Algebra I Workbook For Dummies
Page 22
Here are the standard forms for parabolas and circles:
Parabola:
Circle:
Notice the h and k in both forms. For the parabola, the coordinates (h, k) tell you where the vertex (bottom or top of U-shape) is. For the circle, (h, k) gives you the coordinates of the center of the circle. The r part of the circle’s form gives you the radius of the circle.
Q. Find the vertex of the parabola and sketch its graph.
A. Vertex: (–3, 7); opens downward.
To find the vertex, you need the opposite of 3. The general form has (x – h), which is the same as to deal with the positive sums in the parentheses. Because a is negative, the parabola opens downward; and a being –3 introduces a steepness, much like the slope of a line. You can plot a few extra points to help with the shape.
Q. Find the center and radius of the circle and sketch its graph.
A. Center: (4, –2); radius = 6.
27. Find the vertex and sketch the graph of .
Solve It
28. Find the vertex and sketch the graph of .
Solve It
29. Find the center and radius and sketch the graph of .
Solve It
30. Find the center and radius and sketch the graph of .
Solve It
Graphing with Transformations
You can graph the curves and lines associated with different equations in many different ways. Intercepts are helpful when graphing lines or parabolas; the vertex and center of a circle are critical to the graphing. But many curves can be quickly sketched when they’re just a slight variation on the basic form of the particular graph. Two transformations are translations (slides) and reflections (flips). Neither of these transformations changes the basic shape of the graph; they just change its position or orientation. Using these transformations can save you a lot of time when graphing figures.
When a curve is changed by a translation or slide, it slides to the left, to the right, or up or down. For instance, you can take the basic parabola with the equation y=x2 and slide it around, using the following rules. The C represents some positive number:
y = x2 + C raises the parabola by C units.
y = x2 – C lowers the parabola by C units.
y = (x + C)2 slides the parabola left by C units.
y = (x – C)2 slides the parabola right by C units.
When a curve is changed by a reflection or flip, you have a symmetry vertically or horizontally:
y = –x2 flips the parabola over a horizontal line.
y = (–x)2: flips the parabola over a vertical line.
You can change the steepness of a graph by multiplying it by a number. If you multiply the basic function or operation by a positive number greater than 1, then the graph becomes steeper. If you multiply it by a positive number smaller than 1, the graph becomes flatter. Multiplying a basic function by a negative number results in a flip or reflection over a horizontal line. Use these rules when using the parabola:
y = kx2: Parabola becomes steeper when k is positive and greater than 1.
y = kx2: Parabola becomes flatter when k is positive and smaller than 1.
Q. Use the basic graph of y=x2 to graph y = –3x2 + 1.
A.
The multiplier –3 flips the parabola over the x-axis and makes it steeper. The +1 raises the vertex up one unit.
Q. Use the basic graph of y=x2 to graph .
A.
The –3 lowers the parabola by three units. The +1 inside the parentheses slides the parabola one unit to the left. Compare this graphing technique with the method of graphing using the vertex, explained in the earlier section “Graphing Parabolas and Circles.”
31. Sketch the graph of y = (x – 4)2.
Solve It
32. Sketch the graph of y = –x2 – 3.
Solve It
Answers to Problems on Graphing
The following are the answers (in bold) to the practice problems presented in this chapter.
1. Graph the points (1, 2), (–3, 4), (2, –3), and (–4, –1).
2. Graph the points (0, 3), (–2, 0), (5, 0), and (0, –4).
3. In which quadrant do the points (–3, 2) and (–4, 11) lie? Quadrant II, the upper left quadrant.
4. In which quadrant do the points (–4, –1) and (–2, –2) lie? Quadrant III, the lower left quadrant.
5. Find three points that lie on the line , plot them, and draw the line through them.
6. Find three points that lie on the line , plot them, and draw the line through them.
7. Find three points that lie on the line , plot them, and draw the line through them.
8. Find three points that lie on the line , plot them, and draw the line through them.
9. Use the intercepts to graph the line .
Let to get . The intercept is (4, 0).
Let to get . The intercept is (0, 3).
10. Use the intercepts to graph the line .
Let to get . The intercept is (4, 0).
Let to get . The intercept is (0, –2).
