Pre-Calculus Articles

Cheat Sheet / Updated 04-06-2022

Pre-calculus uses the information you know from Algebra I and II and ratchets up the difficulty level to prepare you for calculus. This cheat sheet is designed to help you review key formulas and functions on the fly as you study. It includes formulas, the laws of logarithmic functions, trigonometric values of basic angles, conic section equations, and interval notation.

Cheat Sheet / Updated 02-24-2022

Pre-calculus draws from algebra, geometry, and trigonometry and combines these topics to prepare you for the techniques you need to succeed in calculus. This cheat sheet provides the most frequently used formulas, with brief descriptions of what the letters and symbols represent. Counting techniques are also here, letting you count numbers of events without actually having to list all the ways to do them. Also, you find a step-by-step description of how to complete the square — most useful when you’re working with conic sections and other equations with specific formats.

Article / Updated 12-21-2021

The fundamental theorem of algebra can help you find imaginary roots. Imaginary roots appear in a quadratic equation when the discriminant of the quadratic equation — the part under the square root sign (b2 – 4ac) — is negative. If this value is negative, you can’t actually take the square root, and the answers are not real. In other words, there is no real solution; therefore, the graph won’t cross the x-axis. Using the quadratic formula always gives you two solutions, because the plus/minus sign means you’re both adding and subtracting and getting two completely different answers. When the number underneath the square-root sign in the quadratic formula is negative, the answers are called complex conjugates. One is r + si and the other is r – si. These numbers have both real (the r) and imaginary (the si) parts. The complex number system consists of all numbers r + si where r and s are real numbers. Observe that when s = 0, you simply have the real numbers. Therefore the real numbers are a subset of the complex number system. The fundamental theorem of algebra says that every polynomial function has at least one root in the complex number system. The highest degree of a polynomial gives you the highest possible number of distinct complex roots for the polynomial. Between this fact and Descartes’s rule of signs, you can get an idea of how many imaginary roots a polynomial has. Here’s how Descartes’s rule of signs can give you the numbers of possible real roots, both positive and negative: Positive real roots. For the number of positive real roots, look at the polynomial, written in descending order, and count how many times the sign changes from term to term. This value represents the maximum number of positive roots in the polynomial. For example, in the polynomial f(x) = 2x4 – 9x3 – 21x2 + 88x + 48, you see two changes in sign (don’t forget to include the sign of the first term!) — from the first term (+2x4) to the second (-9x3) and from the third term (-21x2) to the fourth term (88x). That means this equation can have up to two positive solutions. Descartes’s rule of signs says the number of positive roots is equal to changes in sign of f(x), or is less than that by an even number (so you keep subtracting 2 until you get either 1 or 0). Therefore, the previous f(x) may have 2 or 0 positive roots. Negative real roots. For the number of negative real roots, find f(–x) and count again. Because negative numbers raised to even powers are positive and negative numbers raised to odd powers are negative, this change affects only terms with odd powers. This step is the same as changing each term with an odd degree to its opposite sign and counting the sign changes again, which gives you the maximum number of negative roots. The example equation becomes f(–x) = 2x4 + 9x3 – 21x2 – 88x + 48, which changes signs twice. There can be, at most, two negative roots. However, similar to the rule for positive roots, the number of negative roots is equal to the changes in sign for f(–x), or must be less than that by an even number. Therefore, this example can have either 2 or 0 negative roots. Pair up every possible number of positive real roots with every possible number of negative real roots; the remaining number of roots for each situation represents the number of imaginary roots. For example, the polynomial f(x) = 2x4 – 9x3 – 21x2 + 88x + 48 has a degree of 4, with two or zero positive real roots, and two or zero negative real roots. With this information, you can pair up the possible situations: Two positive and two negative real roots, with zero imaginary roots Two positive and zero negative real roots, with two imaginary roots Zero positive and two negative real roots, with two imaginary roots Zero positive and zero negative real roots, with four imaginary roots The following chart makes the information easier to picture: Positive real roots Negative real roots Imaginary roots 2 2 0 2 0 2 0 2 2 0 0 4 Complex numbers are written in the form r + si and have both a real and an imaginary part, which is why every polynomial has at least one root in the complex number system. Real and imaginary numbers are both included in the complex number system. Real numbers have no imaginary part, and pure imaginary numbers have no real part. For example, if x = 7 is one root of the polynomial, this root is considered both real and complex because it can be rewritten as x = 7 + 0i (the imaginary part is 0). The fundamental theorem of algebra gives the total number of complex roots (say there are seven); Descartes’s rule of signs tells you how many possible real roots exist and how many of them are positive and negative (say there are, at most, two positive roots but only one negative root). Now, assume you’ve found them all: x = 1, x = 7, and x = –2. These roots are real, but they’re also complex because they can all be rewritten. The first two columns in the chart find the real roots and classify them as positive or negative. The third column is actually finding, specifically, the non-real numbers: complex numbers with non-zero imaginary parts.

