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SectionA.2Algebra Review

SubsectionA.2.1Lines and Linear Functions

Lines are perhaps the most important elementary geometric object. A line captures the idea of following a given direction without turning. In ordinary language, we sometimes think of a line as a smooth curve that we could draw. Mathematically, a line would then be a straight line or a straight curve that does not bend. Algebraically, we can define a line using an equation involving two variables. This section reviews the basic principles of the algebraic properties of lines.

DefinitionA.2.1General Equation of a Line

Every line in the \((x,y)\) plane can be described as the set of points \((x,y)\) that satisfy an equation

\begin{equation} Ax + By = C\tag{A.2.1} \end{equation}

where \(A\text{,}\) \(B\) and \(C\) are constants.

There are some special cases that describe horizontal and vertical lines. The \((x,y)\) plane uses \(x\) as the horizontal axis (independent variable) and \(y\) as the vertical axis (dependent variable). So a horizontal line is a line where the dependent variable is constant while a vertical line is a line where the independent variable constant.

DefinitionA.2.2Horizontal Line

A horizontal line in the \((x,y)\) plane is the set of points that satisfy an equation

\begin{equation} y=k\tag{A.2.2} \end{equation}

where \(k\) is a constant.

DefinitionA.2.3Vertical Line

A vertical line in the \((x,y)\) plane is the set of points that satisfy an equation

\begin{equation} x=h\tag{A.2.3} \end{equation}

where \(h\) is a constant.

All other lines have an equation that involves both variables. We often wish to think of the line as describing the dependent variable as a function of the independent variable. These equations involve the calculation of the slope, which represents a rate (or ratio) of change.

DefinitionA.2.4Slope as Rate of Change

Given any two points \((x_1,y_1)\) and \((x_2,y_2)\) on a non-vertical line, the change in the dependent variable \(\Delta y = y_2 - y_1\) is proportional to the change in the independent variable \(\Delta x = x_2 - x_1\text{,}\) written \(\Delta y = m \cdot \Delta x\text{.}\) The proportionality constant \(m\) is called the slope of the line, calculated as the ratio of changes (rate of change)

\begin{equation} m = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}.\tag{A.2.4} \end{equation}

Knowing the slope and one point is enough to quickly find an equation of a line.

DefinitionA.2.5Point–Slope Equation of Line

Given that a line has slope \(m\) and passes through a point \((x,y)=(h,k)\text{,}\) every point on the line satisfies the equation

\begin{equation} y=m \, (x-h)+k.\tag{A.2.5} \end{equation}

We interpret \(k\) as the starting value for \(y\) and the expression \(\Delta y = m\,(x-h)\) as the change in \(y\) given the change in \(x\text{,}\) \(\Delta x = x-h\text{.}\)

A special case of the point–slope equation of the line occurs when the point is on the \(y\)-axis or, in other words, is a \(y\)-intercept.

DefinitionA.2.6Slope–Intercept Equation of Line

Given that a line has slope \(m\) and passes through a \(y\)-intercept \((x,y)=(0,b)\text{,}\) every point on the line satisfies the equation

\begin{equation} y=mx+b.\tag{A.2.6} \end{equation}
RemarkA.2.7

Preparatory mathematics courses often emphasize the slope–intercept equation of a line as if it were the most important. However, the point–slope equation is the preferred equation to use in almost every circumstance.

Another special case of the point–slope equation of a line is when we know the slope and the \(x\)-intercept.

DefinitionA.2.8Slope and X-Intercept Equation of Line

Given that a line has slope \(m\) and passes through an \(x\)-intercept \((x,y)=(a,0)\text{,}\) every point on the line satisfies the equation

\begin{equation} y=m(x-a).\text{.}\tag{A.2.7} \end{equation}

SubsectionA.2.2Quadratic Polynomials

DefinitionA.2.9

A quadratic polynomial in a variable \(x\) is an algebraic function that is equal to a formula of the form

\begin{equation} f(x) = ax^2+bx+c,\tag{A.2.8} \end{equation}

where \(a\text{,}\) \(b\) and \(c\) are constants called coefficients.

The graph of a quadratic function, \(y=ax^2+bx+c\text{,}\) is a parabola. Such a parabola has a mirror symmetry across a vertical line that passes through its vertex \(x=-\frac{b}{2a}\text{.}\) Depending on whether the vertex is above, on or below the \(x\)-axis and whether the parabola opens up or down, the graph can cross the \(x\)-axis twice, once or never. The location of these points are called \(x\)-intercepts, roots or zeros of the function.

