Part I · Mathematical Language, Functions, and Algebraic Models

Chapter 2 — Coordinates, Graphs, and Functions

Why This Matters

When a quantity depends on another quantity, a graph lets us see the relationship before we calculate with it. Voltage may depend on time, current may depend on voltage, and a measured output may depend on a control setting. Functions make such dependencies precise.

2.1 Cartesian coordinates

The Cartesian plane has a horizontal $$x$$ -axis and a vertical $$y$$ -axis. A point $$(x,y)$$ means move $$x$$ units horizontally from the origin and then $$y$$ units vertically. The order matters: $$(2,5)$$ and $$(5,2)$$ are different points.

To visualise a graph of $$y=f(x)$$ , imagine choosing an input $$x$$ , calculating the associated output, and plotting the point $$(x,f(x))$$ . Repeating this process creates a curve or collection of points.

2.2 What a function is

A function assigns exactly one output to each allowed input. We write

f:D\to R,

where $$D$$ is the domain of permitted inputs and $$R$$ is a set containing the outputs. The range is the set of outputs actually attained.

For $$f(x)=x^2$$ , if the domain is all real numbers, then the range is $[0,\infty)$ . The value $$f(-3)=9$$ is a number. The function $$f$$ is the whole rule, not the particular output $$9$$ .

The vertical-line test gives a graphical version of the definition: if a vertical line meets a graph more than once, some input has two outputs, so the graph does not represent a single-valued function of $$x$$ .

Example — Domain before calculation

Consider

g(x)=\frac{\sqrt{x-1}}{x-4}.

The square root requires $x-1\ge0$ , so $x\ge1$ . The denominator requires $x\ne4$ . Therefore

\operatorname{dom}(g)=[1,4)\cup(4,\infty).

Only after finding this domain should we substitute values or sketch the graph.

Read the arrow before calculating. A function is the complete assignment from permitted inputs to outputs, not one value produced by it. A domain restriction is therefore part of the rule: it tells us which plotted horizontal positions may be read, while the range records the vertical positions actually reached.

f:D\to R,\qquad f(-3)=9\text{ for }f(x)=x^2

An input-output reading of a function, with a permitted input mapped to exactly one output.

2.3 Function algebra

Given functions $$f$$ and $$g$$ , we can define

(f+g)(x)=f(x)+g(x),\quad (fg)(x)=f(x)g(x),\quad \left(\frac f g\right)(x)=\frac{f(x)}{g(x)}.

The quotient is defined only where $g(x)\ne0$ . Function algebra is useful for building models from simpler pieces. For example, a sinusoidal input plus a constant offset has the form $v(t)=V_0+A\sin(\omega t)$ .

Combine both values and both restrictions. Function addition and multiplication use the two values at one common input. A quotient has one further condition: the denominator function must not be zero there. Algebra builds a new rule, but it cannot make either original input restriction disappear.

\left(\frac fg\right)(x)=\frac{f(x)}{g(x)},\qquad g(x)\ne0

List inputs permitted by each component function.
Apply the selected operation to their values at a shared permitted input.
For a quotient, exclude every input that makes the denominator zero.

2.4 Graphs as solution tools

To solve an equation $$f(x)=g(x)$$ graphically, plot $$y=f(x)$$ and $$y=g(x)$$ on the same axes. Their intersection points have equal $$y$$ -coordinates, so their $$x$$ -coordinates solve the equation. A graph provides insight and estimates; it does not replace an exact algebraic method when close roots or scale distortions matter.

Example — A graphical model with a check

Suppose current through a resistor is modeled by $$I=V/1000$$ amperes and a load requires $$I=0.003$$ A. The intersection of

y=\frac{x}{1000},\qquad y=0.003

occurs at $$x=3$$ . Algebra confirms the graph:

\frac V{1000}=0.003 \implies V=3\ \mathrm V.

Read an intersection as a simultaneous statement. At a crossing, both graph rules assign the same output to the same input. The horizontal coordinate solves f(x)=g(x); the vertical coordinate is the shared value. A graph is therefore a geometric way to audit an algebraic solution.

f(x)=g(x)\quad\Longleftrightarrow\quad (x,f(x))\text{ lies on both graphs}

Two graphs meeting at the ordered pair that satisfies both rules.

2.5 Transformations and interpretation

Starting from $$y=f(x)$$ :

$$y=f(x)+k$$ shifts the graph upward by $$k$$ .
$$y=f(x-h)$$ shifts it right by $$h$$ .
$$y=af(x)$$ scales output values vertically.
$$y=f(-x)$$ reflects the graph in the $$y$$ -axis.

These are not rules to memorise without meaning. For instance, replacing $$x$$ by $$x-h$$ asks the original function to receive the value $$x-h$$ . To produce the old feature at input $$0$$ , the new graph needs $$x-h=0$$ , hence $$x=h$$ : the feature moves right.

Why this works. The expression inside a function changes the input before the function acts. Thus f(x-h) asks the old graph for a value h units earlier, so the visible graph shifts right by h. The minus sign is therefore an input correction, not a leftward motion.

\[g(x)=a f(x-h)+k\]

A base parabola and its transformed copy, with vertex and shift labels.

Common Mistakes

Confusing $$f$$ with $$f(x)$$ .
Assuming every algebraic expression is defined for every real number.
Reading a graph scale carelessly, especially when one axis is compressed.
Treating graph intersections as exact values when the plot is only an estimate.

