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Article / Updated 08-14-2023

As you work through pre-calculus, adopting certain tasks as habits can help prepare your brain to tackle your next challenge: calculus. In this article, you find ten habits that should be a part of your daily math arsenal. Perhaps you’ve been told to perform some of these tasks since elementary school — such as showing all your work — but other tricks may be new to you. Either way, if you remember these ten pieces of advice, you’ll be even more ready for whatever calculus throws your way. Figure Out What the Problem Is Asking Often, you’ll find that reading comprehension and the ability to work with multiple parts that comprise a whole is an underlying property of a math problem. That’s okay — that’s also what life is all about!! When faced with a math problem, start by reading the whole problem or all the directions to the problem. Look for the question inside the question. Keep your eyes peeled for words like solve, simplify, find, and prove, all of which are common buzz words in any math book. Don’t begin working on a problem until you’re certain of what it wants you to do. For example, take a look at this problem: The length of a rectangular garden is 24 feet longer than the garden’s width. If you add 2 feet to both the width and the length, the area of the garden is 432 square feet. How long is the new, bigger garden? If you miss any of the important information, you may start to solve the problem to figure out how wide the garden is. Or you may find the length but miss the fact that you’re supposed to find out how long it is with 2 feet added to it. Look before you leap! Underlining key words and information in the question is often helpful. This can’t be stressed enough. Highlighting important words and pieces of information solidifies them in your brain so that as you work, you can redirect your focus if it veers off-track. When presented with a word problem, for example, first turn the words into an algebraic equation. If you’re lucky and are given the algebraic equation from the get-go, you can move on to the next step, which is to create a visual image of the situation at hand. And, if you’re wondering what the answer to the example problem is, you’ll find out as you read further. Draw Pictures (the More the Better) Your brain is like a movie screen in your skull, and you’ll have an easier time working problems if you project what you see onto a piece of paper. When you visualize math problems, you’re more apt to comprehend them. Draw pictures that correspond to a problem and label all the parts so you have a visual image to follow that allows you to attach mathematical symbols to physical structures. This process works the conceptual part of your brain and helps you remember important concepts. As such, you’ll be less likely to miss steps or get disorganized. If the question is talking about a triangle, for instance, draw a triangle; if it mentions a rectangular garden filled with daffodils for 30 percent of its space, draw that. In fact, every time a problem changes and new information is presented, your picture should change, too. If you were asked to solve the rectangular garden problem from the previous section, you’d start by drawing two rectangles: one for the old, smaller garden and another for the bigger one. Putting pen or pencil to the paper starts you on the way to a solution. Plan Your Attack — Identify Your Targets When you know and can picture what you must find, you can plan your attack from there, interpreting the problem mathematically and coming up with the equations that you’ll be working with to find the answer: Start by writing a statement. In the garden problem from the last two sections, you’re looking for the length and width of a garden after it has been made bigger. With those in mind, define some variables: Let x = the garden’s width now. Let y = the garden’s length now. Now add those variables to the rectangle you drew of the old garden. But you know that the length is 24 feet greater than the width, so you can rewrite the variable y in terms of the variable x so that y = x + 24. You know that the new garden has had 2 feet added to both its width and length, so you can modify your equations: Let x + 2 = the garden’s new width. Let y + 2 = x + 24 + 2 = x + 26 the garden’s new length. Now add these labels to the picture of the new garden. By planning your attack, you’ve identified the pieces of the equation that you need to solve. Write Down Any Formulas If you start your attack by writing the formula needed to solve the problem, all you have to do from there is plug in what you know and then solve for the unknown. A problem always makes more sense if the formula is the first thing you write when solving. Before you can do that, though, you need to figure out which formula to use. You can usually find out by taking a close look at the words of the problem. In the case of the garden problem from the previous sections, the formula you need is that for the area of a rectangle. The area of a rectangle is A = lw. You’re told that the area of the new rectangle is 432 square feet, and you have expressions representing the length and width, so you can replace A = lw with 432 = (x + 26)(x + 2). As another example, if you need to solve a right triangle, you may start by writing down the Pythagorean Theorem if you know two sides and are looking for the third. For another right triangle, perhaps you’re given an angle and the hypotenuse and need to find the opposite side; in this situation, you’d start off by writing down the sine ratio. Show Each Step of Your Work Yes, you’ve been hearing it forever, but your third-grade teacher was right: Showing each step of your work is vital in math. Writing each step on paper minimizes silly mistakes that you can make when you calculate in your head. It’s also a great way to keep a problem