Zeros are closely related to factoring. If we know the zeros, then we can immediately rewrite the polynomial in a factored form. On the other hand, if we know the factors, then we can quickly solve for the zeros without using the quadratic formula. This is a consequence of the fundamental properties of numbers in Theorem Theorem A.1.6.

A quadratic polynomial that has complex roots is called irreducible because it can not be rewritten in a factored form involving only real roots.

There are some tricks to factoring that can be useful to know. Factoring is the reverse process of multiplying by distribution, so we start by noticing what happens when you multiply out two simple factors:

\begin{equation*} (x+a)(x+b) = x^2+(a+b)x+ab. \end{equation*}

Notice that the coefficient of \(x\) is the sum \(a+b\) and the constant term is the product \(ab\text{.}\) When trying to factor a quadratic, look for numbers that multiply to give the product term and add to give the coefficient of \(x\text{.}\) This is often a matter of trial and error.

Knowing one root \(x=r\) of a quadratic \(f(x)=ax^2+bx+c\) so that \(f(r)=0\text{,}\) we know that \(x-r\) is a factor. The other factor can be determined easily.

Synthetic division is a procedure that works for quadratics as well as higher order polynomials. This procedure uses a table that starts with the coefficients on the first row. For a more thorough discussion for higher-order polynomials, see Algorithm 20.

Every quadratic can be rewritten in a form \(y=a(x-h)^2+k\) where \((h,k)\) is the vertex of the parabola and \(a\) is the leading coefficient and scaling factor. The process of rewriting a quadratic \(y=ax^2+bx+c\) in this vertex form is called completing the square. It is based on noticing what happens with expanding the square of a binomial, \((x+a)^2 = x^2+2ax+a^2\text{.}\) The strategy involves adding a term to form a perfect square and subtracting the same term to guarantee the expression does not change.

ExampleA.2.15

Complete the square for \(3x^2-4x+1\text{.}\)

Solution
  1. Group the non-constant terms and factor out the leading coefficient.

    \begin{align*} 3x^2-4x+1 &= (3x^2-4x) + 1\\ & = 3(x^2 - \frac{4}{3}x) + 1 \end{align*}
  2. Recognize the coefficient \(-\frac{4}{3}\) as double \(-\frac{2}{3}\) and use this to complete the square.

    \begin{align*} 3x^2-4x+1 & = 3\Big(x^2 + 2 \cdot \frac{-2}{3}x \Big) + 1 \\ & = 3\Big(x^2 + 2 \cdot \frac{-2}{3}x + (\frac{-2}{3})^2 - (\frac{-2}{3})^2 \Big) + 1 \\ & = 3 \Big(x^2 - \frac{4}{3} x + \frac{4}{9}\Big) - 3(\frac{4}{9}) + 1\\ & = 3 \Big(x - \frac{2}{3}\Big)^2 - \frac{4}{3} + 1 = 3\Big(x-\frac{2}{3}\Big)^2 - \frac{1}{3}. \end{align*}
  3. Interpret the results as saying \(h=\frac{2}{3}\) (because the completed square is always of the form \((x-h)^2\)) and \(k=-\frac{1}{3}\text{.}\) Thus the vertex of the parabola is at \((\frac{2}{3}, -\frac{1}{3})\text{.}\) The leading coefficient \(a=3\) indicates that the parabola opens up and is three times steeper than the standard parabola \(y=x^2\text{.}\)

SubsectionA.2.3Polynomials

Linear and quadratic formulas are special cases of polynomials. This section gives an overview of principles about polynomials that are likely to appear in calculus. First, we introduce some basic definitions.

DefinitionA.2.16Monomials

A monomial is an expression that is a constant multiple of a variable raised to a non-negative integer power, \(a x^k\text{,}\) where \(k=0,1,2,3,\ldots\) and \(a \in \mathbb{R}\text{.}\)

Examples include \(4x^2\) (with \(a=4\) and \(k=2\)), \(\frac{1}{3}x^7\) (with \(a=\frac{1}{3}\) and \(k=7\)) and \(3\) (where \(a=3\) and \(k=0\)). The following are not monomials: \(3\sqrt{x} = 3x^{1/2}\) (since not an integer power) and \(\frac{3}{x^2} = 3x^{-2}\) since the power is a negative integer.

DefinitionA.2.17Polynomials

An algebraic expression that can be rewritten as a sum of monomials is called a polynomial. The monomials are called the terms of the polynomial. The monomial with the highest power is called the leading term and its power is called the degree of the polynomial. The constant multiples in the monomials are called coefficients and the coefficient in the leading term is called the leading coefficient.

