Understanding Algebra 2
June 2023
Introduction
Algebra 2 has a reputation for being a pile of unrelated tricks: factor this, graph that, memorize a formula for the next quiz. That reputation is unfair, because the course is built around one steady idea. A function is a machine that turns inputs into outputs, and almost everything you study is a way of understanding a particular family of those machines. Linear functions, which you met earlier, grow at a constant rate. In Algebra 2 you meet families that curve, level off, explode, or break apart, and you learn what each one is good for.
This essay walks through the main families in the order most courses present them. Along the way it explains why the standard techniques work, not only how to run them, since a technique you understand is far easier to remember under test pressure. Each section ends with one short worked example so you can see the ideas in action. If you read with a pencil and redo the examples yourself, the essay will serve as a decent review.
Functions and transformations
A function assigns to each input exactly one output. We write $y = f(x)$, and the set of allowed inputs is the domain while the set of resulting outputs is the range. The graph of a function is a picture of every input-output pair at once, and the vertical line test captures the "exactly one output" rule: no vertical line may cross the graph twice.
The most useful habit in this part of the course is to think of complicated functions as simple ones that have been moved around. Starting from a parent function such as $f(x) = x^2$, $f(x) = |x|$, or $f(x) = \sqrt{x}$, you can build a whole family with the general form
$$g(x) = a\,f(x - h) + k.$$The number $h$ slides the graph horizontally, and it does so in the direction opposite to its sign: $f(x-3)$ moves the graph three units to the right, because the input must be larger by three to produce the same output as before. The number $k$ slides the graph vertically, in the direction you would expect. The factor $a$ stretches the graph away from the $x$-axis when $|a| > 1$, compresses it when $0 < |a| < 1$, and reflects it across the $x$-axis when $a$ is negative. Changes made inside the parentheses affect the horizontal direction and tend to behave backwards, while changes outside affect the vertical direction and behave as you would guess.
Functions can also be combined. The composition $(f \circ g)(x) = f(g(x))$ feeds the output of $g$ into $f$, and the inverse function $f^{-1}$ undoes $f$, which is the same as reflecting its graph across the line $y = x$. Only a function that passes the horizontal line test has an inverse that is itself a function.
Worked example. Describe $g(x) = 2(x-3)^2 + 1$ as a transformation of $f(x) = x^2$. The graph is shifted 3 units right and 1 unit up, and stretched vertically by a factor of 2, so its vertex is $(3, 1)$ and it still opens upward. As a check, $f(2) = 4$, and the matching point on $g$ is at $x = 5$, where $g(5) = 2(2)^2 + 1 = 9$.
Quadratics and complex numbers
A quadratic function has the form $f(x) = ax^2 + bx + c$ with $a \neq 0$, and its graph is a parabola. Quadratics describe anything with a single peak or valley, such as the arc of a thrown ball or the profit of a business that earns less when prices are set too low or too high. The same parabola can be written in several forms, and each form answers a different question.
Vertex form, $f(x) = a(x-h)^2 + k$, shows the turning point $(h, k)$ at a glance. You can convert from standard form by completing the square, or by using the fact that the vertex has $x$-coordinate $h = -\dfrac{b}{2a}$ and then substituting to find $k$. Factored form, $f(x) = a(x - r_1)(x - r_2)$, shows the $x$-intercepts $r_1$ and $r_2$ directly, thanks to the zero product property: if a product is zero, one of its factors must be zero. Factoring is the fastest way to solve a quadratic when the numbers cooperate.
When they do not, the quadratic formula always works. Solving $ax^2 + bx + c = 0$ by completing the square in general gives
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$The expression under the radical, $D = b^2 - 4ac$, is called the discriminant, and it tells you what kind of answers to expect before you finish the work. If $D > 0$ there are two distinct real roots and the parabola crosses the $x$-axis twice. If $D = 0$ there is one repeated real root and the vertex sits on the axis. If $D < 0$ there are no real roots, because you cannot take a real square root of a negative number.