organized and clear. And it helps to have your work written down when you get interrupted by a phone call or text message — you can pick up where you left off and not have to start all over again. It may take some time to write every single step down, but it’s well worth your investment. Know When to Quit Sometimes a problem has no solution. Yes, that can be an answer, too! If you’ve tried all the tricks in your bag and you haven’t found a way, consider that the problem may have no solution at all. Some common problems that may not have a solution include the following: Absolute-value equations This happens when the absolute value expression is set equal to a negative number. You may not realize the number is negative, at first, if it’s represented by a variable. Equations with the variable under a square-root sign If your answer has to be a real number, and complex numbers aren’t an option, then the expression under the radical may represent a negative number. Not allowed. Quadratic equations When a quadratic isn’t factorable and you have to resort to the quadratic formula, you may run into a negative under the radical; you can’t use that expression if you’re allowed only real answers. Rational equations Rational expressions have numerators and denominators. If there’s a variable in the denominator that ends up creating a zero, then that value isn’t allowed. Trig equations Trig functions have restrictions. Sines and cosines have to lie between –1 and 1. Secants and cosecants have to be greater than or equal to 1 or less than or equal to –1. A perfectly nice-looking equation may create an impossible answer. On the other hand, you may get a solution for some problem that just doesn’t make sense. Watch out for the following situations: If you’re solving an equation for a measurement (like length or area) and you get a negative answer, either you made a mistake or no solution exists. Measurement problems include distance, and distance can’t be negative. If you’re solving an equation to find the number of things (like how many books are on a bookshelf) and you get a fraction or decimal answer, then that just doesn’t make any sense. How could you have 13.4 books on a shelf? Check Your Answers Even the best mathematicians make mistakes. When you hurry through calculations or work in a stressful situation, you may make mistakes more frequently. So, check your work. Usually, this process is very easy: You take your answer and plug it back into the equation or problem description to see if it really works. Making the check takes a little time, but it guarantees you got the question right, so why not do it? For example, if you go back and solve the garden problem from earlier in this chapter by looking at the equation to be solved: 432 = (x + 26)(x + 2), you multiply the two binomials and move the 432 to the other side. Solving x2 + 28x – 380 = 0 you get x = 10 and x = –38. You disregard the x = –38, of course, and find that the original width was 10 feet. So, what was the question? It asks for the length of the new, bigger garden. The length of the new garden is found with y = x + 26. So, the length (and answer) is that the length is 36 feet. Does this check? If you use the new length or 36 and the new width of x + 2 = 12 and multiply 36 times 12 you get 432 square feet. It checks! Practice Plenty of Problems You’re not born with the knowledge of how to ride a bike, play baseball, or even speak. The way you get better at challenging tasks is to practice, practice, practice. And the best way to practice math is to work the problems. You can seek harder or more complicated examples of questions that will stretch your brain and make you better at a concept the next time you see it. Along with working along on the example problems in this book, you can take advantage of the For Dummies workbooks, which include loads of practice exercises. Check out Trigonometry Workbook For Dummies, by Mary Jane Sterling, Algebra Workbook and Algebra II Workbook For Dummies, both also by Mary Jane Sterling, and Geometry Workbook For Dummies, by Mark Ryan (all published by Wiley), to name a few. Even a math textbook is great for practice. Why not try some (gulp!) problems that weren’t assigned, or maybe go back to an old section to review and make sure you’ve still got it? Typically, textbooks show the answers to the odd problems, so if you stick with those you can always double-check your answers. And if you get a craving for some extra practice, just search the Internet for “practice math problems” to see what you can find! For example, if you search the Internet for “practice systems of equations problems” you’ll find more than a million hits. That’s a lot of practice! Keep Track of the Order of Operations Don’t fall for the trap that always is lying there by performing operations in the wrong order. For instance, 2 – 6 × 3 doesn’t become –4 × 3 = –12. You’ll reach those incorrect answers if you forget to do the multiplication first. Focus on following the order of PEMDAS every time, all the time: Parentheses (and other grouping devices) Exponents and roots Multiplication and Division from left to right Addition and Subtraction from left to right Don’t ever go out of order, and that’s an order! Use Caution When Dealing with Fractions Working with denominators can be tricky. It’s okay to write: But, on the other hand: Also, reducing or cancelling in fractions can be performed incorrectly. Every term in the numerator has to be divided by the same factor — the one that divides the denominator. So, it’s true that: because But , because the factor being divided out is just 4, not 4x. And, again, it has to be the same factor that’s dividing, throughout. Can you spot the error here? Some poor soul reduced each term and the term directly above it — separately. Nope, doesn’t work that way. The correct process is to factor the trinomials and then divide by the common factor:

View ArticleArticle / Updated 05-03-2023

Once you have used the rational root theorem to list all of the possible rational roots of any polynomial, the next step is to test the roots. One way is to use long division of polynomials and hope that when you divide you get a remainder of 0. Once you have a list of possible rational roots, you then pick one and assume that it’s a root. For example, consider the equation f(x) = 2x4 – 9x3 – 21x2 + 88x + 48, which has the following possible rational roots: If x = c is a root, then x – c is a factor. So if you pick x = 2 as your guess for the root, x – 2 should be a factor. You can use long division to test if x – 2 is actually a factor and, therefore, x = 2 is a root. Dividing polynomials to get a specific answer isn’t something you do every day, but the idea of a function or expression that’s written as the quotient of two polynomials is important for pre-calculus. If you divide a polynomial by another and get a remainder of 0, the divisor is a factor, which in turn gives a root. In math lingo, the division algorithm states the following: If f(x) and d(x) are polynomials such that d(x) isn’t equal to 0, and the degree of d(x) isn’t larger than the degree of f(x), there are unique polynomials q(x) and r(x) such that In plain English, the dividend equals the divisor times the quotient plus the remainder. You can always check your results by remembering this information. Remember the mnemonic device Dirty Monkeys Smell Bad when doing long division to check your roots. Make sure all terms in the polynomial are listed in descending order and that every degree is represented. In other words, if x2 is missing, put in a placeholder of 0x2 and then do the division. (This step is just to make the division process easier.) To divide two polynomials, follow these steps: Divide. Divide the leading term of the dividend by the leading term of the divisor. Write this quotient directly above the term you just divided into. Multiply. Multiply the quotient term from Step 1 by the entire divisor. Write this polynomial under the dividend so that like terms are lined up. Subtract. Subtract the whole line you just wrote from the dividend. You can change all the signs and add if it makes you feel more comfortable. This way, you won’t forget signs. Bring down the next term. Do exactly what this says; bring down the next term in the dividend. Repeat Steps 1–4 over and over until the remainder polynomial has a degree that’s less than the dividend’s. The following list explains how to divide 2x4 – 9x3 – 21x2 + 88x + 48 by x – 2. Each step corresponds with the numbered step in the illustration in this figure. The process of long division of polynomials. (Note that using Descartes’s rule of signs, you find that this particular example may have positive roots, so it’s efficient to try a positive number here. If Descartes’s rule of signs had said that no positive roots existed, you wouldn’t test any positives!) Divide. What do you have to multiply x in the divisor by to make it become 2x4 in the dividend? The quotient, 2x3, goes above the 2x4 term. Multiply. Multiply this quotient by the divisor and write it under the dividend. Subtract. Subtract this line from the dividend: (2x4 – 9x3) – (2x4 – 4x3) = –5x3. If you’ve done the job right, the subtraction of the first terms always produces 0. Bring down. Bring down the other terms of the dividend. Divide. What do you have to multiply x by to make it –5x3? Put the answer, –5x2, above the –21x2. Multiply. Multiply the –5x2 times the x – 2 to get –5x3 + 10x2. Write it under the remainder with the degrees lined up. Subtract. You now have (–5x3 – 21x2) – (–5x3 + 10x2) = –31x2. Bring down. The +88x takes its place. Divide. What to multiply by to make x become –31x2? The quotient –31x goes above –21x2. Multiply. The value –31x times (x – 2) is –31x2 + 62x; write it under the remainder. Subtract. You now have (–31x2 + 88x) – (–31x2 + 62x), which is 26x. Bring down. The +48 comes down. Divide. The term 26x divided by x is 26. This answer goes on top. Multiply. The constant 26 multiplied by (x – 2) is 26x – 52. Subtract. You subtract (26x + 48) – (26x – 52) to get 100. Stop. The remainder 100 has a degree that’s less than the divisor of x – 2. Wow . . . now you know why they call it long division. You went through all that to find out that x – 2 isn’t a factor of the polynomial, which means that x = 2 isn’t a root. If you divide by c and the remainder is 0, then the linear expression (x – c) is a factor and that c is a root. A remainder other than 0 implies that (x – c) isn’t a factor and that c isn’t a root.