We usually write a polynomial with terms ordered by decreasing powers, called standard form. An abstract representation of a polynomial with degree \(n\) is written

\begin{equation*} p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \end{equation*}

where the symbols \(a_n, a_{n-1}, \ldots, a_0\) represent the coefficients. A missing term is represented by a coefficient zero.

ExampleA.2.18

\(x^4-2x^2+3x+1\) is a polynomial with degree \(n=4\text{.}\) The coefficients are \(a_4=1\text{,}\) \(a_3=0\) (since no \(x^3\) term), \(a_2=-2\text{,}\) \(a_1=3\) and \(a_0=1\text{.}\)

\(2x^2(3x+1)(x-2)\) is a polynomial, but must be expanded (multiply out) to find the coefficients.

\begin{align*} 2x^2(3x+1)(x-2) &= 2x^2(3x^2-6x+x-2) \\ &= 2x^2(3x^2-5x-2) \\ & = 6x^4-10x^3-4x^2 \end{align*}

We see that the polynomial has degree \(n=4\) and coefficients \(a_4=6\text{,}\) \(a_3=-10\text{,}\) \(a_2=-4\) and \(a_1=a_0=0\text{.}\)

Every polynomial \(p(x)\) is a function whose domain is all real numbers \((-\infty,\infty)\text{.}\) Values of \(x\) for which \(p(x)=0\) are called zeros or roots of the polynomial. These roots are related to factors.

Synthetic division is an algorithm that can both test if a value \(x=r\) is a root and determine the coefficients of the factored polynomial \(q(x)\) at the same time. Synthetic division for quadratic polynomials (degree \(n=2\)) is a special case of this process, described in Algorithm 13.

ExampleA.2.21

Use synthetic division with the polynomial \(p(x)=x^3-6x+2\) with the test value \(x=2\) and interpret the result.

Solution

Start by identifying the coefficients. Any missed terms have a coefficient of zero,

\begin{equation*} p(x) = x^3 + 0 x^2 + -6x + 2. \end{equation*}

We start the synthetic division table using the coefficients in the first row.

\begin{equation*} \begin{matrix} 1 & 0 & -6 & 2 \\ 0 & \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} \\ \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} \end{matrix} \end{equation*}

We then finish filling the table. To find values in the second row, we use the previous result in the third row and multiply by the test value 2. To find the values in the third row, we add the values in the column. The first value in the second row is always 0. The completed table is shown below.

\begin{equation*} \begin{matrix} 1 & 0 & -6 & 2 \\ 0 & 2 & 4 & -4 \\ 1 & 2 & -2 & -2 \end{matrix} \end{equation*}

Once the table is complete, we interpret the values in the third row as coefficients and a remainder. The last value is the remainder, \(r=-2\text{,}\) and the other values are the coefficients of a polynomial whose degree is one smaller than the original, in this case \(n-1=2\text{.}\) That is, the quotient polynomial is \(q(x) = x^2 + 2x - 2\text{.}\) The original polynomial can be written

\begin{align*} p(x) &= (x-2)q(x) + r\\ x^3-6x+2 &= (x-2)(x^2+2x-2) + -2 \end{align*}

The non-zero remainder means that \(x-2\) is not a factor and also tells us that \(p(2) = -2\text{.}\)

How do we know which numbers to try? If you have access to a graph of the polynomial, you should use the values for roots that you see. If you do not have access to a graph, then you might be able to use the results of the Integer Root Theorem or Rational Root Theorem so long as all of the coefficients of your polynomial are integers.

ExampleA.2.23

The polynomial \(p(x) = x^3-6x+2\) has only integer coefficients and a constant coefficient \(a_0=2\text{.}\) The only factors of \(a_0\) are \(\pm 1\) and \(\pm 2\text{.}\) So the Integer Root Theorem guarantees that these four integers are the only four numbers that we need to check if they are roots. A quick test of each of those values (below) shows that \(p(x)\) has no integer roots. (Without the theorem, we wouldn't know how many points we had to check.)

\(x\) \(p(x)\)
1 -3
-1 -5
2 -2
-2 6

The Integer Root Theorem is a special case of the Rational Root Theorem where \(s=1\) (which is always a factor of \(a_n\text{.}\)

SubsectionA.2.4Exponentials and Logarithms

Exponential and logarithmic functions are related to the algebraic ideas of exponents and powers. An exponential function is a function involving powers where the power is a variable while the base is a constant. A logarithmic function is the inverse of an exponential function.

DefinitionA.2.25Exponential Function

An exponential function is a function \(f(x)\) that can be written in the form

\begin{equation*} f(x)=A\cdot b^x \end{equation*}

where \(A\) is a real number (called the coefficient) and \(b\) is a positive number (called the base). A base \(b=1\) is usually not considered an exponential function because then \(f(x)=A\) is a constant function.

