i Information Page Class Website: http://padlet.com/william_kaden/algebra2stem Has PDFs of assignments. Additionally they can sometimes be found on Skyward. Desmos Graphing Calculator http://www.desmos.com/ IXL http://www.ixl.com/ Username: jkrejcha1@mountain Password: gothunder Justin s Algebra II STEM Webpage http://www.justinkrejcha.com/algebra-2-stem/ This webpage catalogs a digital copy of the notebook, including many of the assignments as well (both homework and not). Kaden s Goats Group Chat http://www.messenger.com/t/1567434643569968 Notes and other pages are posted on there sometimes, courtesy of yours truly.
iii Table of Contents Page Title 1 Modeling with Mathematics (Katy Pary) 2 Semester 1 Final Rational Functions 3 Fraction Rules and Parent Function 4 Laws of Exponents 5 Multiplying Fractions and Rational Expressions 6 Excluded Values 7 Dividing Rational Functions 8 Rational Operations Quiz 9 Graphing Rationals Notes 10 Graphing Rational Functions 11 Horizontal and Vertical Asymptotes (Part 1) 12 Horizontal and Vertical Asymptotes (Part 2) 13 Holes of Rational Functions 14 Oblique (Slant) Asymptote 15 X and Y Intercepts of Graphs 16 Graphing Complex Rational Functions 17 Steps for Graphing Rational Functions 18 Graphing Rational Functions Practice 19 Solving Rational Functions 20 Graphing Rational Functions Quiz 21 Rational Functions Quiz 22 Context and Rational Functions 23 Review/Test Rational Functions Radical Functions 24 Parent Function / Solving Radical Functions 25 Rational Exponents Worksheet 26 Solving / Simplifying Radical Functions 27 Graphing and Transformations of Radical Functions 28 Transformations of Radical Functions 29 Quiz / Simplifying Radicals 30 Words / Graphs Transformations 31 Review Sheets 32 Radicals Test Exponential Functions and Logarithmic Functions 33 Exploration of Exponential Functions 34 Exploring Exponential Functions (page 1 and 2) 35 Exploring Exponential Functions (page 3 and 4) 36 Transformations of Exponential Functions
iv Table of Contents Page Title 37 Growth Factors and Rates (S.10 and S.11 Notes) 38 Inverse Functions 39 Exploring Logarithmic Functions 40 Logarithmic Functions 41 Logarithmic Functions, Practice (Again!) 42 Common Logarithms vs. Natural Logarithms 43 Properties of Logarithms 44 Solving Logarithms 45 Compounding Interest 46 Exponential and Logarithmic Test Regression 47 Guessing Ages 48 Cases of Flu / Correlation Coefficient 49 Modeling with Technology Geometry and Circles 50 Right Triangles / Trig Functions 51 Final Review (Rationals) 52 Circles 53 Final Review (Radicals) 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73
v Page 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 Title Table of Contents
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2, I Semester 1 Final
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3, I Parent Function: f(x) = 1 x Rational Functions Fraction Rules: Adding and Subtracting: Make a common denominator (or LCD) and simplify. Multiplying Multiply top and bottom straight across. 3 4 5 6 15 24 = 5 8 Dividing Flip the denominator and multiply. 3 4 3 4 5 6 5 3 4 6 5 = 18 20 = 9 10 6
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4, I Zero Rule Laws of Exponents x 0 = 1 $1,000,000,000,000 0 = 1 Product Rule x 5 + x 2 = x 5+2 = x 7 You can t combine x 5 y 2 because they have a different base. Division Rule x 5 Negative Exponent Rule x 2 x 2 = x5 2 = x 3 x 5 = x2 5 = x 3 = 1 x 3 Additionally, the reverse becomes true. 1 x 4 = 1 1 x 4 = 1 1 x4 1 = x4
4, O Assignment. Simplifying exponents assignment, there are 8 problems.
