IB Applications and Interpretation SL · Functions · Section 3 of 7

Quadratic functions and quadratic models

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. This shape describes what is thrown and falls, areas that depend on one side, and profits with an optimum, which is why it comes up again and again in the exam.

The quadratic function and its parabola

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

If $a > 0$ , the parabola opens upwards (like a U) and its vertex is a minimum.
If $a < 0$ , it opens downwards and its vertex is a maximum.
The greater $∣ a ∣$ is, the narrower the parabola.
The parameter $c$ 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, with equation

$x = - \frac{b}{2 a}$

The vertex lies on that line, so its $x$ -coordinate is $- \frac{b}{2 a}$ , and its $y$ -coordinate is found by substituting that value into $f$ .

Intercepts with the axes and the discriminant

The $y$ -intercept is $(0, c)$ . The $x$ -intercepts are the solutions of $a x^{2} + b x + c = 0$ : they are called the zeros of the function. The quadratic formula gives

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

The number $Δ = b^{2} - 4 a c$ is the discriminant, and it tells you how many $x$ -intercepts the parabola has:

When there are two intercepts, $x_{1}$ and $x_{2}$ , the axis of symmetry passes exactly through the midpoint between them: $x = \frac{x _{1} + x _{2}}{2}$ . It is a quick way to find the $x$ -coordinate of the vertex.

Three ways to write a quadratic

Each form suits one type of question best. To find the axis or the vertex from the general form, use $x = - \frac{b}{2 a}$ . To go from the factored form to the general form, expand the product. In all three, the parameter $a$ is the same.

Parabola f(x) = x² − 6x + 5 with intercepts (0; 5), (1; 0) and (5; 0), minimum (3; −4), axis of symmetry x = 3 and symmetric point (6; 5)

Finding the expression from data

To determine $a$ , $b$ and $c$ , or the parameters of another form, write one equation for each piece of data:

Quadratic models

Many problems are modelled with a quadratic: the height of an object thrown as a function of time, the area of a figure as a function of one of its sides, or the profit as a function of the number of units. The typical questions are:

What is the maximum (or minimum) and where is it reached? It is the vertex. Give both of its coordinates and say what each one represents.
When does it take a particular value? Set the function equal to that value and solve. If you get two solutions, discard the ones that make no sense in context (a negative time, a negative length).
When does it hit the ground? It is the positive zero, that is, the solution of $h (t) = 0$ with $t > 0$ .

Intersection of functions

Two functions $f$ and $g$ intersect where $f (x) = g (x)$ . For a quadratic and a line, this equation reduces to a second-degree one, with zero, one or two solutions. Solve it by hand or with the GDC by drawing both graphs. Then find the $y$ -coordinate by substituting into either of them.

Worked examples

Example 1 Basic

Features of a parabola

Consider the function $f (x) = x^{2} - 6 x + 5$ .

a) Find the equation of the axis of symmetry.

b) Find the coordinates of the vertex.

c) Find the $x$ -intercepts.

d) Find the $y$ -intercept.

Solution

a) With $a = 1$ and $b = - 6$ : $x = - \frac{- 6}{2 \cdot 1} = 3$ . The axis of symmetry is $x = 3$ .

b) Substituting $x = 3$ : $f (3) = 9 - 18 + 5 = - 4$ . The vertex is $(3, - 4)$ . As $a > 0$ , it is a minimum.

c) Solve $x^{2} - 6 x + 5 = 0$ : $(x - 1) (x - 5) = 0$ , so $x = 1$ and $x = 5$ . Intercepts: $(1, 0)$ and $(5, 0)$ . This confirms that $x = 3$ is the midpoint: $\frac{1 + 5}{2} = 3$ .

d) The $y$ -intercept is $(0, c) = (0, 5)$ .

Example 2 Medium

Parabola from its intercepts

A parabola cuts the $x$ -axis at $(- 3, 0)$ and $(5, 0)$ , and cuts the $y$ -axis at $(0, 15)$ . Its equation is $f (x) = a (x - p) (x - q)$ .

a) Find the value of $a$ .