Sometimes an exponential function is not written in its standard form. You need to know how to rewrite formulas involving exponents to see if they are exponential functions or not. This requires applying the properties of powers (or exponents) Theorem 28.

ExampleA.2.26

Show that \(f(x) = 3^{2x-1}\) is an exponential function.

Solution

If we think of \(2x-1\) as a sum \(2x-1=2x+-1\text{,}\) then we can apply the property that adding exponents is equivalent to a product.

\begin{equation*} f(x) = 3^{2x-1} = 3^{2x+-1} = 3^{2x} \cdot 3^{-1} \end{equation*}

Because \(3^{-1} = \frac{1}{3}\) and we can apply the property that multiplying exponents is equivalent to consecutive powers, we can rewrite this as

\begin{equation*} f(x) = \frac{1}{3} (3^2)^{x} = \frac{1}{3} \cdot 9^x. \end{equation*}

This is recognized as the standard form of an exponential function.

DefinitionA.2.27Elementary Exponential Functions

An elementary exponential function is an exponential function with coefficient \(A=1\text{,}\) characterized only by the base \(b \gt 0\text{.}\) We write

\begin{equation*} \exp_b(x) = b^x. \end{equation*}

Exponential functions inherit properties from the properties of powers or exponents.

Elementary exponential functions with \(b \gt 1\) show exponential growth and are increasing functions. Elementary exponential functions with \(b \lt 1\) show exponential decay and are decreasing functions. (When \(b=1\text{,}\) the function is constant.) That exponential functions are either increasing or decreasing means that they are one-to-one. For \(b \ne 1\text{,}\) an elementary exponential function has a range of \((0,\infty)\) so that for any value \(y \gt 0\text{,}\) there is a unique value \(x\) such that \(b^x = y\text{.}\) This value of \(x\) is how we define a logarithm.

DefinitionA.2.29Logarithm Functions

For any base \(b \gt 0\) with \(b \ne 1\text{,}\) we can define the logarithm with base \(b\) as the function whose input is a positive number \(u \gt 0\) and whose output is the number \(x\) such that \(b^x=u\text{.}\) We write

\begin{equation} \log_b(u) = x \qquad \Leftrightarrow \qquad \exp_b(x) = b^x = u.\tag{A.2.18} \end{equation}

The equivalence of equations that defines the logarithm (Equation (A.2.18)) allows us to interpret logarithms in terms of powers. Logarithms are the inverse functions of elementary exponential functions, which allows us to solve some equations where the variable is in a power.

ExampleA.2.31

Find the exact value of \(\log_4(32)\text{.}\)

Solution

The value of \(\log_4(32)\) is unknown, so we use a variable and write the exponential equivalent equation (see Equation (A.2.18)).

\begin{equation*} x = \log_4(32) \quad \Leftrightarrow \quad 4^x = 32 \end{equation*}

The next step is to recognize that 4 and 32 are both powers of 2, so we rewrite the equation in terms of powers of 2:

\begin{equation*} (2^2)^x = 2^5 \quad \Leftrightarrow \quad 2^{2x} = 2^5. \end{equation*}

Now that we have an equation with the same base, because exponentials are one-to-one, the powers must be equal, \(2x = 5\text{,}\) allowing us to solve for \(x=\frac{5}{2}\text{,}\) which is the exact value for the logarithm.

ExampleA.2.32

Solve the equation

\begin{equation*} 2^{3x+1} = 5 \end{equation*}

for \(x\) using a logarithm.

Solution

The equation we are solving can be rewritten in terms of exponential functions as

\begin{equation*} \exp_2(3x+1) = 5. \end{equation*}

This new equation emphasizes that the variable expression \(3x+1\) is the input of an exponential. (It is not necessary to write this fact, just to recognize it.) The logarithm \(\log_2\) is the inverse of \(\exp_2\text{,}\) which in composition with one another yields the identity

\begin{equation*} \log_2(2^{3x+1}) = 3x+1. \end{equation*}

Applying this operation to both sides of the original equation, we have

\begin{equation*} \log_2(2^{3x+1}) = \log_2(5) \quad \Leftrightarrow \quad 3x+1 = \log_2(5). \end{equation*}

The expression \(\log_2(5)\) is just a value, so we can solve for \(x\text{,}\)

\begin{equation*} x = \frac{\log_2(5) - 1}{3}, \end{equation*}

giving us our desired solution.

The properties of powers translate to corresponding properties of logarithms.

