Introduction

Welcome to unit 1.14! In this unit, you’ll learn how to construct a linear, quadratic, cubic, quartic, polynomial of degree $n$, or related piecewise-defined function model; construct a rational function model based on a context; and apply a function model to answer questions about a data set or contextual scenario.

What is a function model?

A function model is an equation (or occasionally a set of equations, as is the case with piecewise functions) that describes how one quantity/variable depends or affects another given some situation. Put simply, it is a regular function that describes some sort of context (within a problem)

Polynomial Models 

If you’ll remember, polynomial functions include linear, quadratic, cubic, quartic, and higher-degree polynomial functions. In the previous subunit, you learned how to decide which of these to use. Now, we will dive deeper into actually constructing the function.

Linear models

In general, you will use a linear model when the rate of change is (about) constant.

There are two forms of a linear function.

The first is slope-intercept form: $y=mx+b$

The second is point-slope form: $y-y_1=m\left(x-x_1\right)$

The advantages of each form will be clear later.

Before you begin constructing the model, you must recognize how the problem gives you data. 

Tabular data

If they give you tabular data (data points in a table), first check if the rate of change is roughly constant. Deciding which model to use is shown in more detail in 1.13; this article will not go into such depth.

If the ROC (rate of change) is about constant, then choose two points from the table (can be random points). Say we chose the points $(x_1,y_1)$ and $(x_2,y_2)$.

Then, we find the slope: $m=\frac{y_2-y_1}{x_2-x_1}$

Then we can either: Use point-slope form with one of these points and our slope:

$y-y_1=m\left(x-x_1\right)$ or $y-y_2=m\left(x-x_2\right)$

Or, convert to slope-intercept form by solving for $b$:

$y_1=m{x}_1+b$ or $y_2=m{x}_2+b$

Point-slope form is typically the faster way to construct an equation from a model. However, slope-intercept form can sometimes be easier to interpret. Ultimately, it is up to you which form you use. 

Two data points are given

If the problem instead provides two data points in non-tabular form (like in a word problem), we can actually treat it the same as if we were given two tabular points.

A graph is given

Occasionally, you’ll be asked to create a function that is depicted in a graph. If the graph looks linear (a straight line), find two points that clearly lay on the graph (on the linear function depicted in the graph) and treat it like two points given in a table or in a word problem. 

There are shortcuts: for example, if the $y$-intercept is shown in the graph, then it’s a lot easier to just use slope-intercept form, as $b$ is already known. Then, just find two points and find the slope to find $m$. Plug both of these into slope-intercept form and you have your function.

Slope is given with a data point

Sometimes, the problem will explicitly tell you the rate of change (or sometimes they’ll implicitly tell you) and one data point (typically the $y$-intercept). 

An example could be: “A population of bunnies is increasing at a constant rate of $300$ bunnies per year; in $2020$, it was $6,700$”

Here, the slope is given directly (once again, this is elaborated more in 1.13 and will not be discussed in detail in this article): “increasing by $300$ bunnies per year) tell us that $m=300$ (assuming that we are graphing population of bunnies vs time in years). 

Then, we identify a point. Here, our point is $\left(2020,6700\right)$. 

Then, we can use point-slope form (it’s usually easier when we’re given a slope and point directly, as we don’t even need to do any work) and plug in our values: $y-6700=300\left(x-2020\right)$

If required (by the question or teacher or whatever), we can convert this into slope-intercept form through algebraic expansion and manipulation, although it's not necessary.

However, here, we can see that point-slope is incredibly convenient when a problem tells us a slope and point. You can just plug in the rate ($m$) and a point and get your function.

Quadratic Models

As a summary of what was discussed in 1.13, you will typically use a quadratic model when there is one maximum/minimum in the data or the ROC of the ROC is constant.

There are three main forms of a quadratic function:

Standard form:

$y=a{x}^2+bx+c$

Vertex form:

$y=a{\left(x-h\right)}^2+k$

Factored (intercept) form:

$y=a\left(x-r_1\right)\left(x-r_2\right)$

Here, each form is better suited for different types of situations.

A vertex (maximum/minimum) and one additional point is given

In this situation, there is an “objectively” best form to use: vertex form.

If you know a vertex $\left(h,k\right)$ (usually something like max/min height, max/min cost, etc. etc.)