2.6 Domain, range, and transformations as graph-reading tools

The domain answers “which inputs make sense?”; the range answers “which outputs can the rule actually produce?” To find a domain, exclude zero denominators, negative even-root inputs, and invalid logarithms. To find a range, use a graph or rewrite the expression to expose a lowest, highest, or excluded value.

Example — range from a transformed parabola

Let $$h(x)=-2(x-3)^2+5$$ . Since $(x-3)^2\ge0$ ,

-2(x-3)^2\le0,\qquad h(x)\le5.

The maximum $$5$$ is achieved at $$x=3$$ , so the range is $(-\infty,5]$ while the domain is all real numbers. The negative factor reflects the basic parabola, the $$-3$$ inside moves its vertex right, and $$+5$$ moves it upward.

Example — composition carries an input restriction

Let $f(x)=\sqrt{x}$ and $$g(x)=4-x$$ . Then

(f\circ g)(x)=\sqrt{4-x}.

Although $$g$$ accepts every real input, the composite requires $4-x\ge0$ . Therefore its domain is $(-\infty,4]$ . Composition is not multiplication: it asks one function to supply the input to the other.

Example — inverse intuition from calibration

If a sensor model is $$V=0.02T$$ , then its inverse is $$T=50V$$ . The inverse swaps the roles of known input and inferred output. Graphically, inverse graphs are reflections in $$y=x$$ once compatible axis scales are understood.

[Figure:

\(y=x^2\)

, then

\(y=-2(x-3)^2+5\)

with vertex/range marked; a restricted one-to-one branch reflected in

\(y=x\)

to show an inverse.]

Keep a domain ledger while transforming. A horizontal shift changes which input reaches a restricted inner rule. A vertical shift changes outputs but cannot repair an input that was never permitted. Reading the inside first is a reliable way to carry exclusions through a transformed graph.

A shifted rational graph with an excluded input and its vertical asymptote marked.

Chapter Summary

Coordinates locate points; a graph displays input-output pairs. A function gives one output for each permitted input. Domain and range are part of the definition. Function algebra and graph transformations build useful models, while graph intersections give a visual interpretation of equations.

2.7 Reading a graph as a mathematical statement

A graph simultaneously communicates domain, range, zeros, intercepts, increasing and decreasing portions, symmetry, and limiting behaviour. Read its axes before reading a numerical value: a plotted “3” may mean $$3$$ V, $$3$$ ms, or $3\times 10^{-3}$ depending on the scale. Algebraic form often predicts the graph before any points are plotted.

Example — rational form predicts a graph

For

r(x)=\frac{x+1}{x-2},

the denominator excludes $$x=2$$ , so there is a vertical asymptote there. Division or rearrangement gives

r(x)=1+\frac{3}{x-2}.

Thus the graph is the reciprocal curve $$1/x$$ , shifted right by $$2$$ and up by $$1$$ , with horizontal asymptote $$y=1$$ . Its range excludes $$1$$ , because $$3/(x-2)$$ cannot be zero. This conclusion comes from structure, not a dense table of points.

Example — graphical solutions are estimates with a meaning

To solve $\sqrt{x+1}=3-x$ , first require $x+1\ge0$ and $3-x\ge0$ because a square root is nonnegative. A graph of the two sides suggests an intersection near $$x=2$$ . Squaring is then safe under those restrictions:

\[x+1=(3-x)^2=x^2-6x+9,\]

so $$x^2-7x+8=0$$ . Of its two algebraic roots, only $(7-\sqrt{17})/2$ satisfies the original equation. The graph locates the solution and alerts us to the one-intersection expectation; substitution rejects the extraneous root created by squaring.

Example — a horizontal-line test is a range test in disguise

For $$y=(x-1)^2+2$$ , a horizontal line at $$y=1$$ does not meet the graph, so $$1$$ is not in the range. Every horizontal line $$y=k$$ with $$k>2$$ meets twice, so the all-real function is not one-to-one. Restricting to $x\ge1$ keeps one branch and makes an inverse possible.

[Figure: A labelled rational-function graph with vertical/horizontal asymptotes; then a parabola with horizontal test lines and its restricted branch reflected in

\(y=x\)

Read structure before points. In r(x)=(x+1)/(x-2)=1+3/(x-2), the denominator records the excluded input x=2, and the separated constant records the horizontal asymptote y=1. The two branches approach those lines without supplying a value at the excluded input.

r(x)=\frac{x+1}{x-2}=1+\frac3{x-2},\qquad x\ne2

Find excluded inputs from the unsimplified denominator.
Rewrite the rational expression to expose its shifted reciprocal form.
Use the exclusion and the constant term to identify the vertical and horizontal asymptotes before estimating values from the graph.

Exercises

Find the domain and range of $f(x)=\sqrt{9-x^2}$ .
Is the relation $$x^2+y^2=1$$ a function of $$x$$ ? Explain using the vertical-line test.
Describe the transformations taking $$y=x^2$$ to $$y=-2(x-3)^2+5$$ .
A sensor has output $v(t)=2+0.5\sin(100\pi t)$ . State its offset, amplitude, and period.
Find the domain and range of $$q(x)=1/(x-2)^2$$ and describe its asymptotes.
Given $$f(x)=x^2+1$$ and $$g(x)=3-x$$ , find $f\circ g$ and $g\circ f$ .
Explain why $$y=x^2$$ has no all-real inverse, then give a restriction that does.
Without plotting points, state the domain, range, and asymptotes of $$(x-4)/(x+1)$$ by rewriting it.
Solve $\sqrt{x+2}=4-x$ , checking all algebraic candidates in the original equation.