Solution. Using the point-slope form of the equation we have the answer immediately: y = 4 5 (x ( 2)) + 9 = 4 (x +2)+9

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1 Chapter Review. Lines Eample. Find the equation of the line that goes through the point ( 2, 9) and has slope 4/5. Using the point-slope form of the equation we have the answer immediately: y = 4 5 ( ( 2)) + 9 = 4 ( +2)+9 5 If we use the slope-intercept form, then we have to find b. There are two ways to do this: distribute 4 5 in the equation above and simplify, or use y = 4 5 get + b, plugin = 2, y =9andsolveforb. Ineithercase,youshould y = Eample 2. Find the equation of the line that goes through the following points (, 7) and (5, 8) We start by finding m; oncewefindm, thenthefastestwayto finish is to use the point-slope form of the equation. m = y = ( ) = 6 Now we plug in either of the given points to the point-slope equation: y = 6 ( 5) + 8

2 CHAPTER. REVIEW 2.2 Fractions Eample. Simplify the following (in each case get a single fraction, with no compound fractions) 2 (a) (b) (c) (d) + 2 (a) = = (b) = = 6 2 = 2 2 (c) 3 + = = = (d) I ll do this one in a fair amount of detail: + = + + = + + = + + = + + To cancel the last we multiply the top and the bottom of this last fraction by and simplify: + + = ( +) = + ( +) Rules of eponents Eample. Using the above properties, simplify the following. (a) ( 2) 5 (b) 7 22 (c) 4 3/2 (d) 36 4 (a) ( 2) 5 =( 2)( 2)( 2)( 2)( 2) = (b) 7 22 = 7 22 = 5 = 5 (c) 4 3/2 = (d) 36 4 = 36 4 = = 3/2 ( 4) = 3 2 = 3 8

3 CHAPTER. REVIEW 3 Eample 2. Simplify the following ( 2 4 y 6 ) 8 (3 3 y 3 ) 2 so that your final answer has no fractions, and each base, 2, 3, and y, appears only once. Note: the best way to do this problem is to to apply algebraic rules to the powers here. Don t rewrite negative powers as reciprocals like 6,etc. I ll simplify the inside first, so that some things cancel before we need to deal with the 2 ontheoutside. ( 2 4 y 6 ) 8 2 ( 2) 8 32 y 48 2 = 2 (3 3 y 3 ) 2 (3) 2 6 y 6 = ( 2) 32 ( 6) y = y 54 2 = y 08.4 Factoring quadratics Eample. Factor and use this to solve = 0. We look for a common factor in both terms, and see that 2 is common. Thus: = 2 (3 +2). Now we solve the equation =0 2 (3 +2)=0 2 =0or(3 +2)=0 =0or3 = 2 =0, 2/3 Eample 2. Solve = 0. We use the quadratic formula. = 4 ± 4 2 4(3)( 5) 2 3 = 4 ± ± 6 = 6 = 30 6, 2 6 =5, /3

4 CHAPTER. REVIEW 4.5 Function Notation Eample. Let f be the function given by f() = 2. (a) What is f(2)? (b) In the equation f() = 2, what is the input? What is the output? (c) In the following description of this function, fill the blank in (with a verb): f is the function that takes an input and it. (d) In the epression f(3), what is the input? What do you get when you take this input and do to it the verb that you filled in the blank with in the previous part? (e) In the epression f(3 + ), what is the input? What do you get when you take this input and do to it the verb that you filled in the blank with in the previous part? (a) f(2) = 4. (b) is the input, and 2 is the output. (c) f is the function that takes an input and squares it. (d) In f(3) the input is 3. Ifwesquarethisweget(3) 2. (e) In f(3 + ) the input is 3 +. Ifwesquarethisweget(3+) 2. Eample 2. Let f() = (a) Find f() (b) Solve f() = (c) Find f(h) (d) Find f( + ) (a) f() = 2 ++=3. (b) We set the output equal to and solve 2 + += 2 + =0 ( +)=0 =0, (c) f(h) =h 2 + h +. (d) f( +)=( +) 2 +( +)+= Piecewise functions Eample. Let f() be defined by the following formulas, each applying to just one piece of f(). 2 if 0 f() = 2 if 0 < 3 e 2 if 3 <. Find f( ), f(2) and f(4), and make a graph of f().

5 CHAPTER. REVIEW 5 For f( ) we plug = intothefirstformula. Thusf( ) = ( ) 2 =. For f(2) we use the second formula, f(2) = 2 2 = 4. For f(4) we use the third formula, f(4) = e 2. The graph looks like 2 on the left (i.e. for 0); it looks like 2 in the middle (for 0 < 3) and it looks like e on the right (for >3). Notice that the graph looks unnatural, especially at =3whereitisdiscontinuous. graphics/piecewise_function-eps-converted-to.pdf.7 Function Combinations Eample. Let f() =3 2 + and g() =sin(2). (a) Find a formula for f()+g(). (b) Find f() g() (c) Find a formula for f()/g() (d) Find f(g(2)). (e) Find a formula for f(g()). (a) sin(2) (b) 3+ sin(2). (c) 32 + sin(2). (d) f(sin(4)) = 3 sin(4) 2 + (e) 3 sin(2) Atomic Functions Eample. Match each of the functions below with it s graph. (a) y = 2 (b) y = 3 (c) y = (d) y = 2 (e) y = (f) y = (g) y = e