Article / Updated 12-21-2021

Sometimes your geometry teacher may spice things up a bit with complicated polar coordinates — points with negative angles and/or radii. The following list shows you how to plot in three situations — when the angle is negative, when the radius is negative, and when both are negative. When the angle is negative: Negative angles move in a clockwise direction. This figure shows an example point, D. To locate the polar coordinate point D at first locate the angle and then find the location of the radius, 1, on that line. When the radius is negative: When graphing a polar coordinate with a negative radius, you move from the pole in the direction opposite the given positive angle (on the same line as the given angle but in the direction opposite to the angle from the pole). For example, check out point F at in the figure. Some teachers prefer to teach their students to move right along the x- (polar) axis for positive numbers (radii) and left for negative. Then you do the rotation for the angle in a positive direction. You’ll get to the same spot with that method. For example, take a look at point F in the figure. Because the radius is negative, move along the left x-axis 1/2 of a unit. Then rotate the angle in the positive direction (counterclockwise) pi/3 radians. You should arrive at your destination, point F. When both the angle and radius are negative: To express a polar coordinate with a negative radius and a negative angle, locate the terminal side of the negative angle first and then move in the opposite direction to locate the radius. For example, point G in the figure has these characteristics at Indeed, except the origin, each given point can have the following four types of representations: Positive radius, positive angle Positive radius, negative angle Negative radius, positive angle Negative radius, negative angle For example, point E in the figure can have three other polar coordinate representations with different combinations of signs for the radius and angle: When polar graphing, you can change the coordinate of any point you’re given into polar coordinates that are easy to deal with (such as positive radius, positive angle).

Article / Updated 12-21-2021

Depending on how many times you must multiply the same binomial — a value also known as an exponent — the binomial coefficients for that particular exponent are always the same. The binomial coefficients are found by using the combinations formula. If the exponent is relatively small, you can use a shortcut called Pascal's triangle to find these coefficients. If not, you can always rely on algebra! Pascal's triangle, named after the famous mathematician Blaise Pascal, names the binomial coefficients for the binomial expansion. It is especially useful when raising a binomial to lower degrees. For example, if a sadistic teacher asked you to find (3x + 4)10, you probably wouldn't want to use Pascal's triangle; instead, you'd just use the algebraic formula described shortly. The figure illustrates this concept. The top number of the triangle is 1, as well as all the numbers on the outer sides. To get any term in the triangle, you find the sum of the two numbers above it. Each row gives the coefficients to (a + b)n, starting with n = 0. To find the binomial coefficients for (a + b)n, use the nth row and always start with the beginning. For instance, the binomial coefficients for (a + b)5 are 1, 5, 10, 10, 5, and 1 — in that order. If you need to find the coefficients of binomials algebraically, there is a formula for that as well. The rth coefficient for the nth binomial expansion is written in the following form: You may recall the term factorial from your earlier math classes. If not, here is a reminder: n!, which reads as "n factorial," is defined as You read the expression for the binomial coefficient as "n choose r." You usually can find a button for combinations on a calculator. If not, you can use the factorial button and do each part separately. To make things a little easier, 0! is defined as 1. Therefore, you have these equalities: For example, to find the binomial coefficient given by substitute the values into the formula:

Article / Updated 12-21-2021

Knowing whether a function is even or odd helps you to graph it because that information tells you which half of the points you have to graph. These types of functions are symmetrical, so whatever is on one side is exactly the same as the other side. If a function is even, the graph is symmetrical about the y-axis. If the function is odd, the graph is symmetrical about the origin. Even function: The mathematical definition of an even function is f(–x) = f(x) for any value of x. The simplest example of this is f(x) = x2 because f(x)=f(-x) for all x. For example, f(3) = 9, and f(–3) = 9. Basically, the opposite input yields the same output. Odd function: The definition of an odd function is f(–x) = –f(x) for any value of x. The opposite input gives the opposite output. These graphs have 180-degree symmetry about the origin. If you turn the graph upside down, it looks the same. The example shown above, f(x) = x3, is an odd function because f(-x)=-f(x) for all x. For example, f(3) = 27 and f(–3) = –27.

