IGCSE · Practice questions · Sketching a quadratic and finding its range

Shape, vertex, range.

Six original Cambridge-style questions on sketching \(y = f(x)\) from its maximum or minimum point, writing the range in correct notation for a given domain, and working back from a sketch to the equation.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Complete the square, then read the graph.

Complete the square: \(a(x - p)^{2} + q\) gives the vertex \((p, q)\) and the line of symmetry \(x = p\).
Shape: \(a > 0\) U-shape with a minimum; \(a < 0\) upside-down U with a maximum.
Intercepts: \(y\)-intercept from \(x = 0\); roots from \(a(x - p)^{2} + q = 0\), if there are any.
Range on a domain: vertex inside? Then work out both ends; use \(\le\) for values reached and \(<\) for values not reached.
From a sketch: start from \(a(x - p)^{2} + q\) or \(a(x - \alpha)(x - \beta)\) and use one more point to find \(a\).

Recall

[4 marks]

Write \(x^{2} + 6x + 5\) in the form \((x + p)^{2} + q\). Hence sketch \(y = x^{2} + 6x + 5\), showing the coordinates of the vertex and of the points where the curve meets the axes.

\(x^{2} + 6x + 5 = (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4\). ✓
U-shape with minimum point \((-3, -4)\) and line of symmetry \(x = -3\). ✓
\(y\)-intercept: \((0, 5)\). ✓
\((x + 3)^{2} = 4 \Rightarrow x + 3 = \pm 2\), so the curve meets the \(x\)-axis at \((-1, 0)\) and \((-5, 0)\). ✓

Calculation

[3 marks]

Express \(2x^{2} - 4x + 5\) in the form \(a(x - p)^{2} + q\). Hence sketch \(y = 2x^{2} - 4x + 5\) and explain why the curve does not meet the \(x\)-axis.

\(2x^{2} - 4x + 5 = 2(x^{2} - 2x) + 5 = 2\left[(x - 1)^{2} - 1\right] + 5 = 2(x - 1)^{2} + 3\). ✓
U-shape (\(a = 2 > 0\)) with minimum point \((1, 3)\) and \(y\)-intercept \((0, 5)\). ✓
The least value of \(y\) is \(3 > 0\), so the curve lies entirely above the \(x\)-axis and has no \(x\)-intercepts. ✓

Calculation

[3 marks]

The function f is defined by \(f(x) = x^{2} - 8x + 10\) for \(1 \le x \le 6\). Find the range of f.

\(f(x) = (x - 4)^{2} - 16 + 10 = (x - 4)^{2} - 6\). The vertex \(x = 4\) is inside the domain, so the least value is \(-6\). ✓
Ends: \(f(1) = 1 - 8 + 10 = 3\) and \(f(6) = 36 - 48 + 10 = -2\). The greatest value is \(3\). ✓
Range: \(-6 \le f(x) \le 3\). ✓

Analysis

[4 marks]

The function g is defined by \(g(x) = 9 - 6x - x^{2}\) for \(-5 \le x < 1\).

(a) Write \(g(x)\) in the form \(q - (x + p)^{2}\).

(b) Find the range of g.

(a) \(9 - 6x - x^{2} = 9 - (x^{2} + 6x) = 9 - \left[(x + 3)^{2} - 9\right] = 18 - (x + 3)^{2}\). ✓
(b) The maximum is at \(x = -3\), inside the domain, so the greatest value is \(18\). ✓
Ends: \(g(-5) = 18 - 4 = 14\) and \(g(1) = 18 - 16 = 2\), but \(x = 1\) is not in the domain, so \(2\) is not reached. ✓
The least end value is the one at \(x = 1\), so the range is \(2 < g(x) \le 18\). ✓

Analysis

[4 marks]

A sketch of a quadratic curve shows that it crosses the \(x\)-axis at \(x = 1\) and \(x = 7\) and crosses the \(y\)-axis at \((0, -14)\).

(a) Find the equation of the curve in the form \(y = ax^{2} + bx + c\).

(b) State the coordinates of its maximum point.

(a) \(y = a(x - 1)(x - 7)\). At \(x = 0\): \(-14 = a(-1)(-7) = 7a\), so \(a = -2\). ✓
\(y = -2(x^{2} - 8x + 7) = -2x^{2} + 16x - 14\). ✓
(b) The line of symmetry is \(x = \dfrac{1 + 7}{2} = 4\), and \(y = -2(3)(-3) = 18\). ✓
Maximum point \((4, 18)\); so \(y = -2(x - 4)^{2} + 18\). ✓

Analysis

[5 marks]

The function f is defined by \(f(x) = 2x^{2} - 12x + c\) for \(x \in \mathbb{R}\), where \(c\) is a constant. The range of f is \(f(x) \ge -3\).

(a) Find the value of \(c\).

(b) Find the range of f when its domain is restricted to \(0 \le x \le 5\).

(a) \(2x^{2} - 12x + c = 2(x^{2} - 6x) + c = 2\left[(x - 3)^{2} - 9\right] + c = 2(x - 3)^{2} + c - 18\). ✓
The minimum value is \(c - 18 = -3\), so \(c = 15\). ✓
(b) The vertex \(x = 3\) is inside \(0 \le x \le 5\), so the least value is \(-3\). ✓
Ends: \(f(0) = 15\) and \(f(5) = 2(2)^{2} - 3 = 5\). The greatest value is \(15\). ✓
Range: \(-3 \le f(x) \le 15\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.