11. Find the slope of the line through the points (3, 2) and (–4, –5) and graph the line.
Using the slope formula, .
12. Find the slope of the line through the points (–1, 7) and (1, 3) and graph the line.
Using the slope formula, .
13. Find the slope of the line through (3, –4) and (5, –4). Slope = 0.
This fraction is equal to 0, meaning that the line through the points is a horizontal line with the equation .
14. Find the slope of the line through (2, 3) and (2, –8). No slope.
This fraction has no value; it’s undefined. The line through these two points has no slope. It’s a vertical line with the equation .
15. Graph the line , using the y-intercept and slope.
The slope is and the y-intercept is 4. Place a point at (0, 4). Then count three units to the right of that intercept and two units down. Place a point there and draw a line through the intercept and the new point.
16. Graph the line , using the y-intercept and slope.
The y-intercept is –2, and the slope is 5. Place a point at (0, –2). Then count one unit to the right and five units up from that intercept. Place a point there and draw a line.
17. Change the equation to the slope-intercept form. The answer is . Subtract 8x from each side: . Then divide every term by 2: . The slope is –4, and they-intercept is .
18. Change the equation to the slope-intercept form. The answer is . Add y to each side to get . Then use the symmetric property to turn the equation around: . (This is easier than subtracting 4x from each side, adding 3 to each side, and then dividing by –1.)
19. Find the equation of the line with a slope of that goes through the point (0, 7). The answer is .
You’re given the slope and the y-intercept, so you can just put in the 7 for the value of b.
20. Find the equation of the line that goes through the points (–3, –1) and (–2, 5). The answer is y = 6x + 17.
First find the slope by using the two points: . Using the point-slope form and the coordinates of the point (–3, –1), you get y – (–1) = 6(x – (–3)) or y + 1 = 6(x + 3). Distributing and subtracting 1 from each side gives you y = 6x + 17.
21. What is the slope of a line that’s parallel to the line ? The slope is . You can find the slope of a line that’s parallel to the line by changing the equation to the slope-intercept form and the subtracting 2x from each side and dividing each term by –3. The equation is then . The slope is , and the y-intercept is .
22. What is the slope of a line that’s perpendicular to the line ? The slope is . Change this equation to the slope-intercept form to find the slope of the line: , . The negative reciprocal of –2 is , so that’s the slope of a line perpendicular to this one.
23. Find the distance between (3, –9) and (–9, 7). 20 units. Using the distance formula, .
24. Find the distance between (4, 1) and (–2, 2). Round the decimal equivale
nt of the answer to two decimal places. The answer is 6.08 units.
. Use a calculator and round to two decimal places.
25. Find the intersection of the lines y = –4x + 7 and y = 5x – 2. The answer is (1, 3). Set –4x + 7 = 5x – 2. Solving for x, you get –9x = –9 or x = 1. Substituting back into either equation, y = 3.
26. Find the intersection of the lines 3x – y = 1 and x + 2y + 9 = 0. The answer is (–1, –4). First write the equations in slope-intercept form. 3x – y = 1 becomes y = 3x – 1 and x + 2y + 9 = 0 becomes . Setting , you solve for x and get or x = –1. Substitute back into either equation to get y = –4.
27. Find the vertex and sketch the graph of . Vertex: (3, –2)
The multiplier of in front of the parentheses makes the graph open wider (more flat).
28. Find the vertex and sketch the graph of . Vertex: (–2, 0)
The vertex is on the x-axis. The multiplier of 2 makes the parabola steeper.
29. Find the center and radius and sketch the graph of (x + 3)2 + (y – 4)2 = 25. Center: (–3, 4); radius: 5.
30. Find the center and radius and sketch the graph of . Center: (5, 2); radius: 3.
31. Sketch the graph of .
The parabola has been translated 4 units to the right.
32. Sketch the graph of .
The parabola has been lowered by 3 units and flipped over to face downward.
Part V
The Part of Tens
In this part . . .
Not everything comes in tens. You find five Great Lakes, seven wonders of the world, eight tiny reindeer, and three men in a tub. Luckily, you’ve now found two lists of ten.
This last part has two chapters that provide quick lists of ten items in For Dummies fashion. Specifically these two chapters give you ten common pitfalls or errors to avoid as well as ten quick tips to make your algebra problem solving a little easier.