View ArticleArticle / Updated 10-06-2022

At some point, your pre-calculus teacher will ask you to find the general formula for the nth term of an arithmetic sequence without knowing the first term or the common difference. In this case, you will be given two terms (not necessarily consecutive), and you will use this information to find a1 and d. The steps are: Find the common difference d, write the specific formula for the given sequence, and then find the term you're looking for. For instance, to find the general formula of an arithmetic sequence where a4 = –23 and a22 = 40, follow these steps: Find the common difference. You have to be creative in finding the common difference for these types of problems. a.Use the formula an = a1 + (n – 1)d to set up two equations that use the given information. For the first equation, you know that when n = 4, an = –23: –23 = a1 + (4 – 1)d –23 = a1 + 3d For the second equation, you know that when n = 22, an = 40: 40 = a1 + (22 – 1)d 40 = a1 + 21d b.Set up a system of equations and solve for d. The system looks like this: You can use elimination or substitution to solve the system. Elimination works nicely because you can multiply either equation by –1 and add the two together to get 63 = 18d. Therefore, d = 3.5. Write the formula for the specific sequence. This step involves a little work. a.Plug d into one of the equations to solve for a1. You can plug 3.5 back into either equation: –23 = a1 + 3(3.5), or a1 = –33.5. b.Use a1 and d to find the general formula for an. This step becomes a simple three-step simplification: an = –33.5 + (n – 1)3.5 an = –33.5 + 3.5n – 3.5 an = 3.5n – 37 Find the term you were looking for. In this example, you weren't asked to find any specific term (always read the directions!), but if you were, you could plug that number in for n and then find the term you were looking for.

View ArticleArticle / Updated 09-22-2022

You can use the sum and difference formulas for cosine to calculate the cosine of the sums and differences of angles similarly to the way you can use the sum and difference formulas for sine, because the formulas look very similar to each other. When working with sines and cosines of sums and differences of angles, you're simply plugging in given values for the variables (angles). Just make sure you use the correct formula based on the information you're given in the question. Here are the sum and difference formulas for cosines: The sum and difference formulas for cosine (and sine) can do more than calculate a trig value for an angle not marked on the unit circle (at least for angles that are multiples of 15 degrees). They can also be used to find the cosine (and sine) of the sum or difference of two angles based on information given about the two angles. For such problems, you'll be given two angles (call them A and B), the sine or cosine of A and B, and the quadrant(s) in which the two angles are located. Use the following steps to find the exact value of cos(A + B), given that cos A = –3/5, with A in quadrant II of the coordinate plane, and sin B = –7/25, with B in quadrant III: Choose the appropriate formula and substitute the information you know to determine the missing information. then substitutions result in this equation: To proceed any further, you need to find cos B and sin A. Draw pictures representing right triangles in the quadrant(s). Drawing pictures helps you visualize the missing pieces of info. You need to draw one triangle for angle A in quadrant II and one for angle B in quadrant III. Using the definition of sine as opp/hyp and cosine as adj/hyp, this figure shows these triangles. Notice that the value of a leg is missing in each triangle. To find the missing values, use the Pythagorean theorem. The length of the missing leg in Figure a is 4, and the length of the missing leg in Figure b is –24. Determine the missing trig ratios to use in the sum or difference formula. You use the definition of cosine to find that cos B = –24/25 and the definition of sine to find that sin A = 4/5. Substitute the missing trig ratios into the sum or difference formula and simplify. You now have this equation: Follow the order of operations to get this answer: This equation simplifies to cos(A + B) = 4/5.

View ArticleArticle / 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:

View ArticleCheat 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.