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

c) Find the coordinates of the vertex.

Solution

a) The $x$ -intercepts give $p = - 3$ and $q = 5$ , so $f (x) = a (x + 3) (x - 5)$ . Since $f (0) = 15$ : $a \cdot 3 \cdot (- 5) = 15 \Rightarrow - 15 a = 15 \Rightarrow a = - 1$

b) Expanding: $f (x) = - (x + 3) (x - 5) = - (x^{2} - 2 x - 15) = - x^{2} + 2 x + 15$ .

c) The axis of symmetry is the midpoint of the intercepts: $x = \frac{- 3 + 5}{2} = 1$ . Then $f (1) = - 1 + 2 + 15 = 16$ . The vertex is $(1, 16)$ , a maximum, since $a < 0$ .

Example 3 Hard

Trajectory of a ball

A player throws a ball from a height above the ground. The height of the ball, in metres, $t$ seconds after the throw is $h (t) = - 5 t^{2} + 12 t + 2, t \geq 0$

a) Write down the height from which the ball is thrown.

b) Find the maximum height and the instant at which it is reached.

c) Find the instant at which the ball reaches the ground.

d) Find the length of time during which the ball is more than $8$ metres above the ground.

Height h(t) = −5t² + 12t + 2 from t = 0 to the ground (t ≈ 2.56), with maximum (1.2; 9.2), initial height 2 m and orange section where h > 8 m between t ≈ 0.71 and t ≈ 1.69

Solution

a) It is the value at $t = 0$ : $h (0) = 2$ . The ball is thrown from a height of $2$ m.

b) The maximum height is at the vertex. The instant is $t = - \frac{12}{2 \cdot ( - 5 )} = \frac{6}{5} = 1.2 s$ and the height is $h (1.2) = - 5 \cdot 1.44 + 14.4 + 2 = 9.2$ . The maximum height is $9.2$ m, reached at $1.2$ s. Using the GDC, the same maximum is read at the point $(1.2; 9.2)$ .

c) Solve $h (t) = 0$ , that is, $- 5 t^{2} + 12 t + 2 = 0$ : $t = \frac{- 12 \pm 144 + 40}{- 10} = \frac{6 \mp 46}{5}$ The solutions are $t \approx - 0.156$ and $t \approx 2.556$ . The negative one is discarded because time cannot be negative in the model. The ball reaches the ground at $t = \frac{6 + 46}{5} \approx 2.56$ s.

d) Solve $h (t) = 8$ : $- 5 t^{2} + 12 t - 6 = 0$ , which is equivalent to $5 t^{2} - 12 t + 6 = 0$ . Then $t = \frac{12 \pm 144 - 120}{10} = \frac{6 \pm 6}{5} \approx 0.710 and 1.690$ The ball is above $8$ m between these two instants. The elapsed time is the difference: $\frac{6 + 6}{5} - \frac{6 - 6}{5} = \frac{2 6}{5} \approx 0.98 s$

In the exam: when you find the instant at which the ball reaches the ground, state why you reject one of the two solutions. In part d), do not subtract the two solutions automatically: first interpret that between the two roots the height is greater than $8$ .

Common mistakes

Summary

$f (x) = a x^{2} + b x + c$ has a parabola as its graph: opening upwards if $a > 0$ (minimum) and downwards if $a < 0$ (maximum).
Axis of symmetry: $x = - \frac{b}{2 a}$ . The vertex is found by substituting that value into $f$ .
$y$ -intercept: $(0, c)$ . $x$ -intercepts: solutions of $f (x) = 0$ ; the discriminant $Δ = b^{2} - 4 a c$ tells you how many there are.
Forms: general $a x^{2} + b x + c$ , factored $a (x - p) (x - q)$ and vertex $a (x - h)^{2} + k$ .
In models, the vertex gives the maximum or minimum; solutions that make no sense in context are discarded.
Two functions intersect where $f (x) = g (x)$ ; the GDC gives the points of intersection.

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