Math 152: Blitzer quick guide and summary 1.1 : Algebraic expressions, Real numbers, and interval notation Algebraic Expressions Ex: 2x, 1 2 + x, x 1 Ex: Evaluate when t = 3: (a) 2t 3 (b) 2(1 t) + 7 Order of operations! P - start inside, work your way out E MD - left to right! AS - left to right! Ex: Evaluate when x = 2, y = 5. (1) 2x 7y + 3 2y x (2) x 3(y + 1) 2 Translating to Algebraic Expressions and equations. Ex: Write the associated algebraic expression (1) The product of 2 and a number. (2) The sum of a number and 3. (3) One less than a number. (4) The product of 4 and a number, subtracted from 3. (5) Two less than three times a number is 13. (6) The quotient of 3 and a number is 2. Sets of numbers. (1) The set of Natural numbers and the set of Whole numbers: (2) The set of Integers: (3) The set of Rational numbers: (4) The set of Irrational numbers: (5) The set of Real numbers: Set builder notation and interval notation for SETS. Ex: Convert between notations 1
(1) {x x > 5} (2) (, 0) (3) {x 5 < x < 8} (4) {x x 1} 1.2 : Operations with numbers and simplifying expressions. The number line and absolute value. Ex: Compute (1) 5 (2) 5 (3) 5 (4) ( ( 1 )) Adding an subtracting with the number line. Ex: Compute using number line. (1) 3 + 2 (2) 5 2 (3) 3 7 (4) 2 5 Add/Sub fractions and decimals. Ex: Compute (1) 1 7 9 7 (2) 2 3 + 1 5 (3) 5 1 2 2 (4) 2.34 + 17.5 (5) 1 5 +.7 + 3 Mult/div real numbers, powers, and why is ( 7)( 3) = 21? Ex: (1) ( 1)(3)( 5) (2) 0 2 (3) 5 0 (4) ( ) 2 3 (5) 3 2 5 (6) ( 3) 2 (7) ( 2) 5 (8) ( 1) 831 Mult/Div fractions Ex: Simplify (1) 1 2 3 4 (2) 1 2 3 4 Why flip and multiply?? (3) 1 2 3 4 2
Properties of Real numbers Commutative Associative Distributive Combining like terms Why is it true that 4x + 5x = 9x?? 4x + 5x = 4x + 5x = x(4 + 5) = x(9) = 9x!! Ex: Simplify (1) 4x 7a + 5ab + 6 5x + a 1 (2) 3x 2 5yx + 7x 2 y + 2x 5xy + yx 2 Simplifying expressions Ex: Simplify (1) 2(7 x) + 5x (1) 6 2(s + 3) + 2s 1.3 : Graphing Vocab: x-axis, y-axis, origin, quadrants Plotting points Ex: (2, 3), (5, 0), ( 1, 2), (0, 2) Graphing a few things with points Ex: y = 4 x 2, y = 2x + 1, y = x Note: The graph is visual representation of all the points that are solutions for an equation. 1.4 : Solving Linear Equations Ex: Linear equation in one variable has the form ax + b = 0. Addition and multiplication properties of equality. 3
Ex: Solve (1) 2x + 3 = 17 (2) 2x 7 + x = 3x + 1 + 2x (3) 4(2x + 1) 29 = 3(2x 5) (4) 2x + 5 5 + x 7 2 = 3x + 1 2 (5) x + 1 3 = 5 x + 2 7 (6) x + 3 = x + 4 1 and x = x + 7 (7) Suppose a hippopotamus is floating down a river at 3 miles-per-hour, then the distance floated by the hippo after t hours is D = 3t. If the hippos favorite grass patch is exactly 18 miles downstream, how long before the hippo gets there? (show work) 1.5 : Application questions and manipulating formulas (1) I want to mark off an area for a garden which is three times longer than it is wide. If I have a spool of fencing that is 100 feet long, what dimensions should my garden be? (2) Solve the equation 2l + 2w = p for l. This equation gives the relation between the length, width, and perimeter of a rectangle. (3) Solve the equation V = 4 3 πr3 for r. This equation gives the relation between the radius and the volume of a sphere. (4) Circle: formulas for area and circumference. Formulas for volume of sphere, cube, cylinder. 1.6 : Exponents (1) x a x b =?? (2) pf p g =?? (3) (bm ) n =?? (4) (rs) l =?? (5) ( ) p r =?? (6) a n =?? q (7) 1 a n =?? (8) x 0 =?? (9) 0 5 =?? 4
( 25x (10) 0 0 =?? (11) a 0 2 y 4 ) 2 =?? (12) 5x 6 y 8 SECOND WEEK: 1.7 : Scientific notation Ex: 2.7 10 3, 3.0019 10 7 To take a number out of scientific notation you (i) Move the decimal point one place for each exponent of ten. Move to the right for positive exponents and to the left for negative exponents. Ex: Write as a decimal (1) 2.7 10 3 (2) 3.0019 10 7 (3) 2.001 10 2 (4) 5 10 4 To put a number in scientific notation you (i) Put a decimal to the right of the first non-zero digit. (ii) Count how many places you moved the decimal point (this is the exponent on the ten). (iii) Drop all zeros after the last non-zero digit