5, I Multiplying Fractions: Rational Functions The rule for multiplying fractions is to simply multiply straight across. 3 5 3 4 = 9 20
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6, I Excluded Values in Rational Functions Rational functions are fractions. Because of this, the denominator cannot be zero. Consider the following equation: x = 1 0 There is no way to get an answer for x. This affects the domain of the function. For example: x + 7 x + 7 The equation simplifies to one, but we cannot use a value of -7 in our equation because this puts a zero in the denominator. This example is a hole.
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7, I Diving Rational Functions The rule to divide rational functions is to take the reciprocal of the denominator. For example: x + 7 1 x + 7 x 2 x + 7 x + 7 x 2 x + 7 1 x 2 x + 7 = x 2 1 The domain for this is x 7. = x 2
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8, I Rational Expressions Quiz
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10, I Rational Functions Extension
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11, I Finding Asymptotes Vertical Asymptotes Simplify. Factor completely and cancel common factors. Solve the equation for the denominator. For example, for the rational function: (x + 3)(x 3)(x + 2) (x + 5)(x 3) The vertical asymptote would be at -5. Horizontal Asymptotes Degree(numerator) Degree(denominator) Asymptote Example Example Asymptote Less than 0 y = 0 f(x) = 1 x 2 y = 0 + 1 Equal to 0 y = ratio of leading f(x) = 2x2 + 7 y = 2 3 coefficients Equal to 1 Divide the numerator by the denominator. 3x 2 + x + 12 f(x) = x2 + x + 1 y = x + 1 x More than 1 None f(x) = 2x4 None 3x 2 + 1
11, O Integrated Algebra 3 Rational Functions Asymptotes Exploration.
12, I Horizontal Asymptote page.
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13, I Holes in Rational Functions Holes occur whenever the numerator and denominator have common factors. The hole is at x 3. (x + 3)(x 3) (x + 3) (x 3) 1 Another example perhaps? (x + 4)(x 5)(x + 6) (x + 7)(x + 4)(x + 6) x 5 x + 7 The vertical asymptote is at x 7, the horizontal asymptote at y 1 and there are two holes. They are at x 4, 6.
13, O Holes in Rational Functions
14, I Oblique Asymptotes Rule 3 If the degree of the numerator is greater than that of the denominator, there is no horizontal asymptote. However, if the numerator degree is greater by one, there exists an oblique asymptote. The expression 4x4 x2 has no horizontal asymptote, however, the expression 4x3 x2 has a oblique asymptote. Example f(x) = 3x3 4x 2 + 1 x 2 2 The vertical asymptote is the solution to x 2 2, or ± 2. You can get the oblique asymptote through long division. The oblique asymptote is 3x 4.
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15, I Finding Intercepts The y-intercept is (x, 0), or when y is equal to zero. To do that, solve the numerator substituting for 0. (You don t need to worry about the denominator, because 0 divided by anything is 0. Be sure to check for asymptotes or holes however.) The x-intercept is (0, y), or when x is equal to zero. To solve this, set the numerator equal to zero and solve.
15, O X-intercepts of Rational Functions
16, I Practice with Graphing Rational Functions
16, O Rational Functions matching (not the match game). This one has A-K on the graphs.
17, I Steps to Graphing Rational Functions 1) Simplify Factor Holes 2) Vertical Asymptotes (solve for denominator) 3) Horizontal/Oblique Asymptotes (see page 11) 4) x-intercept (set numerator equal to 0) 5) Y-intercept (set all x s to 0 and solve) 6) Input and output to find other points.
17, O Rational Asymptotes Worksheet
18, I Graphing Rational Functions
18, O Algebra 2 STEM Worksheet Rational Functions
19, I By the way, there was a volleyball game at 6:30 April 21 st in the main gym.
