# MAT 101: Absolute Value Equations and Absolute Value Inequalities

Absolute Value Equations

First let us review the definition of absolute value $|\cdot |$.

Definition. The absolute value of $|a|$ of a number $a$ is defined by
$$|a|=\left\{\begin{array}{ccc} a & \mbox{if} & a\geq 0\\ -a & \mbox{if} & a<0 \end{array}\right.$$
Interestingly a lot of students get confused with the definition while they have no problem with getting the absolute of a particular number correctly like $|2|=2$ and $|-2|=2$. That’s because of the way they were taught. Many teachers teach absolute value like some sort of magic: “Hey guys what absolute value does is whatever number you put into between the two vertical lines it becomes positive.” While this may be easy to understand for students this is not an action performed in mathematics. In algebra whatever action you do should be carried out by operations such as $+$, $-$, $\times$, or $\div$. In terms of an operation what happens to $|-2|=2$ is in fact that $|-2|=-(-2)=2$. It is not abracadabra. For a letter $a$ representing a number, if $a<0$, to make it positive $|a|=-a$. Students get confused because $-a$ looks like negative but it is not. Remember the condition $a<0$. So $-a$ is actually positive. So don’t be deceived by its look.

Now we are ready to discuss absolute value equations. All you need to know is

\label{eq:abs}
|x|=k\ \mbox{if and only if}\ x=\pm k

for $k>0$. If $k=0$, $x=0$. If $k<0$, obviously there is no solution.

Example. Solve the equation
$$|2x-5|=3$$

Solution. By \eqref{eq:abs}, we get the two linear equations $2x-5=\pm 3$. Solving these equation, we find $x=4$ or $x=1$.

Absolute value equations are related to quadratic equations. If you square the above equation, we get the quadratic equation
$$x^2-5x+4=0$$
Solving this quadratic equation, we of course obtain the same solutions. In practice why bother? Solving absolute value equations directly is easier. Note though that absolute value equations are actually obtained when you solve quadratic equations by the method of completing the square.

Absolute Value Inequalities

For absolute value inequalities all you have to know is the following picture.

Figure 1. Absolute Value Inequalities

From Figure 1 we can read for $k>0$

1.  $|x|<k$ if and only if $-k<x<k$.
2. $|x|\leq k$ if and only if $-k\leq x\leq k$.
3. $|x|>k$ if and only if $x<-k$ or $x>k$.
4. $|x|\geq k$ if and only if $|x|\leq -k$ or $|x|\geq k$.

Example. Solve the inequality $|x-5|<2$.

Solution. $|x-5|<2$ implies that $-2<x-5<2$ i.e. $3<x<7$.

Example. Solve the inequality $|3x+2|\geq 4$.

Solution. $|3x+2|\geq 4$ implies that $3x+2\leq -4$ or $3x+2\geq 4$ i.e. $x\leq-2$ or $x\geq\frac{2}{3}$.

# MAT 101: Lines

A line in the plane is determined by two points meaning there is only one line passing through two given points in the plane. But it could be determined by some other quantities. An important such quantity is slope. Slope measures steepness of a line and it is defined by $\frac{\mbox{rise}}{\mbox{run}}$. If two points $(x_1,y_1)$ and $(x_2,y_2)$ are known, the slope $m$ of the line through the two points is

\label{eq:slope}
m=\frac{y_2-y_1}{x_2-x_1}

Example. Find the slope of the line passing through the points $P(2,1)$ and $Q(8,5)$.

Solution. Note that it really doesn’t matter which ones you label as $(x_1,y_1)$ and $(x_2,y_2)$. Here we choose $(x_1,y_1)=(2,1)$ and $(x_2,y_2)=(8,5)$. Using \eqref{eq:slope} we find the slope
$$m=\frac{5-1}{8-2}=\frac{4}{6}=\frac{2}{3}$$

A cool thing to see is that certain geometric objects can be described by equations so we can study geometry in terms of algebra. Such objects include lines, circles, parabolas, elipses, and so on so forth. Why is this cool? Because algebra is much easier than geometry. A branch of mathematics that studies geometric objects in terms of algebra is analytic geometry and this is further generalized into another branch called algebraic geometry.