And if you know a second point $\left(x_1,y_1\right)$ on the parabola,

Then: Plug in the known vertex $\left(h,k\right)$

$y=a{\left(x-h\right)}^2+k$

And then, using the other known point $\left(x_1,y_1\right)$, solve for $a$:

$y_1=a{\left(x_1-h\right)}^2+k$

This is common in physics problems, where you’re modeling the height of a projectile.

Zeros (x-intercepts) and an additional point are given)

If you are given both zeros ($x$-intercepts) and some other point on the graph, then you can use factored form.

First, plug in your two roots (say, you have the roots $\left(r_1,0\right)$ and $\left(r_2,0\right)$)

$y=a\left(x-r_1\right)\left(x-r_2\right)$

Then, plug in your additional point (say, $\left(x_1,y_1\right)$):

$y_1=a\left(x_1-r_1\right)\left(x_1-r_2\right)$

Three or more points (tabular or in a word problem)

With three exact points, you can either use systems of equations or technology.

Say the three random points are $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$, and $\left(x_3,y_3\right)$.

To solve a system of equations (usually for integer points that can be solved without causing a migraine)

Plug each point into standard form:

$y=a{x}^2+bx+c$

Then, you can solve the system with substitution or elimination.

This is more algebra-heavy, but it’s fairly simple if the numbers are also simple.

However, if your data is only approximate or just hard data to work with by hand, you can use regressions. Using regressions is covered in the previous article.

Graph

When a graph is very clearly a single parabola, look for a vertex, zeros, or the intercept.

If the vertex is a clear point (on the grid), use vertex for and find $a$ from some other point on the graph.

If zeros are obvious, use factored form and find $a$ from another point.

If the intercept is obvious, use standard form (the $y$-value of the intercept will just be $c$) and solve a systems of equations (this is also what you do if none of the features are available)

Cubic, Quartic, and higher-degree polynomial models

When dealing with functions that have a degree of 3, 4, or even higher, there are a small number of tips for constructing the function. Usually, all you can do is create a system of equations by plugging in points.

The only scenario in which it would be easier is if they give you the roots, in which you can write a factored form using zeros. This will usually be the case when you’re given a graph.

$y=a\left(x-r_1\right)\left(x-r_2\right)\left(x-r_3\right)...\left(x-r_{n-1}\right)\left(x-r_n\right)$

Then, you can solve for $a$ by picking a point that is not a zero and solving for $a$ through a regression (which is further elaborated on in 1.13).

Deciding when to use these types of functions is explained in the previous article 1.13.

Piecewise Functions

There are no “piecewise” specific tips. That’s because a piecewise truly is just a mix of different functions. Simply split the piecewise function into the separate types of functions and use the tips above. Identifying where a piecewise split is elaborated more in 1.13.

Rational Functions

A rational function is a ratio of two polynomials. It takes the form of:

$f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}$, where $h\left(x\right)\ne 0$

When modeling a situation, the most common use of Rational Functions as models is modeling inverse relationships.

Inverse Variation

When something like $y$ is inversely proportional to $x$, it means that $y=\frac{k}{x}$.

This takes on the form of a rational function, where $g\left(x\right)=k$ and $h\left(x\right)=x$. It is absolutely possible for $h\left(x\right)$ to be something other than $x$, such as $x^2$.

Sometimes, the phrasing will be more vague. For example, a problem could say “$y$ decreases with the square of $x$”

To solve for $k$, all you need is one point. Simply plug that point in and solve for $k$. From there, you have a complete function to model with.

Graph

Recognizing a rational function is actually fairly simple. If both sides of the graph approach some value (a horizontal asymptote), then you have a rational function. If there’s a vertical asymptote (where the function approaches a vertical line to infinity), then you have a rational function. If you have a hole in your function with these traits, you absolutely have a rational function.

Technically, it is possible to see a slant/oblique asymptote. However, the graph will still show the behaviors listed. Additionally, it will still “follow” an asymptote, it is just that the asymptote is a linear function. It will very clearly follow a straight line.

To create the function, first look at the vertical asymptotes. Remember, the vertical asymptote is a vertical line that the graph will seemingly approach but never cross.

If the asymptote is at, say, $x=h$, then one of the factors in the polynomial in the denominator is $\left(x-h\right)$. 