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21, I Do Now! (Rational functions quiz on output side.) 1 x 2 + 3 x + 3 = 4 x 2 + x 6 1(x 2)(x + 3) 3(x 2)(x + 3) 4(x 2)(x + 3) + = x 2 x + 3 x 2 + x 6 x + 3 + 3(x 2) = 4 x + 3 + 3x 6 = 4 4x 3 = 4 4x 4 = 7 4 x = 7 4
21, O Rational Functions Quiz (titled Unit 5 Quiz)
22, I STEMersion Anesthesiologist
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23, I Rational Functions Test
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24, I Graphing Radical Functions Exploration
24, O Solving Radical Equations
25, I Rational Exponents
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26, I Simplify. Solving/Simplifying Radicals 3 64x 2 y 5 z 3 Our first step is to find things we can eliminate. To do so, (for this problem) we need to find er pairs of three. 64 does not have any pairs that fit the criteria, so we can create this branch to simplify. We re left with 6 2s (2 + 2 + 2 = 6) or 2 6 z. We can now apply an exponent rule x y = x y z. For the purposes of our problem that leaves us with 2 6 3 or 2 2 or 4. We can now put the 4 on the outside, giving us a new problem of: 3 4 x 2 y 5 z 3 We can get rid of one more thing: the z 3. This easily is removable into pairs of three. We ll put it on the outside as well, giving us a final answer of: 3 4z x 2 y 5 Solve. k 9 k = 1 First we need to isolate giving us a new equation of k 9 = k 1. Next we need to square both sides, resulting in a new equation of k 9 = ( k 1) 2. Foiling this right side gives you a new right side equation of: k 9 = k 2 k + 1 Only now can we manipulate the equation. Firstly, remove the k, and then subtract the one from the other side, giving an isolated equation. Rearranging the terms (for simplicity) gives: 2 k = 10 Divide the -2 out ( k = 5). Square both sides to get your answer. k = 25
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27, I Radicals Transformations Worksheet
27, O Math Lab: Investigating Radical Functions
28, I Transformations of Radical Functions Quadratic y = x 2 y = a(x h) 2 + k Square Root y = x y = a x h + k A change in a is a dilation (if a > 1 it is steeper, if 0 < a < 1 it is shallower). a = reflection A change in h is a horizontal shift left or right (x h = 0) and a change in k is a vertical shift up or down. These are both translations.
28, O Graphing using Transformations
29, I Simplifying Radicals (Half-sheet)
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30, O Writing Equations Using [sic] Transformations
31, I Can I Do This? (Radicals Edition)
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32, I Radicals Unit Test
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33, I Exponential Exploration
33, O Exponential Exploration
34, I Transformations of Exponential Functions Parent Function y = 2 x Answer the question: Where do you see transformations? y = a(b) x This isn t even my final form! If negative, it exponentially decays, instead of grows unless b is also negative. Moves the y-intercept. If negative, it reflects over the x-axis. This means you can have up to two reflections over the x- axis. When more than 0, but less than 1, a higher value means a shallower exponential decay. When more than 1, a higher value means a steeper exponential growth. When less than 0 but more than -1, a higher value means a shallower exponential decay with a reflection over the x-axis.
34, O Exploring Exponential Equations
35, I Exponential Explorations
35, O Exponential Explorations
36, I Transformations of Exponential Functions Parent Function y = 2 x Answer the question: Where do you see transformations? y = a(b) x This isn t even my final form! If negative, it exponentially decays, instead of grows unless b is also negative. Moves the y-intercept. If negative, it reflects over the x-axis. This means you can have up to two reflections over the x- axis. When more than 0, but less than 1, a higher value means a shallower exponential decay. When more than 1, a higher value means a steeper exponential growth. When less than 0 but more than -1, a higher value means a shallower exponential decay with a reflection over the x-axis.