Chapter 22
Ten Common Errors That Get Noticed
Most of this book involves methods for correctly doing algebra procedures. When working in algebra, you want to take the positive approach. But sometimes it helps to point out the common errors or pitfalls that occur. Many of these slip-ups involve the negative sign, fractions, and/or exponents. Mistakes happen in algebra because people often think what they’re doing works — or that two wrongs make it right! This chapter gives you a chance to spot some of those errors and sets you straight if you’re making them yourself.
Squaring a Negative or Negative of a Square
The two expressions –x2 and (–x)2 are not the same (except in that single case where x = 0). For instance, let x = –3 and indicate which two of the following are correct.
A) If x = –3, then . B) If x = –3, then .
C) If x = –3, then . D) If x = –3, then .
Are your choices B and C? Then good for you! You remember the rules for order of operations (found in Chapter 6) where you perform all powers first, then multiplication or division, and then addition or subtraction. The parentheses always interrupt the order and tell you to perform whatever is inside first. So in the expression , if you let x = –3, you square the –3 first to get 9 and then find the opposite, making the final answer –9. In the expression , you would find the opposite of –3 first, giving you +3 in the parentheses, and then you’d square the 3 to get 9.
Squaring a Binomial
A binomial is two terms separated by addition or subtraction. Squaring a binomial involves more than just squaring each term; the terms also have to interact with one another.
Can you spot which expansion is correct below?
A) B)
Of course, you chose A as the correct expansion. The most common error when doing these expansions is squaring the first and last term of the binomial and forgetting about the middle term.
The pattern for squaring a binomial is . When you square a binomial, remember that you’re multiplying it times itself: (a + b)2 = (a + b)(a + b). You find the products of the first terms, the outer terms, the inner terms, and the last terms, and then add them up. The right way to expand is (x + 5)2 = x2 + 10x + 25 or (3y – 2z)2 = 9y2 – 12yz + 4z2.
You can find more on squaring binomials in Chapter 7.
Operating on Radicals
The square root of a product is the product of the roots, and the square root of a quotient is the quotient of the roots, but this “reversal” doesn’t apply when you have a sum or difference.
It’s very helpful when you can simplify radicals — using the correct rules, of course. Which two of the following are correct?
A) B)
C) D)
Had you going there, didn’t I? But you spotted the fact that only A and B can be true. The rules that apply are and . Finishing the simplifications of the four expressions:
A) B)
C) D)
Refer to Chapter 5 for more on working with radicals.
Distributing a Negative Throughout
Distributing a number by multiplying it times every term within parentheses is a fairly straightforward process. However, some people can’t seem to finish the job. Can you spot the good algebra?
A) –1(4 – 3x + 2y) = –4 –3x + 2y B) –1(4 – 3x + 2y) = –4 +3x – 2y
Of course, you chose B! The distributive property of multiplication over addition is a(b + c) = ab + ac. The pitfall or problem comes in when distributing a negative (the same as multiplying by –1) over terms. Often, people distribute –1 over the first term and not the rest. Remember to distribute the negative over every term. Basically, all the signs change. Chapter 2 has even more on the distribution rule.
Fracturing Fractions
A fraction line is like a grouping symbol. Everything in the denominator (bottom) divides into everything in the numerator (top). You need to remember that . But this setup is too obvious. Why do people mess up with the fractions? Let me show you. Which of the following is correct?
A) B)
Yes, I know that B looks much nicer, but it’s wrong, wrong, wrong. The error in B is that the person broke up the fraction, putting the 4 under the –8 only.
Chapter 3 covers fractions completely.
Raising a Power to a Power
The rule for raising a power to a power is that you multiply the exponents. Sounds simple enough, but can you spot which of the following are correct?
A) B)
C) D)
You were looking for something wrong, weren’t you? This time they’re all correct. Applying the rule sounds easy, but putting variables or negative numbers in for the exponents can make things more complicated. In the end, you just multiply the exponents together.
Refer to Chapter 4, if you want to see more on exponents.
Making Negative Exponents Flip
The general rule for negative exponents is . Negative exponents are very handy when combining terms with the same base. A common pitfall is to forget about number multipliers when applying this rule. Which of the following is correct?