It is most useful to remember the properties of logarithms conceptually. For example, Equation (A.2.21) states that the logarithm of a product is equivalent to the sum of the logarithms of the factors. Similarly, Equation (A.2.22) states that the logarithm of a quotient is equivalent to the difference of the logarithms (and the logarithm of the denominator is subtracted). Finally, Equation (A.2.23) states that the logarithm of a value raised to a power is equivalent to the product of the power times the logarithm of the value.

Although there are valid logarithm functions for every base \(b \gt 0\) with \(b \ne 1\text{,}\) calculators generally have only two specific bases. The common logarithm refers to the logarithm with base \(b=10\) and is chosen to be conveniently related to scientific notation. The natural logarithm refers to the logarithm with base \(b=e\) (a special transcendental number). These two logarithms have their own special notation. The common logarithm (base 10) is often written without any base,

\begin{equation*} \log(x) \equiv \log_{10}(x). \end{equation*}

The natural logarithm (base \(e\)) is often written

\begin{equation*} \ln(x) \equiv \log_{e}(x). \end{equation*}

The properties of logarithms allow us to express the logarithm of any base in terms of the logarithm of a base of your choice. This is accomplished through the change of base formula for logarithms.

The strategy to prove these formulas is the same as would be used to solve an equation with logarithms in terms of the natural or common logarithm.

ExampleA.2.35

Solve the equation \(2^{3x} = 5\) in terms of the natural logarithm. (Note that this is equivalent to solving \(8^x=5\) or \(x=\log_8(5)\text{.}\))

Solution

Starting with the equation involving the exponential, we will apply the natural logarithm (as asked in the problem) to both sides of the equation.

\begin{equation*} 2^{3x}=5 \quad \Leftrightarrow \quad \ln(2^{3x}) = \ln(5) \end{equation*}

The left side of the equation is the logarithm of a power, so we can use the corresponding logarithm property (Equation (A.2.23)) to say

\begin{equation*} \ln(2^{3x}) = 3x \cdot \ln(2) = \ln(5). \end{equation*}

Dividing both sides of the equation by the product \(3 \ln(2)\) we find

\begin{equation*} x = \frac{\ln(5)}{3 \ln(2)} \end{equation*}

which matches what we would get using the change of base formula since \(3 \ln(2) = \ln(2^3) = \ln(8)\) (using Equation (A.2.23).)

SubsectionA.2.5Absolute Value

The absolute value operation takes a number and finds its magnitude (or distance from zero). Because magnitude is a non-negative value and positive and negative pairs are the same distance from zero, we often imagine that the role of absolute value is to remove a negative sign, \(|-3| = 3\text{.}\) However, when a variable is involved, a negative sign means finding the inverse of a value for which we may not know if it is positive or negative. So it is incorrect to say that \(|-x| = x\) (FALSE). The proper definition of absolute value uses a piecewise formula.

DefinitionA.2.36Absolute Value

\begin{equation*} |x| = \begin{cases} x, & x \ge 0 \\ -x, & x \lt 0 \end{cases} \end{equation*}

As a function, the graph of absolute value \(y=|x|\) gives two lines: \(y=x\) when \(x \ge 0\) and \(y=-x\) when \(x \lt 0\text{.}\)

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FigureA.2.37Graph of the absolute value, \(y=|x|\)

It is sometimes useful to take advantage of an identity between the square root of a square and the absolute value. This is the source of the plus/minus when solving an equation with a square.

ExampleA.2.39

Solve the equation \(x^2 = 16\text{.}\)

Solution

Applying a square root to both sides of the equation, we then get to use the absolute value identity.

\begin{equation*} \sqrt{x^2} = \sqrt{16} \quad \Leftrightarrow \quad |x| = \sqrt{16} = 4 \end{equation*}

The source of plus/minus is that there are two numbers with magnitude 4,

\begin{equation*} x = \pm 4. \end{equation*}

The absolute value splits nicely with multiplication (and division). However, addition of two values with opposite signs shows that absolute values do not add: \(|3+-4| = |-1| \ne |3| + |-4| = 7\text{.}\) Instead, we have an inequality called the triangle inequality.

The triangle inequality is used to show that the absolute value of a sum (or difference) is bounded by the sum of the magnitudes of the individual terms.

Occasionally we need to apply the triangle in reverse, showing that the absolute value of a sum (or difference) must be bigger than the difference in magnitudes of the parts.

Absolute value and subtraction is often used to describe the distance between two values. For example, the graph \(y=|x-2|\) represents a shift of the graph \(y=|x|\) two units to the right, so that instead of measuring the distance of \(x\) from 0 it measure the distance of \(x\) from 2.

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Note that \(|x+3| = |x--3|\) so that it represents the distance between \(x\) and the value \(-3\text{.}\)