That last case is what motivates complex numbers. We define $i$ by $i^2 = -1$, and a complex number is anything of the form $a + bi$ with $a$ and $b$ real. They add and multiply like ordinary binomials, with $i^2$ replaced by $-1$ whenever it appears. For example, $(3+2i)(1-i) = 3 - 3i + 2i - 2i^2 = 5 - i$. With complex numbers in play, every quadratic has exactly two roots, and when the coefficients are real, a negative discriminant produces a conjugate pair $p \pm qi$. The equation $x^2 + 2x + 5 = 0$, for instance, has $D = 4 - 20 = -16$ and roots $-1 \pm 2i$.
Worked example. Solve $2x^2 - 3x - 5 = 0$. By factoring, $(2x - 5)(x + 1) = 0$, so $x = \tfrac{5}{2}$ or $x = -1$. By the formula, $D = (-3)^2 - 4(2)(-5) = 49$, so $x = \dfrac{3 \pm 7}{4}$, which gives $\tfrac{10}{4} = \tfrac52$ and $-1$ again. Because $D$ is a positive perfect square, the roots are rational, which is exactly why the quadratic factored.
Polynomials
A polynomial is a sum of terms of the form $a_k x^k$ with whole-number exponents, such as $p(x) = x^3 - 6x^2 + 11x - 6$. Its degree is the highest exponent, and its leading coefficient is the number attached to that term. Quadratics are just the degree-two members of a much larger family.
The first thing to know about a polynomial's graph is its end behavior, which depends only on the leading term, because for very large $|x|$ that term dominates all the others. An even-degree polynomial points the same way at both ends: both up if the leading coefficient is positive, both down if it is negative. An odd-degree polynomial points in opposite directions, rising to the right and falling to the left when the leading coefficient is positive. A polynomial of degree $n$ has at most $n$ real roots and at most $n-1$ turning points, and the Fundamental Theorem of Algebra says that, counting multiplicity and allowing complex values, it has exactly $n$ roots.
Finding those roots is where two connected theorems help. The Remainder Theorem states that when $p(x)$ is divided by $x - c$, the remainder equals $p(c)$. The Factor Theorem is the special case that matters most: $x - c$ is a factor of $p(x)$ exactly when $p(c) = 0$. So to factor a polynomial you hunt for a root, perhaps by testing the candidates suggested by the Rational Root Theorem (fractions whose numerator divides the constant term and whose denominator divides the leading coefficient), then divide it out with long division or synthetic division and continue with the smaller quotient.
Worked example. Factor $p(x) = x^3 - 6x^2 + 11x - 6$. Since $p(1) = 1 - 6 + 11 - 6 = 0$, the Factor Theorem says $x - 1$ is a factor. Dividing gives the quotient $x^2 - 5x + 6$, which factors as $(x-2)(x-3)$. So $p(x) = (x-1)(x-2)(x-3)$ with roots $1$, $2$, and $3$. The leading coefficient is positive and the degree is odd, so the graph falls on the left and rises on the right. As a check on the Remainder Theorem, $p(4) = 64 - 96 + 44 - 6 = 6$, which matches $(3)(2)(1) = 6$.
Rational functions
A rational function is a ratio of two polynomials, $f(x) = \dfrac{p(x)}{q(x)}$. Because division by zero is impossible, these functions are undefined wherever the denominator vanishes, and the interesting behavior of their graphs happens near those places and far out at the ends.
Start by factoring both the numerator and the denominator. A factor that appears in both cancels, and it leaves a hole in the graph at that $x$-value, since the original expression is still undefined there even though the simplified one is not. A factor in the denominator that survives cancellation produces a vertical asymptote, a line the graph approaches but never touches as the function shoots toward positive or negative infinity. The $x$-intercepts come from the zeros of whatever is left in the numerator.
Horizontal behavior is governed by comparing degrees. If the numerator has lower degree than the denominator, the graph flattens toward $y = 0$. If the degrees are equal, it flattens toward the ratio of the leading coefficients. If the numerator has higher degree, there is no horizontal asymptote, and when the degree is exactly one higher you get a slant asymptote that polynomial division will reveal. Arithmetic with rational expressions mirrors arithmetic with fractions: multiply across after factoring, flip and multiply to divide, and find a common denominator to add or subtract. Solving a rational equation usually means multiplying through by the common denominator, and afterwards you must discard any answer that makes an original denominator zero.