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37, I Exponential Functions and S.10/S.11 Notes Growth Factors and Growth Rates We know a linear function is linear because its growth rate is constant, or to put it another way y = 2x has a growth rate of 2. In exponential, we multiply by our rate of change, instead of adding like in linear. For example: EXPONENTIAL LINEAR y = 3 x y = 3x y = 0. 5 x y = 0. 5x We multiply by 3 each term. We multiply by 0.5 each term. We add by 3 each term. We add by 0.5 each term. What are the growth factors and growth rates? y = 5(4) x y = 100(0. 3) x This function is a growth function (it has a b value of more than 1). Growth Factor: 4 (400%) Growth Rate: 3 (300%) Because we start with 1 (100%), we have to subtract that from 4 (400%). 4 1 = 3 (400% 100% = 300%) The function is a decay function (it has a b value of less than 1 (but more than 0)). Growth Factor: 0.3 (30%) Growth Rate: 0.7 (70%) Because this is a decay function, we subtract our growth factor from 1, rather than the other way around. Exponential is about what you have left, not what they lose. 1 0.3 = 0.7 (100% 30% = 70%) Side note: we call it the growth factor and growth rate even in decay functions. Why? there isn t a reason.
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38, I Inverse Functions Here s a function, and it s inverse. The inverse of a function is just swapping the inputs and outputs. See these two tables below. X Y 0 70 1 80 2 90 3 100 X Y 70 1 80 2 90 3 100 4 It can also be expressed as an equation. Here s a different function: y = x x = y Write the inverse of this function: 7x + 18 y = 2 The first thing to do is flip the variables. 7y + 18 x = 2 To re-write in y = mx + b form, solve for y (side note: also called algebraic stuff by Mr. Kaden). 2x = 7y + 18 2x 18 = 7y 2x 18 y = 2 Write the inverse of this function: (Remember the inverse of the functions of x 2 is x, x 3 3 has an inverse of x, and so on and so forth. The inverse of exponential function y = b x is x = log b y. See the exploration for more details.) y = 2x 3 + 3 x = 2y 3 + 3 x 3 = 2y 3 x 3 = y 3 2 3 y = x 3 2
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39, I Introduction to Logarithms Logarithms are really exponents in disguise. The following examples will explain this idea: x y = log x 1 0 10 1 100 2 1000 3 10,000 4 100,000 5 If you input each into your calculator, substituting x, you ll get each of these values. This is a good time to point out that like how y = x hides values so that the equation is actually y = 1x + 0, log x also hides something: in this case a 10. Our expanded function looks like: y = log 10 x Let s continue the example to see where we end up. x log x log 10 x 10 y = x 1 0 0 1 10 1 1 10 100 2 2 100 1000 3 3 1000 The rule is log b x = y if and only if, b y = x Calculating Logarithms of Different Bases The calculator has a log key, but that calculates logarithms in base 10. If you want to calculate logartihms in different bases, use this in your calculator (the example here is calculating a base-2 logarithm, but it works for any base): log 2 32 = log 10 32 log 10 2 Or you could just go into Math and click on LogBASE(, but you can t do that on the SBAC.
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40, I Logarithmic Functions b y = x Inverse b x = y Inverse log b x = y log b y = x Rewrite the equation. log 11 121 = x 11 2 = 121 4 3 = 64 log 4 64 = 3 Evaluate. log 4 64 = 3 (4 3 = 64) log 2 16 = 4 (2 4 = 16) log 343 log 7 343 = log 7 = 3 (73 = 343) IXL assignments are R.1 and R.3.
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42, I Common Logs vs. Natural Logs The common logarithm is a logarithm that has a base of 10. log 64 1.806 This is where that hiding comes in, the log 64 hides a 10 in the base. Expanded, it d look like this: log 10 64 1.806 Remember, we were solving for the exponent. If written as an equation where we solve for x, it d look like this in the exponential form: 10 x = 64 Now, we are introduced to the uh number e. e 2.71 A natural logarithm is a logarithm that have a base of e. It is written like this: ln 7 1.9959 Written in exponential form, it looks like: And yeah, that s it. e 1.9959 = 7
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43, I Properties of Logarithms Like exponents, logarithms have properties as well. Illustrated below are some of the properties. P of L: To illustrate the properties, some variables are defined here. Variable Defined Constraints m and n Real numbers greater than 0. b Real numbers greater than 1. p Any real number. If all of this is true, than the following properties hold: Property Condensed Form Expanded Form Product Property log b mn log b m + log b n Quotient Property m log log b m log b n b n Power Rule log b m p plog b m Property of Equality log b m = log b n m = n Remember, the bases have to be the same number.
