IB Analysis and Approaches HL · Functions · Section 3 of 15

Quadratic function

3 worked examples · By Álvaro López

Key idea

A quadratic function $f (x) = a x^{2} + b x + c$ has a parabola as its graph, with a vertex that is its highest or lowest point. Choosing the right one of its three forms gives you, with no effort, the intercepts, the vertex and the maximum or minimum value: it is the key tool for the optimisation questions in the IB exam.

The parabola and its three forms

A quadratic function is a function of the form $f (x) = a x^{2} + b x + c$ , with $a \neq = 0$ . Its graph is a parabola.

If $a > 0$ , it opens upwards and the vertex is the minimum.
If $a < 0$ , it opens downwards and the vertex is the maximum.
The greater $∣ a ∣$ is, the narrower the parabola.
The number $c$ is the value at $x = 0$ : the parabola cuts the $y$ -axis at $(0, c)$ .

The parabola is symmetric about a vertical line through the vertex, the axis of symmetry.

All three describe the same function; only what you see at first glance changes. The value $a$ is the same in all three. The canonical form comes from completing the square: $x^{2} - 4 x - 5 = (x - 2)^{2} - 9$ .

Parabola y=x^2-4x-5 with intercepts (-1,0), (5,0), (0,-5), vertex (2,-9) and axis of symmetry x=2

Intercepts with the axes, axis of symmetry and vertex

The intercept with the $y$ -axis is $(0, c)$ . The zeros (intercepts with the $x$ -axis) are the solutions of $a x^{2} + b x + c = 0$ : $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

The axis of symmetry passes through the midpoint of the two zeros (if they exist), and is always at $x = - \frac{b}{2 a}$

The vertex has that value as its $x$ -coordinate, and its $y$ -coordinate is obtained by substituting into $f$ .

Three parabolas x^2-2, x^2 and x^2+2: with two, one and no intercepts with the x-axis

Finding the equation from data

Depending on the data you are given, one form or another is more convenient:

If you are given the two zeros $p$ and $q$ and another point, start from $f (x) = a (x - p) (x - q)$ and calculate $a$ using the point.
If you are given the vertex $(h, k)$ and another point, start from $f (x) = a (x - h)^{2} + k$ and calculate $a$ .
If you are given three arbitrary points, substitute each one into $a x^{2} + b x + c$ and solve the resulting system of three equations.

Optimisation problems

When a quantity depends on a variable through a quadratic function, the maximum or minimum is at the vertex. Follow this order: write the function in one variable, decide its domain (in a context problem this is usually an interval), find the vertex and check that it falls within the domain.

Worked examples

Example 1 Basic

Complete study of a parabola

Let $f (x) = x^{2} - 4 x - 5$ .

a) Find the intercept with the $y$ -axis.

b) Find the zeros of $f$ .

c) Find the vertex and the equation of the axis of symmetry.

Solution

a) $f (0) = - 5$ , so the intercept is $(0, - 5)$ .

b) $x^{2} - 4 x - 5 = 0$ factorises as $(x + 1) (x - 5) = 0$ , so $x = - 1$ and $x = 5$ .

c) The axis of symmetry is $x = - \frac{- 4}{2 \cdot 1} = 2$ (it is also the midpoint of $- 1$ and $5$ ). Substituting: $f (2) = 4 - 8 - 5 = - 9$ . The vertex is $(2, - 9)$ , a minimum because $a = 1 > 0$ .

Example 2 Medium

Equation of a parabola from its zeros

A parabola cuts the $x$ -axis at $x = - 2$ and $x = 4$ , and passes through the point $(0, - 16)$ .

a) Find its equation in the form $f (x) = a x^{2} + b x + c$ .

b) Find its vertex.

c) Find the range of $f$ .

Solution

a) With the zeros $- 2$ and $4$ : $f (x) = a (x + 2) (x - 4)$ . As it passes through $(0, - 16)$ : $- 16 = a \cdot 2 \cdot (- 4) = - 8 a ⟹ a = 2$ Expanding, $f (x) = 2 (x^{2} - 2 x - 8) = 2 x^{2} - 4 x - 16$ .

b) The axis of symmetry passes through the midpoint of the zeros: $x = \frac{- 2 + 4}{2} = 1$ . Then $f (1) = 2 - 4 - 16 = - 18$ . The vertex is $(1, - 18)$ .

c) As $a = 2 > 0$ , the vertex is the minimum, so the range is $[- 18, + \infty [$ .

Example 3 Hard

Maximum profit of a shop

A shop sells $60 - 2 x$ T-shirts per day when it sets the price at $x$ euros each. Each T-shirt costs the shop $4$ euros. The daily profit is $B (x)$ euros.

a) Write $B (x)$ as $a x^{2} + b x + c$ .

b) Find the price that gives the maximum profit and the value of that profit.

c) Find the set of prices for which the profit is positive.

d) Write $B (x)$ in canonical form.

Solution

a) The profit per T-shirt is $x - 4$ , and $60 - 2 x$ are sold: $B (x) = (x - 4) (60 - 2 x) = - 2 x^{2} + 68 x - 240$

b) As $a = - 2 < 0$ , the maximum is at the vertex: $x = - \frac{68}{2 \cdot ( - 2 )} = 17$ . The profit is $B (17) = (13) (26) = 338$ The price is set at $17$ euros and the maximum profit is $338$ euros per day.

c) The zeros come directly from the factorised form: $x - 4 = 0$ and $60 - 2 x = 0$ , that is, $x = 4$ and $x = 30$ . The parabola opens downwards, so $B > 0$ between the zeros: $4 < x < 30$ , that is, $] 4, 30 [$ .

d) With vertex $(17, 338)$ and $a = - 2$ : $B (x) = - 2 (x - 17)^{2} + 338$ . Check by expanding: $- 2 (x^{2} - 34 x + 289) + 338 = - 2 x^{2} + 68 x - 240$ .

In the exam: in an optimisation problem the marks are in the set-up ( $B (x)$ correct) and in justifying that the maximum is at the vertex. Always write the maximum value with its unit, and answer what is asked: the price, the profit, or both.

Common mistakes

Summary

$f (x) = a x^{2} + b x + c$ is a parabola: opening upwards if $a > 0$ (minimum), downwards if $a < 0$ (maximum).
Three forms: polynomial (intercept with $y$ ), factorised $a (x - p) (x - q)$ (zeros) and canonical $a (x - h)^{2} + k$ (vertex).
Axis of symmetry $x = - \frac{b}{2 a}$ ; vertex: substitute that $x$ -value into $f$ .
$Δ = b^{2} - 4 a c$ : two zeros if $Δ > 0$ , one double zero if $Δ = 0$ , none if $Δ < 0$ .
With two zeros and a point, use the factorised form; with the vertex and a point, the canonical form.
In optimisation, the maximum or minimum is at the vertex if this falls within the domain.

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