Worked example. Analyze $f(x) = \dfrac{x^2 - 1}{x^2 - x - 2}$. Factoring gives $\dfrac{(x-1)(x+1)}{(x-2)(x+1)}$. The factor $x+1$ cancels, so there is a hole at $x = -1$; the simplified function $\dfrac{x-1}{x-2}$ gives a height of $\dfrac{-2}{-3} = \tfrac23$, so the hole is at $(-1, \tfrac23)$. The remaining denominator factor gives a vertical asymptote at $x = 2$. The degrees are equal with leading coefficients $1$ and $1$, so the horizontal asymptote is $y = 1$. The only $x$-intercept is at $x = 1$.
Exponents and radicals
Exponent rules are the grammar behind much of the course. When you multiply powers of the same base you add the exponents, $x^m \cdot x^n = x^{m+n}$. When you divide you subtract them, $\dfrac{x^m}{x^n} = x^{m-n}$. A power raised to a power multiplies exponents, $(x^m)^n = x^{mn}$. A zero exponent gives $x^0 = 1$ for nonzero $x$, and a negative exponent means a reciprocal, $x^{-n} = \dfrac{1}{x^n}$. Each of these rules is forced on us by wanting the earlier ones to keep working.
The same logic extends to fractions. If $(x^{1/2})^2$ is to equal $x^{1}$ under the power rule, then $x^{1/2}$ must be a number whose square is $x$, which is the square root. In general, $x^{1/n} = \sqrt[n]{x}$ and $x^{m/n} = \left(\sqrt[n]{x}\right)^m$. This means radicals and rational exponents are two notations for one idea, and converting a radical to an exponent is often the easiest way to simplify it.
Simplifying a radical means pulling out perfect powers: $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$. Radicals should not be left in a denominator, so we rationalize by multiplying top and bottom by a suitable radical, as in $\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$. When solving a radical equation, isolate the radical, square both sides, and then check every answer in the original equation, because squaring can introduce extraneous solutions.
Worked example. Simplify $(27x^6)^{2/3}$. Apply the exponent to each factor: $27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$, and $(x^6)^{2/3} = x^{6 \cdot 2/3} = x^4$. The result is $9x^4$. In the same spirit, for $x \ge 0$ we have $\sqrt{72x^5} = \sqrt{36x^4 \cdot 2x} = 6x^2\sqrt{2x}$.
Exponential and logarithmic functions
In a polynomial the variable sits in the base, as in $x^2$. In an exponential function the variable sits in the exponent, as in $f(x) = a \cdot b^x$ with $b > 0$ and $b \neq 1$. That small change produces a very different kind of growth. When $b > 1$ the function models growth, because each step of one unit multiplies the output by $b$. When $0 < b < 1$ it models decay. A population that grows by 5 percent per year has $b = 1.05$, and a medicine that loses 20 percent of its strength each hour has $b = 0.8$. Continuous growth is written with the special constant $e \approx 2.718$, as in $A = Pe^{rt}$.
The logarithm is the inverse of the exponential. The statement $\log_b x = y$ means exactly the same thing as $b^y = x$, so a logarithm is simply an exponent in disguise: it answers the question "what power of $b$ gives $x$?" For example, $\log_2 8 = 3$ because $2^3 = 8$. The two most common bases are 10, written $\log$, and $e$, written $\ln$. Because logarithms are exponents, their rules mirror the exponent rules:
$$\log_b(xy) = \log_b x + \log_b y, \qquad \log_b\!\left(\frac{x}{y}\right) = \log_b x - \log_b y, \qquad \log_b(x^n) = n\log_b x.$$There is also a change-of-base formula, $\log_b x = \dfrac{\ln x}{\ln b}$, which lets a calculator evaluate any base. The power rule is the one that makes logarithms so useful for equations, since it drags a variable out of the exponent. To solve an exponential equation, isolate the exponential expression, take a log of both sides, and use the power rule. To solve a logarithmic equation, combine the logs into one, rewrite it in exponential form, solve, and then check the answers, because the argument of a logarithm must be positive.
