2.6 The quadratic function (IB Maths: Analysis and Approaches SL): Revision notes

Section 1

The quadratic function and its graph

A quadratic function has the form $f(x) = ax^2 + bx + c$ with $a \ne 0$ . Its graph is a parabola.

If $a > 0$ the parabola opens upwards and the vertex is a minimum; if $a < 0$ it opens downwards and the vertex is a maximum.
The $y$ -intercept is $(0, c)$ .
The axis of symmetry is the vertical line $x = -\frac{b}{2a}$ (given in the formula booklet). The vertex lies on it.

For $f(x) = 2x^2 - 12x + 10$ : opens upwards, $y$ -intercept $(0, 10)$ , axis $x = 3$ , vertex $(3, f(3)) = (3, -8)$ .

Key termsquadratic functionparabolaaxis of symmetry

Common mistake

Dropping the minus sign: for $2x^2 - 12x + 10$ , $-\frac{b}{2a} = -\frac{-12}{4} = +3$ .

Section 2

Factorised form: a(x − p)(x − q)

In the form $f(x) = a(x - p)(x - q)$ the $x$ -intercepts are $(p, 0)$ and $(q, 0)$ : the zeros can be read straight off.

The axis of symmetry is midway between the zeros: $x = \frac{p + q}{2}$ . For $-2(x - 2)(x + 4)$ the zeros are $2$ and $-4$ , so the axis is $x = -1$ .

If you know the zeros and one other point, write $f(x) = a(x - p)(x - q)$ and substitute the point to find $a$ . An arch meeting the ground at $x = 0$ and $x = 40$ with height 12 at $x = 20$ gives $a(20)(-20) = 12$ , so $a = -0.03$ .

Key termsfactorised formzeros

Common mistake

Reading signs wrongly: $(x + 4)$ gives the zero $x = -4$ , not $x = 4$ .

Section 3

Vertex form: a(x − h)² + k

In the form $f(x) = a(x - h)^2 + k$ the vertex is $(h, k)$ and the axis of symmetry is $x = h$ .

Watch the sign of $h$ : $-3(x + 2)^2 + 12 = -3(x - (-2))^2 + 12$ has vertex $(-2, 12)$ .

Since $(x - h)^2 \ge 0$ , when $a < 0$ the greatest value of $f$ is $k$ , and when $a > 0$ the least value is $k$ . So the range is $f(x) \le k$ or $f(x) \ge k$ .

Key termsvertex formvertex

Exam tip

To get the $y$ -intercept from vertex form, still substitute $x = 0$ : $g(0) = -3(2)^2 + 12 = 0$ , not 12.

Section 4

Changing from one form to another

Vertex or factorised form to $ax^2 + bx + c$ : expand the brackets. $-2(x + 1)^2 + 18 = -2x^2 - 4x - 2 + 18 = -2x^2 - 4x + 16$ .

$ax^2 + bx + c$ to vertex form (completing the square): take out $a$ from the $x$ terms, complete the square, then multiply back: $2x^2 - 12x + 10 = 2[(x - 3)^2 - 9] + 10 = 2(x - 3)^2 - 8.$ $ax^2 + bx + c$ to factorised form: take out $a$ and factorise: $2(x^2 - 6x + 5) = 2(x - 1)(x - 5)$ .

Vertex form to factorised form: solve $f(x) = 0$ to find the zeros, e.g. $(x + 2)^2 = 4$ gives $x = 0, -4$ , or use symmetry about the axis.

Key termscompleting the square

Common mistake

Forgetting to multiply the $-9$ by $a$ when completing the square: $2[(x - 3)^2 - 9] + 10$ is $2(x - 3)^2 - 8$ , not $2(x - 3)^2 + 1$ .

Exam tip

Check a conversion by expanding back, or by substituting $x = 0$ into both forms.

Section 5

Quadratic models

Parabolas model arches, cables, projectiles and profit. Choose the form that matches the information:

zeros given $\to$ factorised form,
highest or lowest point given $\to$ vertex form,
$y$ -intercept and a general rule $\to$ $ax^2 + bx + c$ .

When testing whether something fits under an arch, find the height at the edges of the object: for a vehicle 8 m wide centred under an arch with axis $x = 20$ , test $x = 16$ (or $x = 24$ ).

Example

$h(x) = -0.03x(x - 40)$ : $h(16) = 11.52$ m, so an 11 m high, 8 m wide vehicle passes under the centre.

Must know

$ax^2 + bx + c$ : $y$ -intercept $(0, c)$ , axis $x = -\frac{b}{2a}$ ; $a > 0$ minimum, $a < 0$ maximum.
$a(x - p)(x - q)$ : zeros $p$ and $q$ , axis midway between them.
$a(x - h)^2 + k$ : vertex $(h, k)$ .
Move between forms by expanding, factorising or completing the square.
Use the given information (zeros, vertex, a point) to choose the form and find $a$ .

That's the notes covered.

Carry on to the next subtopic.