Year 12 Maths - Methods (Unit 3 & Unit 4) Functions, relations and graphs

Applications of functions

Q: What does it mean to model a situation with a function?

Modelling means writing a function that links the variables in a real situation - such as an area, a cost or a height expressed in terms of one input - so that the features of the function describe the situation. Substituting a value into the function then predicts the quantity, and the graph shows the overall behaviour.

Q: How do you choose a suitable model?

Look at the pattern in the relationship or data. A constant rate of change suggests a linear model, a single turning point suggests a quadratic, and more complicated shapes suggest a higher-degree polynomial or another type of function. The model should also match known features such as intercepts and limiting values.

Q: Why do you restrict the domain in an applied problem?

Because only some inputs make physical sense. Lengths, areas and times cannot be negative, and a constraint such as a fixed amount of fencing or a fixed sheet of cardboard puts an upper limit on the input, so the domain is restricted to the realistic values for which the model is valid.

Q: How do you interpret the intercepts of a model?

The y-intercept f(0) is the initial or starting value of the quantity, such as a fixed cost or an initial height. An x-intercept is an input where the modelled quantity is zero, such as the time a projectile returns to the ground or a width that gives zero area.

Q: How do you find a maximum or minimum in context?

Find the turning point of the model - the vertex of a quadratic at x equals minus b over 2a, or a stationary point where the derivative is zero - then check it lies inside the domain and read off the corresponding value. The optimal input and the optimal value should both be stated with units.

Q: What does the rate of change tell you in an application?

The gradient or derivative measures how fast the modelled quantity is changing. The average rate of change over an interval is the change in output divided by the change in input, and the instantaneous rate at a point is the derivative there, for example a velocity or a marginal cost.

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Theory

Applications of functions describe a real situation with a rule $y=f(x)$ - linear, quadratic or a higher-degree polynomial - built from the quantities and constraints in the problem. The domain is restricted to values that make sense in context, and the key features of the graph carry meaning: intercepts give initial or zero values, a turning point gives a maximum or minimum, and the gradient or derivative gives a rate of change.

To model a situation you write one quantity as a function of another, for example an area, cost or height in terms of a single input. The shape of the relationship decides the type of function: a constant rate of change gives a linear model $y=mx+c$; a single turning point gives a quadratic; and a more involved shape, such as a volume that rises then falls, is often a polynomial of higher degree.

A model usually comes from a constraint. If a fixed amount of material or a fixed perimeter links two variables, use that relationship to eliminate one variable so the quantity of interest depends on a single input. The realistic inputs then form the restricted domain - lengths, areas and times cannot be negative, and the constraint puts an upper limit on the input.

Once the model is set up, its features are read in context. The $y$-intercept $f(0)$ is the initial value; an $x$-intercept is where the quantity is zero. A turning point gives the maximum or minimum value, and the gradient (average rate) or derivative (instantaneous rate) says how fast the quantity is changing.

Key idea. Build the rule from the constraint, restrict the domain to realistic inputs, then interpret the intercepts, turning point and rate of change as features of the situation - always checking the answer lies inside the domain.

A thrown ball: $h(t)=20t-5t^{2}$ on the domain $0\le t\le 4$. Intercepts at $t=0,4$ (launch and landing); maximum height $20$ m at $t=2$

A pen with $40$ m of fencing: $A(x)=x(20-x)$ on $0\le x\le 20$. Zero area at the endpoints; maximum $100$ m$^2$ at width $x=10$

Interpreting the intercepts of a model $y=f(x)$:

\[f(0)=\text{initial value (}y\text{-intercept)} \qquad f(x)=0 \Rightarrow \text{zeros (}x\text{-intercepts)}\]

f (0) = initial value

The turning point of a quadratic model $y=ax^{2}+bx+c$ - its maximum or minimum:

\[x=-\dfrac{b}{2a}\]

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

For a general model, a maximum or minimum occurs at a stationary point:

\[f'(x)=0\]

f^{'} (x) = 0

The average rate of change of $f$ over an interval $[a,b]$:

\[\dfrac{f(b)-f(a)}{b-a}\]

\frac{f (b) - f (a)}{b - a}

Domain in context. Always state the restricted domain, e.g. \(0\le t\le 4\) or \(0<x<6\), and confirm the optimal input lies inside it before quoting the maximum or minimum value.

How to model a situation with a function

Define the variables. Name the input (e.g. a width $x$ or time $t$) and the quantity to model (area, volume, cost, height).
Form the rule. Use the constraint (fixed perimeter, fixed material, given ratio) to eliminate any extra variable, so the quantity depends on a single input.
Restrict the domain. Write the realistic values of the input - non-negative, and bounded above by the constraint.
Find the required feature. An intercept (initial or zero value), a maximum or minimum (vertex $x=-\dfrac{b}{2a}$ or $f'(x)=0$), or a rate of change (gradient or derivative).
Interpret in context. State the answer with units, and check it lies inside the domain.

Optimisation shortcut. For a quadratic model you do not need calculus - the turning point is exactly halfway between the two \(x\)-intercepts, so if \(A(x)=x(20-x)\) has zeros at \(0\) and \(20\), the maximum is at \(x=10\).

Example 1 — a linear model

A phone plan costs a fixed $\$20$ per month plus $15$ cents per call. The monthly cost is $C(n)=20+0.15n$ dollars for $n$ calls. Interpret the two numbers, and find the cost for $80$ calls.