View Cheat SheetArticle / Updated 07-09-2021

Polar coordinates are an extremely useful addition to your mathematics toolkit because they allow you to solve problems that would be extremely ugly if you were to rely on standard x- and y-coordinates. In order to fully grasp how to plot polar coordinates, you need to see what a polar coordinate plane looks like. A blank polar coordinate plane (not a dartboard) In the figure, you can see that the plane is no longer a grid of rectangular coordinates; instead, it's a series of concentric circles around a central point, called the pole. The plane appears this way because the polar coordinates are a given radius and a given angle in standard position from the pole. Each circle represents one radius unit, and each line represents the special angles from the unit circle. Because you write all points on the polar plane as to graph a point on the polar plane, you should find theta first and then locate r on that line. This approach allows you to narrow the location of a point to somewhere on one of the lines representing the angle. From there, you can simply count out from the pole the radial distance. If you go the other way and start with r, you may find yourself in a pickle when the problems get more complicated. Plotting polar coordinates example For example, to plot point E at which has a positive value for both the radius and the angle — you simply move from the pole counterclockwise until you reach the appropriate angle (theta). You start there in the following list: Locate the angle on the polar coordinate plane. Refer to the figure to find the angle: Determine where the radius intersects the angle. Because the radius is 2 (r = 2), you start at the pole and move out 2 spots in the direction of the angle. Plot the given point. At the intersection of the radius and the angle on the polar coordinate plane, plot a dot and call it a day! This figure shows point E on the plane. Visualizing simple and complex polar coordinates Polar coordinate pairs can have positive angles or negative angles for values of theta. In addition, they can have positive and negative radii. This concept might be new; in past classes, you might've always heard that a radius must be positive. When graphing polar coordinates, though, the radius can be negative, which means that you move in the opposite direction of the angle from the pole. Because polar coordinates are based on angles, unlike Cartesian coordinates, polar coordinates have many different ordered pairs. Because infinitely many values of theta have the same angle in standard position, an infinite number of coordinate pairs describe the same point. Also, a positive and a negative co-terminal angle can describe the same point for the same radius, and because the radius can be either positive or negative, you can express the point with polar coordinates in many ways.

View ArticleArticle / Updated 07-08-2021

In pre-calculus, you may need to find the equation of asymptotes to help you sketch the curves of a hyperbola. Because hyperbolas are formed by a curve where the difference of the distances between two points is constant, the curves behave differently than other conic sections. This figure compares the different conic sections. Cutting the right cone with a plane to get conic sections. Because distances can't be negative, the graph has asymptotes that the curve can't cross over. Creating a rectangle to graph a hyperbola with asymptotes. Hyperbolas are the only conic sections with asymptotes. Even though parabolas and hyperbolas look very similar, parabolas are formed by the distance from a point and the distance to a line being the same. Therefore, parabolas don't have asymptotes. Some pre-calculus problems ask you to find not only the graph of the hyperbola but also the equation of the lines that determine the asymptotes. When asked to find the equation of the asymptotes, your answer depends on whether the hyperbola is horizontal or vertical. If the hyperbola is horizontal, the asymptotes are given by the line with the equation If the hyperbola is vertical, the asymptotes have the equation The fractions b/a and a/b are the slopes of the lines. Now that you know the slope of your line and a point (which is the center of the hyperbola), you can always write the equations without having to memorize the two asymptote formulas. You can find the slope of the asymptote in this example, by following these steps: Find the slope of the asymptotes. The hyperbola is vertical so the slope of the asymptotes is Use the slope from Step 1 and the center of the hyperbola as the point to find the point-slope form of the equation. Remember that the equation of a line with slope m through point (x1, y1) is y – y1 = m(x – x1). Therefore, if the slope is and the point is (–1, 3), then the equation of the line is Solve for y to find the equation in slope-intercept form. You have to do each asymptote separately here. Distribute 4/3 on the right to get and then add 3 to both sides to get Distribute –4/3 to the right side to get Then add 3 to both sides to get