Article / Updated 12-21-2021

When you’re given a problem in radical form, you may have an easier time if you rewrite it by using rational exponents — exponents that are fractions. You can rewrite every radical as an exponent by using the following property — the top number in the resulting rational exponent tells you the power, and the bottom number tells you the root you’re taking: For example, you can rewrite as Fractional exponents are roots and nothing else. For example, 641/3 doesn’t mean 64–3 or In this example, you find the root shown in the denominator (the cube root) and then take it to the power in the numerator (the first power). So 641/3 = 4. The order of these processes really doesn’t matter. You can choose either method: Cube root the 8 and then square that product Square the 8 and then cube root that product Either way, the equation simplifies to 4. Depending on the original expression, though, you may find the problem easier if you take the root first and then take the power, or you may want to take the power first. For example, 643/2 is easier if you write it as (641/2)3 = 83 = 512 rather than (643)1/2, because then you’d have to find the square root of 262,144. Take a look at some steps that illustrate this process. To simplify the expression rather than work with the roots, execute the following: Rewrite the entire expression using rational exponents. Now you have all the properties of exponents available to help you to simplify the expression: x1/2(x2/3 – x4/3). Distribute to get rid of the parentheses. When you multiply monomials with the same base, you add the exponents. Hence, the exponent on the first term is and the exponent of the second term is 1/2+4/3=11/6. So you get x7/6 – x11/6. Because the solution is written in exponential form and not in radical form, as the original expression was, rewrite it to match the original expression. This gives you Typically, your final answer should be in the same format as the original problem; if the original problem is in radical form, your answer should be in radical form. And if the original problem is in exponential form with rational exponents, your solution should be as well.

Article / Updated 12-17-2021

Knowing how to calculate the circumference of a circle and, in turn, the length of an arc — a portion of the circumference — is important in pre-calculus because you can use that information to analyze the motion of an object moving in a circle. An arc can come from a central angle, which is one whose vertex is located at the center of the circle. You can measure an arc in two different ways: As an angle. The measure of an arc as an angle is the same as the central angle that intercepts it. As a length. The length of an arc is directly proportional to the circumference of the circle and is dependent on both the central angle and the radius of the circle. If you think back to geometry, you may remember that the formula for the circumference of a circle is with r representing the radius. Also recall that a circle has 360 degrees. So if you need to find the length of an arc, you need to figure out what part of the whole circumference (or what fraction) you're looking at. You use the following formula to calculate the arc length: The symbol theta (θ) represents the measure of the angle in degrees, and s represents arc length, as shown in the figure: If the given angle theta is in radians, Time for an example. To find the length of an arc with an angle measurement of 40 degrees if the circle has a radius of 10, use the following steps: Assign variable names to the values in the problem. The angle measurement here is 40 degrees, which is theta. The radius is 10, which is r. Plug the known values into the formula. This step gives you Simplify to solve the formula. You first get which multiplies to The figure shows what this arc looks like. Now try a different problem. Find the measure of the central angle of a circle in radians with an arc length of and a radius of 16. This time, you must solve for theta (the formula is s = rθ when dealing with radians): Plug in what you know to the radian formula. Divide both sides by 16. Your formula looks like this: Reduce the fraction. You're left with the solution:

Cheat Sheet / Updated 07-24-2021

When you study pre-calculus, you are crossing the bridge from algebra II to Calculus. Pre-calculus involves graphing, dealing with angles and geometric shapes such as circles and triangles, and finding absolute values. You discover new ways to record solutions with interval notation, and you plug trig identities into your equations.

Article / Updated 07-22-2021

Vertical parabolas give an important piece of information: When the parabola opens up, the vertex is the lowest point on the graph — called the minimum, or min. When the parabola opens down, the vertex is the highest point on the graph — called the maximum, or max. Only vertical parabolas can have minimum or maximum values, because horizontal parabolas have no limit on how high or how low they can go. Finding the maximum of a parabola can tell you the maximum height of a ball thrown into the air, the maximum area of a rectangle, the minimum value of a company's profit, and so on. Finding the max of a parabola For example, say that a problem asks you to find two numbers whose sum is 10 and whose product is a maximum. You can identify two different equations hidden in this one sentence: x + y = 10 xy = MAX If you're like most people, you don't like to mix variables when you don't have to, so you should solve one equation for one variable to substitute into the other one. This process is easiest if you solve the equation that doesn't include min or max at all. So if x + y = 10, you can say y = 10 – x. You can plug this value into the other equation to get the following: (10 – x)x = MAX If you distribute the x on the outside, you get 10x – x2 = MAX. This result is a quadratic equation for which you need to find the vertex by completing the square (which puts the equation into the form you're used to seeing that identifies the vertex). Finding the vertex by completing the square gives you the maximum value. To do that, follow these steps: Rearrange the terms in descending order. This step gives you –x2 + 10x = MAX. Factor out the leading term. You now have –1(x2 – 10x) = MAX. Complete the square. This step expands the equation to –1(x2 – 10x + 25) = MAX – 25. Notice that –1 in front of the parentheses turned the 25 into –25, which is why you must add –25 to the right side as well. Factor the information inside the parentheses. This gives you –1(x – 5)2 = MAX – 25. Move the constant to the other side of the equation. You end up with –1(x – 5)2 + 25 = MAX. The vertex of the parabola is (5, 25). Therefore, the number you're looking for (x) is 5, and the maximum product is 25. You can plug 5 in for x to get y in either equation: 5 + y = 10, or y = 5. Graphing a parabola to find a maximum value from a word problem This figure shows the graph of the maximum function to illustrate that the vertex, in this case, is the maximum point.