((2, 18)\).
  2. Vertex form: \(y = a(x - 2)^{2} + 18\). Through \((5, 0)\): \(0 = 9a + 18\), so \(a = -2\).
  3. \(y = -2(x - 2)^{2} + 18 = -2x^{2} + 8x + 10\).
  4. Check: \(-2x^{2} + 8x + 10 = -2(x - 5)(x + 1)\), which is zero at \(x = 5\) and \(x = -1\).
Answer: \(y = -2(x - 2)^{2} + 18\), that is \(y = -2x^{2} + 8x + 10\)
Examiner trap

The \(y\)-intercept is not \(q\) unless the vertex is on the \(y\)-axis. In the first example the curve crosses at \(4\) but \(q = -8\). Substitute the extra point into the vertex form, and square the bracket before multiplying by \(a\).

Stage 2 · Exam

Exam-style questions

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Find the range of \(f(x) = 2x^{2} - 8x + 3\), \(x \in \mathbb{R}\).

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Find the range of \(g(x) = x^{2} + 4x - 1\) for \(-3 < x \le 1\).

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A quadratic curve has maximum point \((-1, 6)\) and passes through \((1, -2)\). Find its equation.

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Find the range of \(h(x) = 3 + 2x - x^{2}\) for \(x \ge 2\).

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Quadratics sketched and ranges written.

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Six original Cambridge-style questions
Sketching quadratics from the completed square, ranges over all reals and over restricted domains with strict and non-strict ends, and finding a quadratic from its sketch.