View ArticleArticle / Updated 07-08-2021

Gaussian elimination is probably the best method for solving systems of equations if you don't have a graphing calculator or computer program to help you. The goals of Gaussian elimination are to make the upper-left corner element a 1, use elementary row operations to get 0s in all positions underneath that first 1, get 1s for leading coefficients in every row diagonally from the upper-left to the lower-right corner, and get 0s beneath all leading coefficients. Basically, you eliminate all variables in the last row except for one, all variables except for two in the equation above that one, and so on and so forth to the top equation, which has all the variables. Then you can use back substitution to solve for one variable at a time by plugging the values you know into the equations from the bottom up. You accomplish this elimination by eliminating the x (or whatever variable comes first) in all equations except for the first one. Then eliminate the second variable in all equations except for the first two. This process continues, eliminating one more variable per line, until only one variable is left in the last line. Then solve for that variable. You can perform three operations on matrices in order to eliminate variables in a system of linear equations: You can multiply any row by a constant (other than zero). multiplies row three by –2 to give you a new row three. You can switch any two rows. swaps rows one and two. You can add two rows together. adds rows one and two and writes it in row two. You can even perform more than one operation. You can multiply a row by a constant and then add it to another row to change that row. For example, you can multiply row one by 3 and then add that to row two to create a new row two: Consider the following augmented matrix: Now take a look at the goals of Gaussian elimination in order to complete the following steps to solve this matrix: Complete the first goal: to get 1 in the upper-left corner. You already have it! Complete the second goal: to get 0s underneath the 1 in the first column. You need to use the combo of two matrix operations together here. Here's what you should ask: "What do I need to add to row two to make a 2 become a 0?" The answer is –2. This step can be achieved by multiplying the first row by –2 and adding the resulting row to the second row. In other words, you perform the operation which produces this new row: (–2 –4 –6 : 14) + (2 –3 –5 : 9) = (0 –7 –11: 23) You now have this matrix: In the third row, get a 0 under the 1. To do this step, you need the operation With this calculation, you should now have the following matrix: Get a 1 in the second row, second column. To do this step, you need to multiply by a constant; in other words, multiply row two by the appropriate reciprocal: This calculation produces a new second row: Get a 0 under the 1 you created in row two. Back to the good old combo operation for the third row: Here's yet another version of the matrix: Get another 1, this time in the third row, third column. Multiply the third row by the reciprocal of the coefficient to get a 1: You've completed the main diagonal after doing the math: You now have a matrix in row echelon form, which gives you the solutions when you use back substitution (the last row implies that 0x + 0y + 1z = 4, or z = –4). However, if you want to know how to get this matrix into reduced row echelon form to find the solutions, follow these steps: Get a 0 in row two, column three. Multiplying row three by the constant –11/7 and then adding rows two and three gives you the following matrix: Get a 0 in row one, column three. The operation gives you the following matrix: Get a 0 in row one, column two. Finally, the operation gives you this matrix: This matrix, in reduced row echelon form, is actually the solution to the system: x = –1, y = 3, and z = –4.

View ArticleArticle / Updated 07-08-2021

If you’re asked to graph the inverse of a function, you can do so by remembering one fact: a function and its inverse are reflected over the line y = x. This line passes through the origin and has a slope of 1. When you’re asked to draw a function and its inverse, you may choose to draw this line in as a dotted line; this way, it acts like a big mirror, and you can literally see the points of the function reflecting over the line to become the inverse function points. Reflecting over that line switches the x and the y and gives you a graphical way to find the inverse without plotting tons of points. The best way to understand this concept is to see it in action. For instance, say that you know these two functions are inverses of each other: To see how x and y switch places, follow these steps: Take a number (any that you want) and plug it into the first given function. Say you pick –4. When you evaluate f(–4), you get –11. As a point, this is written (–4, –11). Take the value from Step 1 and plug it into the other function. In this case, you need to find g(–11). When you do, you get –4 back again. As a point, this is (–11, –4). Whoa! This works with any number and with any function and its inverse: The point (a, b) in the function becomes the point (b, a) in its inverse. But don’t let that terminology fool you. Because they’re still points, you graph them the same way you’ve always been graphing points. The entire domain and range swap places from a function to its inverse. For instance, knowing that just a few points from the given function f(x) = 2x – 3 include (–4, –11), (–2, –7), and (0, –3), you automatically know that the points on the inverse g(x) will be (–11, –4), (–7, –2), and (–3, 0). So if you’re asked to graph a function and its inverse, all you have to do is graph the function and then switch all x and y values in each point to graph the inverse. Just look at all those values switching places from the f(x) function to its inverse g(x) (and back again), reflected over the line y = x. You can now graph the function f(x) = 3x – 2 and its inverse without even knowing what its inverse is. Because the given function is a linear function, you can graph it by using the slope-intercept form. First, graph y = x. The slope-intercept form gives you the y-intercept at (0, –2). Since the slope is 3=3/1, you move up 3 units and over 1 unit to arrive at the point (1, 1). If you move again up 3 units and over 1 unit, you get the point (2, 4). The inverse function, therefore, moves through (–2, 0), (1, 1), and (4, 2). Both the function and its inverse are shown here.

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