Solution

Interpret the intercept and the rate:

$C(0)$	$=$	$\$20$ (fixed monthly cost)
gradient	$=$	$0.15$ ($\$0.15$ per extra call)

Evaluate at $n=80$:

$C(80)$	$=$	$20+0.15\times 80$
	$=$	$20+12=32$

$\therefore$ $80$ calls cost $\$32$; $\$20$ is the base cost and $15$c is the rate per call

C (80) = 32

Example 2 — quadratic area and domain

A rectangular pen is made with $40$ m of fencing. If one side is $x$ m, express the area $A$ as a function of $x$, state the domain, and find the dimensions giving the greatest area.

Solution

Form the model — the two side lengths sum to $20$:

$2x+2y$	$=$	$40 \Rightarrow y=20-x$
$A(x)$	$=$	$x(20-x)$

State the domain — both sides must be positive:

domain

$:$

$0<x<20$

Maximum — the vertex is midway between the zeros $0$ and $20$:

$x$

$=$

$10 \Rightarrow A=10\times 10=100$

$\therefore$ the greatest area is $100$ m$^2$, from a $10\text{ m}\times 10\text{ m}$ square

A (x) = x (20 - x)

Example 3 — forming a volume model

An open box is made from a $12$ cm square of cardboard by cutting a square of side $x$ cm from each corner and folding up the sides. Find the volume $V(x)$, state the domain, and find $x$ giving the maximum volume.

Solution

Form the model — base $(12-2x)$ by $(12-2x)$, height $x$:

$V(x)$

$=$

$x(12-2x)^{2}$

Domain — the cut cannot exceed half the side:

domain

$:$

$0<x<6$

Differentiate and solve $V'(x)=0$:

$V'(x)$	$=$	$(12-2x)(12-6x)$
$12-6x=0$	$\Rightarrow$	$x=2$ ($x=6$ rejected)

Maximum volume at $x=2$:

$V(2)$

$=$

$2(8)^{2}=128$

$\therefore$ the maximum volume is $128$ cm$^3$, when $x=2$ cm

V (x) = x {(12 - 2 x)}^{2}

Example 4 — intercepts, maximum and rate

A ball is thrown so its height is $h(t)=20t-5t^{2}$ metres after $t$ seconds. Find when it lands, its maximum height, and its average velocity over the first $2$ seconds.

Solution

Landing — the $x$-intercept where $h=0$:

$20t-5t^{2}$	$=$	$5t(4-t)=0$
$t$	$=$	$0$ or $t=4$

Maximum height — vertex midway between $t=0$ and $t=4$:

$t=2:\;h(2)$

$=$

$40-20=20$

Average velocity over $[0,2]$:

$\dfrac{h(2)-h(0)}{2-0}$

$=$

$\dfrac{20-0}{2}=10$

$\therefore$ it lands at $t=4$ s; maximum height $20$ m at $t=2$ s; average velocity $10$ m/s

h (2) = 20

Common pitfalls

Forgetting to restrict the domain. A width or a time cannot be negative, and a constraint bounds it above. For the open box, $x$ must satisfy $0<x<6$, not all real $x$; values outside give a meaningless model.

Confusing the input with the maximum value. The optimising input $x=10$ is the width, not the area. Substitute it back to get the maximum value $A=100$ m$^2$; state both, each with units.

Not eliminating the second variable. Use the constraint first, e.g. $y=20-x$, so the model depends on a single input before you optimise. Trying to optimise with two variables still present will not work.

Frequently asked questions

What does it mean to model a situation with a function?

You write one quantity - an area, cost or height - as a function $f$ of a single input, so features of $f$ describe the situation. Substituting a value predicts the quantity, and the graph shows the overall behaviour.

How do you choose a suitable model?

Match the pattern: a constant rate of change is linear, a single turning point is quadratic, and a shape that rises then falls (like a volume) is a higher-degree polynomial. Check that the model fits known intercepts and limits.

Why do you restrict the domain in an applied problem?

Only some inputs make sense: lengths, areas and times are non-negative, and a constraint such as fixed fencing or a fixed sheet limits the input from above. The domain is those realistic values, e.g. $0<x<6$.

How do you interpret the intercepts of a model?

The $y$-intercept $f(0)$ is the initial value, such as a fixed cost. An $x$-intercept is where the quantity is zero, such as when a projectile returns to the ground or a width gives zero area.

How do you find a maximum or minimum in context?

Find the turning point: the vertex of a quadratic at $x=-\dfrac{b}{2a}$, or a stationary point where $f'(x)=0$. Check it lies in the domain, then read off the value and state both with units.

What does the rate of change tell you in an application?

It measures how fast the quantity changes. The average rate over $[a,b]$ is $\dfrac{f(b)-f(a)}{b-a}$, and the instantaneous rate at a point is the derivative there - for example a velocity or a marginal cost.

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Applications of trigonometric functions

\(C(0)\)	\(=\)	\(\$20\) (fixed monthly cost)
gradient	\(=\)	\(0.15\) (\(\$0.15\) per extra call)

\(C(80)\)	\(=\)	\(20+0.15\times 80\)
	\(=\)	\(20+12=32\)

\(2x+2y\)	\(=\)	\(40 \Rightarrow y=20-x\)
\(A(x)\)	\(=\)	\(x(20-x)\)

\(V'(x)\)	\(=\)	\((12-2x)(12-6x)\)
\(12-6x=0\)	\(\Rightarrow\)	\(x=2\) (\(x=6\) rejected)

\(20t-5t^{2}\)	\(=\)	\(5t(4-t)=0\)
\(t\)	\(=\)	\(0\) or \(t=4\)

Applications of functions

📖 Theory

How to model a situation with a function

Common pitfalls

Frequently asked questions

Theory