So how do we write an equation for a given line? Well, we already have it. The equation \eqref{eq:slope}. But in order to write it as an equation we need to tweak it a bit. An equation relates an arbitrary point $(x,y)$ on the line to some known quantities. So let’s say slope $m$ and a point $(x_1,y_1)$ is known. Then by \eqref{eq:slope} we get
$$m=\frac{y-y_1}{x-x_1}$$
This is the equation of the line with slope $m$ passing through a point $(x_1,y_1)$. But to make it look a bit nicer we rewrite it as

\label{eq:line}
y-y_1=m(x-x_1)

Example. Find the equation of the line through $(-1,3)$ with slope $-\frac{1}{2}$ and sketch the line.

Solution. Using \eqref{eq:line} we find
$$y-3=-\frac{1}{2}(x-(-1))$$
Solving this for $y$ we obtain
$$y=-\frac{1}{2}x+\frac{5}{2}$$
There are two ways to sketch the line. One is using the slope and the given point. Slope being $-\frac{1}{2}$ means that when $x$ moves 2 units to the right its corresponding $y$ moves 1 unit downward. Apply this to the point $(-1,3)$ we will land at another point which is $(1,2)$. You draw the line passing through the two points $(-1,3)$ and $(1,2)$ as shown in Figure 1.

Figure 1. Drawing a line

The other way to sketch the line is to find another point. An easy choice is to find the $x$-intercept by setting $y=0$. The $x$-intercept is $(5,0)$. You draw the line through $(-1,3)$ and $(5,0)$.

As a special case if the slope $m$ and the $y$-intercept $(0,b)$ are given, the equation \eqref{eq:line} becomes

\label{eq:line2}
y=mx+b

Even though the $y$-intercept is not known in fact \eqref{eq:line2} can be also used to find the equation of the line in the previous example. Since $m=-\frac{1}{2}$, we set
$$y=-\frac{1}{2}x+b$$
The line is passing through $(-1,3)$ we have
$$3=-\frac{1}{2}(-1)+b$$
Solving this for $b$ we find $b=\frac{5}{2}$.

Parallel and Perpendicular Lines

It’s obvious that two lines with slopes $m_1$ and $m_2$ are parallel if $m_1=m_2$. What’s not so obvious however is the following property.

Two lines with slopes $m_1$ and $m_2$ are perpendicular if $m_1m_2=-1$.

We will not mind the proof of this property here.

Example. Find an equation of the line through $(5,2)$ that is parallel to the line $y=-\frac{2}{3}x-\frac{5}{6}$.

Solution. The slope is $m=-\frac{2}{3}$. Since this line is passing through the point $(5,2)$, by \eqref{eq:line} the equation is
$y-2=-\frac{2}{3}(x-5)$. This simplifies to
$y=-\frac{2}{3}x+\frac{16}{3}$.

Example. Find an equation of the line through $(5,2)$ that is perpendicular to the line $y=-\frac{2}{3}x-\frac{5}{6}$.

Solution. The slope is $m=\frac{3}{2}$. Since the line is passing through $(5,2)$, by \eqref{eq:line} the equation is
$y-2=\frac{3}{2}(x-5)$ which simplifies to $y=\frac{3}{2}x-\frac{11}{2}$.

Figure 2. Two perpendicular lines y=-(2/3)x-5/6 (in blue) and y=(3/2)x-11/2 (in red)

# MAT 101: Graphing Polynomial Functions

Polynomial functions have the following important property:

Every polynomial function of degree $n$ has at most $n$ real zeros.”

This property is called the Fundamental Theorem of Algebra.

As an application of this property, we see that a polynomial function of degree $n$ can have at most $n$ $x$-intercepts and at most $(n-1)$ turning points (local maximum and local minimum values).

Example. The function $f(x)=x^4-7x^3+12x^2+4x-16$ has three turning points that are two local minimum values and one local maximum value.

Graphing a Polynomial Function $p(x)$

1. First find all zeros of $p(x)$
2. Considering the even or odd multiplicity of each factor of $p(x)$, we can see the graph is crossing or touching the $x$-axis at each zero.
3. Use the leading-term test to determine the end behavior.
4. Use the $y$-intercept.