and before the first non-zero digit. Ex: Put into scientific notation (1) 20, 054, 000 (2).00705 (3).0001 Ex: Compute (1) ( 2.7 10 3) ( 5 10 4) (2) 2.1 : Intro to functions.0006.000002 Vocab: relation (ordered pair), domain, codomain, function (each element in domain goes to exactly one element in the codomain), range. Domain: The collection of all objects for which a function is defined i.e. the collection of inputs of a function. Codomain: The collection of objects into which all of the outputs of a function fall. Function: codomain. A rule that assigns each object in the domain to exactly one object in the Range: The collection of all outputs of a function. Ex: (1) The favorite color function, f : {People in class} {colors} 5
Why does this example require that we each have only one favorite color? (2) y = 2x + 1 Also, sometimes written as f(x) = 2x + 1, where f : {x values} {y values} (3) NBA champ function, f : {years since 1947} {NBA teams} {2015, 2014,..., 1975,...} {GSW, SAS,...} Why would we not be able to make a function in the other direction? NOTATION: From Ex: (1) f(lenny) = Red, f(??) = P urple From Ex: (2) f(0) = 1, f(??) = 3 From Ex: (3) f(2015) = GSW, f(??) = SAS We say, f of 0 is 1, i.e. the function evaluated at 0 is eual to 1. Note: This is NOT multiplication, we are NOT multiplying f by 0. 2.2 : Graphs of functions Graph using tables: f(x) = 2x + 1, g(x) = x 2 Vertical line test NOT functions: x 2 + y 2 = 1, y 2 = x Identify domain and range from graph: f(x) = 2x + 1, f(x) = x 2, f(x) = ln x, f(x) = e x, f(x) = x 2.3 : The Algebra of functions (1) (f + g)(x) = f(x) + g(x) (2) (f g)(x) = f(x) g(x) (3) (fg)(x) = f(x) g(x) (4) f f(x) (x) = g g(x) Ex: Perform each operation for f(x) = 2x and g(x) = x 1 Ex: Let p(x) = x 2 3 and q(x) = 4x + 5, find 6
(1) (p + q)(3) (2) q (0) (3) (pq)(1) p Two cool/important items: (1) Whenever you divide functions you may have values which are not allowed in the domain, q e.g. p (x) =?? and f (x) =??, what values are not in the domain of these functions? g (2) Recall adding fractions! Ex: (a) 2 3 + 1 5 Now (b) 5 x 7 x 8 2.4 : Linear functions and slope Graphing using intercepts (c) 3 x(x + 1) + 1 x + 1 (1) To find the x-intercept, set y = 0 and solve for x. (2) To find the y-intercept, set x = 0 and solve for y. (3) Plot the two points and draw your line through them! Ex: Graph (1) 4x 3y = 6 (2) y = 3x 2 (d) 2 x 3 + 1 x(x + 1) The slope of a line through two points (x 1, y 1 ) and (x 2, y 2 ) is m = Rise Run = change in y change in x = y 2 y 1 x 2 x 1 Note: The order you subtract matters! Ex: Find the slope of the line passing through the two points. (1) ( 3, 4) and ( 1, 6) (2) ( 1, 3) and ( 4, 5) Equation of a line in slope intercept form Ex: Graph y = 2x + 1 using intercepts, notice slope and y-intercept The slope intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept. Where did this come from anyways?! I m glad you asked... Start with m = y 2 y 1 x 2 x 1. Now instead of choosing (x 1, y 1 ) and (x 2, y 2 ) as arbitrary points, choose (0, b) and (x, y). Notice (0, b) is the y-intercept. 7
Then m = y b. Solving for y we get our wonderful formula y = mx + b. x 0 Ex: Put into slope intercept form (we want to do this because it makes graphing easy) (1) 4x + 2y = 10 (2) 3x y = 2 (3) 2y x = 1 Graphing using slope and y-intercept How to... (1) Plot the y-intercept (2) Start at the y-intercept and use the slope for find a second point. (3) Draw the line through these two points. Ex: Graph (1) y = 2x + 1 (2) y = x y = 3 2 x 2 Graphing horizontal and vertical lines How to... (1) Write down two points with y = 2 or x = 1 (2) Plot those points (3) Draw your line Ex: Graph (1) y = 2 (2) x = 1 (3) y = 0 (4) x = 0 Slope as rate of change Think of the y-axis as elevation gain and the x-axis as distance traveled. Now, you are climbing up a mountain! The steeper the slope, the faster you change elevation. Ex: Graph and discuss y = x, y = 2x, y = 10x, y = 1 2 x, y = 1 10 x The average rate of change of a function Suppose I throw a basketball into the air. Using the same x and y axis as directly above, we see that the graph of the basketball s path is not a straight line and it is difficult to tell the slope at a given point. We can however figure out the slope of the line connecting any two points on the graph. This slope is the average slope of the graph between these two points. This is the average rate of change of a function. Ex: Find the average rate of change of the function y = (x 4) 2 +16 between x = 1 and x = 4 8