Worked example. How long does it take a quantity growing at 5 percent per year to double? We need $1.05^t = 2$. Taking natural logs, $t \ln 1.05 = \ln 2$, so $t = \dfrac{\ln 2}{\ln 1.05} \approx \dfrac{0.6931}{0.04879} \approx 14.2$ years. For a logarithmic equation, solve $\log_2 x + \log_2 (x-2) = 3$. Combining gives $\log_2\big(x(x-2)\big) = 3$, so $x^2 - 2x = 8$, or $x^2 - 2x - 8 = 0$, which factors as $(x-4)(x+2) = 0$. The value $x = -2$ would put a negative number inside a logarithm, so the only solution is $x = 4$.
Sequences and series
A sequence is an ordered list of numbers, and a series is the sum of the terms of a sequence. Two kinds appear again and again because they mirror the two growth patterns you have already seen. An arithmetic sequence adds the same common difference $d$ each time, so it is the discrete cousin of a linear function, with $a_n = a_1 + (n-1)d$. A geometric sequence multiplies by the same common ratio $r$ each time, so it is the discrete cousin of an exponential function, with $a_n = a_1 r^{\,n-1}$.
Series have tidy formulas. The sum of the first $n$ terms of an arithmetic sequence is the number of terms times the average of the first and last, $S_n = \dfrac{n(a_1 + a_n)}{2}$, a fact Gauss is said to have noticed as a schoolboy when he paired the first and last numbers in a long list. For a geometric sequence with $r \neq 1$,
$$S_n = a_1\,\frac{1 - r^n}{1 - r}.$$If $|r| < 1$, the terms shrink so quickly that an infinite geometric series converges to $S = \dfrac{a_1}{1 - r}$. Sigma notation packages a series compactly: $\displaystyle\sum_{k=1}^{n} a_k$ means "add up $a_k$ for $k$ from 1 to $n$." The index $k$ is a placeholder, much like the variable in a function definition.
Worked example. Evaluate $\displaystyle\sum_{k=1}^{10} (3k + 2)$. This is arithmetic with first term $5$, last term $32$, and ten terms, so the sum is $\dfrac{10(5 + 32)}{2} = 185$. For a geometric case, $2 + 6 + 18 + 54 + 162$ has $a_1 = 2$, $r = 3$, and $n = 5$, so $S_5 = 2\cdot\dfrac{1 - 3^5}{1 - 3} = 2\cdot\dfrac{-242}{-2} = 242$. And the infinite series $\displaystyle\sum_{k=1}^{\infty} 6\left(\tfrac13\right)^{k-1}$ converges to $\dfrac{6}{1 - 1/3} = 9$.
Systems of equations
A system of equations asks for values that satisfy several equations at once. Geometrically, a solution of a two-variable system is a point where the graphs meet. Two lines usually cross at a single point, but they can also be parallel, in which case the system is inconsistent and has no solution, or they can be the same line, in which case there are infinitely many solutions.
There are two main algebraic methods. In substitution you solve one equation for a variable and plug the result into the other. In elimination you add multiples of the equations together so that one variable cancels. Either method extends to three variables, where each equation describes a plane, and the same ideas scale up to the matrices you may see later. Systems are not limited to lines, either: a line and a parabola can meet in zero, one, or two points, and finding them means substituting and solving a quadratic, which brings the discriminant back into the story.
Worked example. Solve $2x + 3y = 12$ and $x - y = 1$. From the second equation, $x = y + 1$. Substituting into the first, $2(y+1) + 3y = 12$, so $5y + 2 = 12$ and $y = 2$, which gives $x = 3$. Checking, $2(3) + 3(2) = 12$ and $3 - 2 = 1$, so $(3, 2)$ works in both equations.
Conclusion
Looking back, the topics of Algebra 2 fit together more tightly than the table of contents suggests. Transformations give you a vocabulary for moving any graph around. Quadratics and polynomials teach you to read a function through its roots, and the Factor Theorem connects the factored form of an expression to the places where its graph meets the axis. Rational functions show what happens when polynomials are divided, and exponents and radicals supply the rules that make exponential and logarithmic functions possible. Sequences and series reappear the same patterns of linear and exponential change in a discrete setting, and systems of equations ask when several of these relationships hold together.
If you carry away one habit, let it be this: whenever you meet a new function, ask what family it belongs to, which features define that family, and what each form of its equation reveals. Intercepts, asymptotes, end behavior, and growth rate are the questions that organize the whole course. Answer them well and the computation becomes routine, and you will be ready for precalculus and beyond.