Example. Consider $f(x)=x^4-7x^3+12x^2+4x-16$. It can be factored as $f(x)=(x+1)(x-2)^2(x-4)$. (At this moment you don’t have to worry about how we get the facotring. We will discuss this in Sections 3.3 and 3.4.) So we find three zeros $-1$, $2$ (with multiplicity 2), and $4$. We can tell that the graph crosses at $-1$ and $4$ and touches the graph without crossing at $2$. Since the degree is $4$, an even number the graph goes up when $x\to -\infty$ and $x\to\infty$. These findings are all featured in the above graph.

The Intermediate Value Theorem

Let $p(x)$ be a polynomial (with real coefficients). Suppose that $p(a)$ and $p(b)$ have different signs for two distinct numbers $a$ and $b$. Then the graph of $p(x)$ must cross the $x$-axis between $a$ and $b$, i.e. $p(x)$ must have a zero between $a$ and $b$. This property is called the Intermediate Value Theorem.

Example. (a) Use the Intermediate Value Theorem to determine if
$$f(x)=x^3+3x^2-9x-13$$
has a zero between $a=1$ and $b=2$.

Solution. All you have to do is to evaluate $f(x)$ at $x=1$ and $x=2$, i.e. calculate $f(1)$ and $f(2)$ and see if they are different.
\begin{align*}
f(1)&=(1)^3+3(1)^2-9(1)-13=-18,\\
f(2)&=(2)^3+3(2)^2-9(2)-13=-11.
\end{align*}
$f(1)$ and $f(2)$ have the same sign, so the Intermediate Value Theorem won’t tell if $f(x)$ has a zero between $1$ and $2$. The following graph shows that it actually does not.

(b) Does $f(x)$ has a zero between $a=-5$ and $b=-4$?

Solution. $f(-5)=-18$ and $f(-4)=7$. Since their signs are different, by the Intermediate Value Theorem, there must be a zero between $-5$ and $-4$. The following graph confirms it.

# MAT 101: Polynomial Functions and Models

A polynomial is a function of the form
$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.$$ The number $n$ is called the degree of the polynomial $P(x)$. The term $a_nx^n$ is called the leading term and $a_n$ is called the leading coefficient. The number $a_0$ is called the constant term. $P(x)$ is called linear if $n=1$, quadratic if $n=2$, cubic if $n=3$, quartic if $n=4$, quintic if $n=5$, sextic if $n=6$, septic if $n=7$, and so on so forth. You don’t really have to worry about memorizing these jargons. Names are less important. However you need to remember at least what the degree is and what the leading coefficient is.

The Leading Term Test

There is a pattern for the long term behavior of a polynomial, i.e. the behavior of a polynomial when $x\to\infty$ or $x\to -\infty$. The behavior can be characterized as follows.

• $n=\mbox{even}$ and $a_n>0$:

Example. $f(x)=3x^4-2x^3+3$

• $n=\mbox{even}$ and $a_n<0$:

Example. $f(x)=-x^6+x^5-4x^3$

• $n=\mbox{odd}$ and $a_n>0$:

Example. $f(x)=x^5+\frac{1}{4}x+1$

• $n=\mbox{odd}$ and $a_n<0$:

Example. $f(x)=-5x^3-x^2+4x+2$

Finding zeros of a polynomial $P(x)$

By factoring, solve the equation $P(x)=0$. The solutions are the zeros of $P(x)$.

Example. Find the zeros of $P(x)=x^3+2x^2-5x-6$.

Solution. \begin{align*}
P(x)&=x^3+2x^2-5x-6\\
&=(x^3+x^2)+(x^2-5x-6)\ \mbox{(grouping)}\\
&=x^2(x+1)+(x-6)(x+1)\\
&=(x+1)(x^2+x-6)\\
&=(x+1)(x+3)(x-2).
\end{align*}
Hence, $P(x)$ has zeros $x=-3,-1,2$.

How do we determine whether $x=a$ is a zero of a polynomial $P(x)$?

To only check whether $x=a$ is a zero of $P(x)$, you don’t really have to factor $P(x)$. This is what you need to know. If $P(a)=0$, then $x=a$ is a zero of the polynomial $P(x)$.

Example. Consider $P(x)=x^3+x^2-17x+15$. Determine whether each of numbers 2 and $-5$ is a zero of $P(x)$.

Solution. $P(2)=(2)^3+(2)^2-17(2)+15=-7$, so $x=2$ is not a zero. $P(-5)=(-5)^3+(-5)^2-17(-5)+15=0$, so $x=-5$ is a zero of $P(x)$.