Then, find any zeros. If a zero is at $x=c$, then one of the factors in the polynomial in the denominator is $\left(x-c\right)$. However, note that this is only true if the polynomial simply “passes through” the $x$-axis. If the polynomial “bounces off” the $x$-axis, the factor $\left(x-c\right)$ has an even degree (the exponent on the factor $\left(x-c\right)$ must be 2, 4, 6, 8, etc.). If it seems to smoothen and flatten out before crossing, then it has an odd degree greater than one (3, 5, etc.)

Then, find the horizontal asymptote. Again, this is a horizontal line the graph approaches as $x$ approaches infinity and negative infinity.

If the horizontal asymptote is $y=k$, where $k$ is a constant, then the lead coefficient of the polynomial in the numerator divided by the lead coefficient of the polynomial in the denominator must be equal to $k$, and the degrees of both polynomials must be equal.

If the horizontal asymptote is $y=0$, then the degree of the polynomial in the denominator must be greater than that of the polynomial in the numerator’s.

If there is a slant or oblique asymptote, then the slant or oblique asymptote is the result of dividing the numerator’s polynomial by the denominator’s, and then ditching (ignoring) the remainder.

Finding how to derive an equation of a rational function from a graph is detailed more in later articles. For now, this portion isn’t too important.

Conceptual ideas of Rational Function Modeling

There is one main takeaway of using Rational Functions to model scenarios. 

Rational models are used when one quantity changes in the denominator, so that as that input grows, the output shrinks towards some value.

Transformations, Compositions, and Inverses 

So far, you’ve been building models from scratch; you’ve had to decide what model to use, solve for constants, etc.

However, instead of creating a new function from scratch, it’s possible to build new models by modifying existing functions. You can do this by utilizing:

Note: more on Compositions and Inverses in Unit 2.

Transformations

A parent function is the simplest member of a function family (examples include $y=x$, $y=x^2$, $y=\frac{1}{x}$. 

If $f\left(x\right)$ is the parent function, then

A vertical shift would be defined as $f\left(x\right)+k$, where the graph moves up or down by $k$ (down if k is negative).

A horizontal shift would be defined as $f\left(x-h\right)$, where the graph moves left or right (left if $h<0$).

A vertical stretch/compression and reflection take the form $af\left(x\right)$.

When $\left|a\right|>1$, you have a stretch

When $0<\left|a\right|<1$, you have a compression

And when $\left|a\right|<0$, you have a reflection of the $x$-axis (you can still have a stretch/compression with a reflection over the $x$-axis.

You can additionally have a reflection over the $y$-axis, which takes the form $y=f\left(-x\right)$

Quite often, these will be used in tandem: $y=af\left(x-h\right)+k$

To obtain a model from context, you have to first choose your parent function (using decision making from above or from article 1.13). Then, use a key feature to determine the shift. For example, let’s look at the vertex form of a quadratic function.

In the equation: $y=a{\left(x-h\right)}^2+k$

We have a stretch/dilation by $a$ (a reflection too if $\left|a\right|<0$), a horizontal shift right $h$, and a vertical shift up $k$.

Compositions

A composition models a situation where one quantity depends on another, which depends on another.

Basically, it takes the form 

$y=f\left(g\left(x\right)\right)$.

Compositions often show up in modeling during unit conversions (converting days into hours, then applying a cost per hour model), geometry and a changing input (radius grows with time $t$, and area/volume depends on $r$), or in general, the output of one process affects another.

Inverses

Sometimes the context can ask for the opposite of a typical function model; they may ask you to swap the inputs and outputs.

Say you have a function that models the total cost of a toy factory given a number of units (toys). But now you’re asked to find how many units you could buy given $$400$. Now, yes, you can set $y=400$. However, the point is that (especially if you specifically need a function) you can find the inverse of this function and simply plug in $400$ as your input.

Applying Models with Context

Once you have a model, whatever type of function it is, you can use it to answer questions, including predicting values and analyzing rates of change. Given any function model $f\left(x\right)$, you can 

Rate of Change in Context

Additionally, you will sometimes be asked to find the rate of change between two points. Simply find those two points using your function and use the Rate of Change formula.

However, the harder part is putting that rate in context. 

The formula for average rate is

$\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Physically, it’s a measure of the change in output per unit change in input, and its units are:

$\frac{\text{units of }f}{\text{units of }x}$

You will often be asked to interpret the meaning of your ROC for a situation.

Changing Rate of Change

You are not expected to do any calculus, but you should recognize when the rate of change itself is changing. You can do this by using concavity, which is also expressed in more detail in 1.13. 