2.5 : The point-slope form of the equation of a line Point-slope form of a line is y y 1 = m(x x 1 ), where m is the slope and the line passes through the point (x 1, y 1 ). Ex: Graph y = 2x + 1 using slope and y-intercept. Use the above equation choosing a point on the line and using the slope (m = 2) and see what you get (try a couple points from the line). When asked to find the equation of a line use the point-slope form of a line. (1) Find the equation of the line with slope 5 passing through the point (2, 3). (2) Find the equation of the line passing through (0, 1) and (3, 7). Final note: This formula also came right from the definition of slope! Again we start with m = y 2 y 1 x 2 x 1. Instead of choosing (x 1, y 1 ) and (x 2, y 2 ) as arbitrary points, choose (x 1, y 1 ) and (x, y). Now (x 1, y 1 ) is the point we are given. Instead of choosing a second point (x 2, y 2 ) and computing the slope, we let the second point vary as (x, y), our independent variable x and our dependent variable y. Then m = y y 1 x x 1. Re-arranging, we get our wonderful formula y y 1 = m(x x 1 ). Parallel and perpendicular lines Slope of parallel lines. Line l 1 has slope m, then any parallel line l 2 also has slope m. Ex: Find another line parallel to y = 2x + 1. Slope of perpendicular lines. Line l 1 has slope m, any perpendicular line l 2 has slope 1 m i.e. it is the negative reciprocal. Another way to check if two lines are perpendicular is if (m 1 )(m 2 ) = 1. Notice that (m) ( 1 m) = 1. Ex: Find another line perpendicular to y = 2x + 1. 4.1 : Solving linear inequalities A linear inequality behaves almost exactly the same as a linear equation, except there are two important differences. (1) If you multiply or divide by a negative number, you have to change the direction of the inequality sign. 9
Check it out: 6 < 10 is a perfectly fine and true statement, however, if we divide both sides by 2 we get: 3 < 5. This statement is false! In fact, 5 < 3. (2) The solution is not a single number, but a set of numbers. If we solve an equation we end up with something that looks like x = 2, however, if we solve an inequality we end up with something that looks like x < 2. Give me a few numbers that make this inequality true. This may seem like a subtle point, but x < 2 is not the answer! This is a set that we can describe using interval notation i.e. (, 2) or set builder notation i.e. {x x < 2}. Ex: Solve (1) 3x 5 > 17 (2) 2x 4 > x + 5 Inequalities with strange solution sets Ex: Solve (1) x > x + 1 (2) x < x + 1 In the first case we write, in the second case we write (, ). 4.2 : Compound Inequalities Ex: x > 2 or x < 0, 2 < x < 1 Intersection of Sets The intersection of two sets A and B is written A B. Something is in the intersection of two sets if it is in both sets. Think of an intersection of two streets, the intersection is on both streets! In set builder notation: A B = {x x A and x B} Ex: Given A = {1, 5, 9, 3, 2, 7} and B = {5, 4, 2, 7, 3}, find Find A B. Compound inequalities involving and i.e. intersection A number is a solution to the compound inequality x 6 and x 0 if it is a solution to both i.e. the number 1 is a solution to the compound inequality since 1 6 and 1 0 are both true statements. Note: You can always rewrite a compound inequality involving and in a shorter and nicer format! How to solve compound inequalities involving and or intersection of sets : (1) Solve each inequality separately. (2) Graph each inequality on separate number lines, one below the other. 10