Even and Odd Multiplicity

Even and odd multiplicity is an important property for sketching the graph of a polynomial function. Suppose that $k$ is the largest integer such that $(x-c)^k$ is a factor of $P(x)$. The number $k$ is called the multiplicity of the factor $x-c$.

1. If $k$ is odd, the graph of $P(x)$ crosses the $x$-axis at $(c,0)$.
2. If $k$ is even, then the graph of $P(x)$ is tangent to the $x$-axis, i.e. touches the $x$-axis without crossing at $(c,0)$.

Example. Consider $f(x)=x^2(x+3)^2(x-4)(x+1)^4$. The factors $x$ and $x+3$ have multiplicity 2 and the factor $x+1$ has multiplicity 4. Hence the graph of $f(x)$ touches the $x$-axis without crossing at $x=0$, $x=-3$ and $x=-1$. The factor $x-4$ has multiplicity 1, so the graph crosses the $x$-axis at $x=4$. This is also shown in the following figure.

# MAT 101: Solving Linear Inequalities

Linear Inequalities:

Solving linear inequalities is as easy as solving linear equations. You only need to know the following principles.

For a given inequality,

1. Adding the same number to each side of the inequality does not change the symbol $<$.
2. Subtracting the same number from each side of the inequality does not change the symbol $<$.
3. Multiplying or dividing the inequality by the same positive number does not change the symbol $<$.
4. Multiplying or dividing the inequality by the same negative number reverses the symbol $<$ from, say, $<$ to $>$.

The principle 4 can be easily understood with a simple example, say everybody would agree that $1<2$. If one multiplies the inequality by $-1$, it is true that $-1>-2$, i.e. the symbol $<$ reverses.

Example. Solve the inequality
$$3x-5<6-2x.$$

Solution. Adding 5 to each side of the inequality results
$$3x<11-2x.\ \ \ \ \ \mbox{(1)}$$
Adding $2x$ to each side of (1) results
$$5x<11.\ \ \ \ \ \mbox{(2)}$$
Dividing the inequality (2) by the same positive number 5 results
$$x<\frac{11}{5}.$$
There are various ways to write the solution. Normally $x<\frac{11}{5}$ would suffice but it can be written more formally as the solution set
$$\left\{x|\ x<\frac{11}{5}\right\}.$$
In interval notation, the solution can be written as
$$\left(-\infty,\frac{11}{5}\right).$$
Graphically it can be represented as

Example. Solve the inequality
$$13-7x\geq 10x-4.$$

Solution. Subtracting 13 from each side of the inequality results
$$-7x\geq 10x-17.\ \ \ \ \ \mbox{(3)}$$
Subtracting $10x$ from each side of (3) results
$$-17x\geq -17.\ \ \ \ \ \mbox{(4)}$$
Dividing each side of (4) by the same negative number $-17$ results
$$x\leq 1.$$

Note that the symbol $\geq$ has been reversed to $\leq$. The solution set can be written as $\{x|\ x\leq 1\}$ or $(-\infty,1]$ in interval notation. Graphically it can be represented as

Compound Inequalities:

A compound inequality is two inequalities joined by a conjunction AND or OR.

Example. Solve $-3<2x+5\leq 7$.

Solution. Notice that the compund inequality is formed by
$$-3<2x+5\ \mbox{and}\ 2x+5\leq 7.$$
Subtract 5 from each side of the given inequality.
$$-8<2x\leq 2.\ \ \ \ \ \mbox{(5)}$$
Devide (5) by 2.
$$-4<x\leq 1.$$
Hence the solution set is $\{x|\ -4<x\leq 1\}$ or $(-4,1]$ in interval notation. The graph of the solution set is given by

Example. Solve $2x-5\leq -7$ or $2x-5>1$.

Solution. Add 5 to eqch side of the two inequalities:
$$2x\leq -2\ \mbox{or}\ 2x>6.\ \ \ \ \ \mbox{(6)}$$
Divide each side of the two inequalities (6) by 2:
$$x\leq -1\ \mbox{or}\ x>3.$$
The solution set is $\{x|\ x\leq -1\ \mbox{or}\ x>3\}$ or $(-\infty,-1]\cup(3,\infty)$ in interval notation. The graph of the solution set is given by