You can discuss more physical ideas (related to the problem) with concavity, such as acceleration.

Using Units and Reasonableness

Whenever you use a model, ask yourself what are the units of my inputs and outputs ($x$ and $y$)? 

Understanding what your function model even models is of utmost importance.

More context about domain and range are covered in 1.13.

Practice Problems

1) A car rental company charges a flat daily fee plus a per-mile cost:

Let $m$ be the number of miles driven in and, and $C\left(m\right)$ models the total daily cost in dollars.

2) A man is attempting to throw a rock at a wasp nest. The man throws the rock from a platform. The rock’s height $h$ (in meters) above the ground as a function of time $t$ (in seconds) is represented by $h\left(t\right)$ 

The rock is launched from a height of $20$ meters and reaches a max height of $44$ meters after $3$ seconds.

3) The intensity ($I$) of a signal is inversely proportional to the square of the distance ($d$) in meters. When a man stands $2$ meters from the source, the intensity is $180$ (of some unit)

Answers

1) A car rental company charges a flat daily fee plus a per-mile cost:

Let $m$ be the number of miles driven in and, and $C\left(m\right)$ models the total daily cost in dollars.

Answer:

Part a:

First, there will always be a base fee of $$30$.

For $0\le m\le 100$, the total cost will be the base fee of $$30$ + cost per mile. So, $C\left(m\right)=0.25m+30$. 

However, for $m>100$, we must have a new function

For the first $100$ miles, we have to pay $100\cdot 0.25+30=55$ dollars. Then, for any extra miles $m-100$, we pay $$0.15$ per mile. So, our function is $C\left(m\right)=0.15\left(m-100\right)+55$

Our final piecewise function is

$C(m)=\{\begin{array}{ll}0.25m+30,& 0\le m\le 100\\ 0.15(m-100)+55,& m>100\end{array}$

Part b:

We know that $60$ miles is less than $100$ miles, so we use the first function and plug in $60$, which is $$45$.

We know that $150$ miles is greater than $100$ miles, so we use the second function and plug in $150$, which is $$62.50$.

Part c:

AROC between $m=50$ and $m=150$ is described by:

$\frac{C\left(150\right)-C\left(50\right)}{150-50}$

This simplifies to $0.2$

What this means is that on average, each additional mile between $50$ miles and $150$ miles will increase the total cost by $$0.20$.

2) A man is attempting to throw a rock at a wasp nest. The man throws the rock from a platform. The rock’s height $h$ (in meters) above the ground as a function of time $t$ (in seconds) is represented by $h\left(t\right)$ 

The rock is launched from a height of $20$ meters and reaches a max height of $44$ meters after $3$ seconds.

Answer:

Part a:

Because this is projectile motion, we know it’s best modeled by a quadratic.

Using vertex form (because we are given a max height), we can obtain the form:

$h\left(t\right)=a{\left(t-3\right)}^2+44$

Then, we can plug in the point $\left(0,20\right)$ to solve for $a$.

Solving for $a$, we get that $a=-\frac{8}{3}$

So, $h(t)=-\frac{8}{3}(t-3{)}^2+44$

Part b:

First, we can see that there was a reflection over the $y$-axis, as $a$ is negative.

Then, we see that there was a vertical stretch by $\frac{8}{3}$.

After, we see that the graph was shifted right $3$ units.

Finally, we see that the graph was shifted up $44$ units.

Part c:

Plugin in $t=5$ into our function, we get that $h\left(5\right)=\frac{100}{3}\approx 33.33$

After $5$ seconds, the rock is (about) $33.33$ meters above the ground

3) The intensity ($I$) of a signal is inversely proportional to the square of the distance ($d$) in meters. When a man stands $2$ meters from the source, the intensity is $180$ (of some unit)

Answer:

Part a:

First, we know that (because it is inversely proportional) $I=\frac{k}{d^2}$

Plugging in the point $\left(2,180\right)$, we see that $k=720$ and our function is $I=\frac{720}{d^2}$

Part b:

We can simply plug in $d=5$ to obtain that Intensity equals $28.8$ units.

Part c)

First off, we know that as the man walks away, the intensity will decrease as distance is increasing. Then, intensity will approach $0$ (never reaching it).

Part d)

We can simply find the inverse of the intensity function: it is $d\left(I\right)=\sqrt{\frac{720}{I}}$