6 CHAPTER. REVIEW 6 (h) y =(/2) (i) y =ln() (j) y =sin() (k) y = cos() (l) y = tan().9 Logarithmic functions Eample. Simplify ln(e 2+ ). If we think in terms of boes, ln(e )=, wehave2 +insideof the bo: ln(e 2 + )= 2 + =2 +

7 CHAPTER. REVIEW 7 Eample 2. Solve using ln(). We have e =2 ln(e )=ln(2) = ln(2) =ln(2)+.0 Trigonometric functions Eample. (a) Use a circle of radius to calculate the radian measure of θ = 360. (b) Multiply or divide your answer to the previous part, to find radian equivalents of the following angles (a) The picture looks like this: 0, 30, 45, 60, 90, 80, 270. θ =360 r = The arc-length s is the whole circumference s =2πr =2π (b) Thus, θ =360 =2π rad 360 =2π rad 2 80 = π rad 80 = π rad 2 90 = π 2 rad

8 CHAPTER. REVIEW 8 80 = π rad 3 60 = π 3 rad 80 = π rad 6 30 = π 6 rad 90 = π 2 rad 2 45 = π 4 rad 90 = π 2 rad = 3π 2 rad 360 = π rad 0 0 =0rad Eample 2. (a) Fill in the missing side of the triangle, and then fill in the table of trig values: π/6 π/3 sin(π/6) = sin(π/3) = cos(π/6) = cos(π/3) = 3 tan(π/6) = tan(π/3) = (b) Fill in the missing side of the triangle, and then fill in the table of trig values: π/4 sin(π/4) = cos(π/4) = π/4 tan(π/4) = (a) We solve for the missing side using the Pythagorean Theorem a 2 + b 2 = c 2 2 +( 3) 2 = c 2 +3=c 2 4=c 2 c =2 Now we can read off any pair of sides for the trig functions: sin(π/6) =/2 sin(π/3) = 3/2 cos(π/6) = 3/2 cos(π/3) =/2 tan(π/6) =/ 3 tan(π/3) = 3

9 CHAPTER. REVIEW 9 (b) We solve for the missing side using the Pythagorean Theorem a 2 + b 2 = c = c 2 +=c 2 2=c 2 c = 2 Now we can read off any pair of sides for the trig functions: sin(π/4) =/ 2 cos(π/4) =/ 2 tan(π/4) = Eample 3. Fill in the following chart: sin cos tan 0 π/2 π 3π/2 2π (, y) =(0, ) θ (, y) =(, 0) sin(0) = 0 cos(0) = tan(0) = 0 sin(π/2) = cos(π/2) = 0 tan(π/2) = undefined

10 CHAPTER. REVIEW 0 θ (, y) =(, 0) θ (, y) =(, 0) sin(π) =0 cos(π) = tan(π) =0 sin(2π) =0 cos(2π) = tan(2π) =0 θ (, y) =(0, ) sin(3π/2) = cos(3π/2) = 0 tan(3π/2) = undefined 0 π/2 π 3π/2 2π sin cos 0 0 tan 0 DNE 0 DNE 0 Eample 4. Summarize the previous eamples by filling in the following chart. See if you can find an easy mnemonic pattern to fill it in (i.e. something that you can remember without having to go back and check all the triangles). sin(θ) cos(θ) 0 π/6 π/4 π/3 π/2

11 CHAPTER. REVIEW 0 π/6 π/4 π/3 π/2 sin(θ) 0 /2 / 2 3/2 cos(θ) 3/2 / 2 /2 0 note this pattern is easy to remember if you write, or think of, these entries using square roots of 0,,2,3,4 on top: 0/2, /2, 2/2, 3/2, 4/2 Eample 5. Fill in the following chart: sin cos tan 6π/3 We start by seeing which quadrant 6π/3 isin. Startbydividing the top and the bottom of the circle into π/3 wedges: 3ontopand3on bottom. π/3 Now start with the first one (the one just on top of the positive -ais) and count 6 of them. You should go all the way around twice, and then count 4 more, finishing in quadrant III: graphics/plot_6_pi_over_3_radians-eps-converted-to.pdf (, y) Furthermore, we can see that and y form two sides of a right triangle in this quadrant; one with a π/3 angle:

12 CHAPTER. REVIEW 2 y (, y) Thus, this is a right triangle, with hypotenuse. In any triangle with 60, we already know the values of sine, cosine, and tangent. Now, we just need to recall that and y can be negative, to get the final result. For instance, sin(θ) =y, wecanseethatinthisquadrant,sinwillbenegative. The eact value for sin comes from the value we usually get in a triangle: sin(6π/3) = sin(4π/3) = sin(π/3) = 3/2 cos(6π/3) = cos(4π/3) = cos(π/3) = /2 tan(960) = tan(240) = tan(60) = 3 Eample 6. Combining () The standard trig values between 0 and π/2, (2) eamples where we ve found trig values in other quadrants, and (3) the rule about which trig functions are positive in which quadrants, we should now be able to fill in the values for sine and cosine all the way around the unit circle. Fill in the following chart, and again look for simple mnemonic patterns. 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 π/6 sin cos tan