43, O Solve the logarithm. Answer log 11 (x 6) + 7 = 9 log 11 (x 6) = 2 11 2 = (x 6) 121 = x 6 x = 127 Solve the logarithm. log 7 4r = log 7 2r 4r = 2r 0 = 6r r = 0 Is r > 0? Because it is not, there is no solution.
44, I Solving Logarithms First things first Remember that you can t take a logarithm of zero or a negative number. For example, take this equation: log 7 4r = log 7 2r 4r = 2r 6r = 0 r = 0 Because when we substitute in 0 for r, we get 0, we cannot use this solution. There is no solution. Solve. Remember that the natural logarithm is just a logarithm to the base of e. ln(x + 5) 3 = 1 Our first step is to put this in solitary confinement (isolate). ln(x + 5) = 2 We know the definition of a logarithm is that the answer is the exponent and the base is the base. We can rewrite the equation to look like this: e 2 = x + 5 Evaluating e 2 ends up with an approximate answer of 7.389 7.389 x + 5 We can move the 5 over giving us a final answer of: x 2.389 Yay, we re done! Side note: If you want to use the precise answer (which you can do), you can skip the evaluation step. The process looks like this instead. ln(x + 5) 3 = 1 ln(x + 5) = 2 e 2 = x + 5 x = e 2 5
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45, I Compounding Interest The Interest Compounding Equation A = P (1 + r n ) nt What are the variables? The variable A is our answer. The variable P is the principal value (the starting value). The variable r is the interest rate. The variable P is the amount of times we are compounding. For example, if it s yearly, it s 1, if it s monthly it s 12, weekly 52, daily 365, etc. The variable t is the time for which we are looking at interest. The Continuous Interest Compounding Equation A = Pe rt In this equation, e is the constant that is about 2.787, not a variable. Example Starting at a principal value of $341,000 (our starter home) at a market rate of 4.5%, what is the value of the home after 10 years? Starter home: $341,000 Time: 10 years Average market rate is 4.5% Using continuous interest A = $341,000 (1 + 0.045 1 ) (1)(10) A = $341,000(1.045) 10 A = $529,600 A = $341000e (0.045)(10) A $341000(2.718) (0.045)(10) A $534,800
45, O Given a home that is $1,500,000, at an interest rate of 10% over 18 years, what is the value of the home? Compounded Quarterly Compound Continuously A = P (1 + r n ) nt A = $1,500,000 (1 + 0.10 4 ) (4)(18) A $8,875,842.09 A = Pe rt A = $1,500,000e (0.10)(18) A $9,074,471.20 Buying a home There is another equation, one to find out how much you ll pay each month. For this example, we will be trying to buy a home for $950,000 with a 3.61% interest rate. We make $5000/month. Here are our equations: Let s plug our numbers in. 1 (1 + i) n P v = P ( ) i [APR] i = [Payments] 360. 0361 1 (1 + P v = P ( 12 ) ). 0361 12 P v P(219.6800583) P v $950,000 219.6800583 P v $4324.47 We can also multiply this by the number of payments (360) to get how much it ll be if you paid interest on it for 30 years, which is about $1,556,809.49.
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49, I Modeling with Technology The correlation coefficient (r) is a measure of how close to the data our regression matches. 1) What kind of function best models this data? Quadratic. (r 0.9054) 2) Equation of Best Fit: f(x) 0.9057x 2 + 18.0513x 0.0072 3) Predict if you study for 5 hours. f(5) 67.6205 4) How many hours did you study if your test score was 90? f(x) = 90 x = No Solution
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50, I B Right Triangles and Trig Functions Consider the following Our right triangle is defined this way a c With a side length and an angle, we can figure out everything else of a right triangle. How? We can use the trigonometric functions. The important ones we will use are the sine, cosine, and tangent functions. C b A B Function Definition Inverse Function Sine (sin) Cosine (cos) Tangent (tan) Cotangent (cot) Secant (sec) sin θ = opposite hypotenuse sin 1 opposite hypotenuse = θ cos θ = adjacent hypotenuse cos 1 adjacent hypotenuse = θ tan θ = opposite adjacent tan 1 opposite adjacent = θ cot θ = adjacent opposite cot 1 adjacent opposite = θ sec θ = hypotenuse adjacent sec 1 hypotenuse adjacent = θ Cosecant (csc) csc θ = hypotenuse opposite csc 1 hypotenuse opposite = θ We can use algebra to figure out each side. For example, if we have an angle of 39.2 (for A), and a side of 2.1 (for b), we can figure out each side. Let s redraw and solve our triangle. a c First, let s solve for angle B. A triangle has 180, so we can remove the angles we already know from this to get angle B. (180 90 39.2 = 50.8 ) Next, we can either solve for a or c (I m solving for a here). I m going to use A. Because of this, 2.1 is our adjacent and a the opposite. We need to use the tangent function. 90 2.1 39.2 Use the Pythagorean Theorem to solve for c. tan 39.2 = a 2.1 (2.1) tan 39.2 = a a 1.7 c = 1.71 2 + 2.1 2 c 2.7 Why multiply? Because the tangent of an angle is angle a divided by angle b. Or generally: tan θ = a b We know angle b, so we can use algebra to solve for angle a.
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51, I Final Review Rational Functions Find all parts of the function below f(x) = x2 3x 4 2x 2 2x + 4 Vertical Asymptotes: x = 1, 2 = (x 4)(x + 1) 2(x + 2)(x 1) Horizontal Asymptotes: y = 1 2 Holes: None X-intercepts: x = 4, 1 Solve v + 2 4v 3 + 1 4v 3 = 1 v = 2 4v 3 ( v + 2 4v 3 + 1 4v 3 = 1) v + 2 + 1 = 4v 3 v + 3 = 4v 3 6 3 = 3v 3 v = 2
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52, I Circles Triangles and Circles on the Coordinate Grid A Right Triangle a 2 + b 2 = c 2 Distance Formula x 2 + y 2 = D 2 D = x 2 + y 2 With two points, it s like this D = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 Example 1 Find the distance between (1,6) and (3, 5). Relate to a Circle D = (3 1) 2 + ( 5 3) 2 D = 4 + 64 D = 125 D = 5 5 r = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 Equation of a Circle x 2 + y 2 = r 2 The center of a circle is at (h, k). The radius is r. (x h) 2 + (y k) 2 = r 2
52, O Example 2 Write the equation of a circle with the center of (4, 2) and a radius 6. (x h) 2 + (y k) 2 = r 2 (x 4) 2 + (y + 2) 2 = 36 Unit Circle The unit circle is a circle at the origin (0,0) with a radius of 1. (x 0) 2 + (y 0) 2 = 1 2 x 2 + y 2 = 1 Example 3 Find the x-intercepts of a circle with radius 6 centered at (2,4). (x h) 2 + (y k) 2 = r 2 (x 2) 2 + (y 4) 2 = 36 x =? y = 0 (x 2) 2 + (0 4) 2 = 36 (x 2) 2 + 16 = 36 (x 2) 2 = 20 x 2 = ± 20 x = 2 ± 20 x = 2 ± 2 5
53, I Final Review (Radicals) Simplify. 1. 2 = 2 3 18 3(2) 2 9 2 2 8 2(2) 2 4 2 2. 3. 3 6( 6 2 2) 18 + 6 12 18 + 12 3 3 512x 4 3 x 512x 3 8x x