(3) See where the graphs overlap, the overlap is your solution set. Ex: Solve (1) x 3 < 5 and 2x + 4 < 14 (2) 2x 7 > 3 and 5x 4 < 6 Union of Sets The union of two sets A and B is written A B. Something is in the union of two sets if it is in either sets. Think of a union of two families (through a wedding or such), a person is in the new bigger family if they are in either family! In set builder notation: A B = {x x A or x B} Ex: Given A = {1, 5, 9, 3, 2, 7} and B = {5, 4, 2, 7, 3}, find Find A B. Compound inequalities involving or i.e. union A number is a solution to the compound inequality x 6 or x 0 if it is a solution to either i.e. the number 1 is a solution to the compound inequality since 1 0, even though 1 is not a solution to the other inequality x 6. How to solve compound inequalities involving or i.e. union of sets : (1) Solve each inequality separately. (2) Graph each inequality on separate number lines, one below the other. (3) See what part of the number line is covered by either graph, this is your solution set. Ex: Solve (1) 2x 7 > 3 or 5x 4 < 6 (2) x 3 < 5 or 2x + 4 < 14 4.3 : Equations and inequalities involving absolute value. Ex: Solve (a) x = 2 (b) x < 2 (c) x 2 Absolution Value Equalities Ex: (a) x = 2, here we have two solutions (b) x = 2, here we get??? Ex: Solve (1) 2x 3 = 11 (2) 5 1 4x 15 = 0 (3) 3x 1 = x+5 Absolution Value Inequalities type 1 11
Ex: (a) x < 2 (b) x < 2, here we get??? Ex: Solve (1) x 4 < 3 (2) 2 3x + 5 + 7 13 Absolution Value Inequalities type 2 Ex: (a) x > 2 (b) x > 2, here we get??? Ex: Solve (1) 2x + 3 5 Beware these weird solution sets: (1) x + 1 < 2 (2) x + 1 > 2 5.1 : Intro to polynomials Vocab: term, coefficient, degree of a polynomial, leading term, leading coefficient, monomial, binomial, trinomial, polynomial End behavior of polynomial functions Ex: Graph f(x) = x 2, f(x) = x 3, f(x) = x 2, f(x) = x 3 The leading coefficient criteria: As x increases or decreases without bound, the graph of a polynomial function eventually shoots upward or plummets downward. (a) For polynomials whose degree is even: *If the leading coefficient is positive then (i) the graph will eventually shoot upward as x increases and (ii) the graph will eventually shoot upward as x decreases. *If the leading coefficient is negative then (i) the graph will eventually plummet downward as x increases and (ii) the graph will eventually plummet downward as x decreases. (b) For polynomials whose degree is odd: *If the leading coefficient is positive then (i) the graph will eventually shoot upward as x increases and (ii) the graph will eventually plummet downward as x decreases. *If the leading coefficient is negative then (i) the graph will eventually plummet downward as x increases and (ii) the graph will eventually shoot upward as x decreases. Adding/subtracting polynomials (1) (5x 3 y 4x 2 y 7y) + (2x 3 y + 6x 2 y 4y 5) (2) (6x 2 + 5x 3) (4x 2 2x 7) 12
5.2 : Multiplication of polynomials Ex: Multiply (1) (5x 3 y 4 )( 6x 7 y 8 ) (2) 4x 3 (6x 5 2x 2 + 3) (3) (7x + 2)(4x + 5) (4) (x 2 + 4x + 5)(3x + 7) Special Products (1) (a + b) 2 (2) (a b) 2 (3) (a + b)(a b) (4) (a + b)(a 2 ab + b 2 ) (5) (a b)(a 2 + ab + b 2 ) Ex: Multiply polynomial functions (1) Given f(x) = x 5 and g(x) = x 2, find (fg)(x) Ex: Given f(x) = x 2 7x + 3 find (1) f(a + 4) (2) f(a + h) f(a) 5.3 : Greatest common factor and factor by grouping The goal of factoring is to turn a sum of terms into a product of factors. Since we can cancel factors in a fraction, this will be super useful! **The first thing to look for ALWAYS when factoring, is the greatest common factor. Ex: Factor (1) 21x 2 + 28x (2) 9x 2 + 15x 3 (3) 12x 5 y 4 4x 4 y 3 + 2x 3 y 2 Ex: Factor by grouping (1) 2(x 7) + 9a(x 7) (2) 5y(a b) (a b) (3) x 3 5x 2 + 3x 15 (4) 3x 2 + 12x 2xy 8y 5.4 : Factor trinomials Ex: x 2 + 2x + 1 = (x + 1)(x + 1) How to... factor trinomial with leading coefficient one (1) Factor out GCF (2) Determine sign of factors (3) List factors of last term, find factors that add up to middle term 13
Ex: Factor (1) x 2 + 5x + 6 (2) x 2 14x + 24 (3) y 2 + 7y 60 (4) x 2 4xy 21y 2 (5) 8x 3 40x 2 48x (6) 6y 4 +12y 2 +6 (7) y 4 12y 3 +35y 2 (8) x 2 4x+45 (9) x 3 y 2x 2 y 2 3xy 3 How to... Factor trinomials by grouping (leading coefficient not one) (1) Multiply leading coefficient and constant. (2) Find factors of ac whose sum is b. (3) Re-write middle term and factor by grouping! Ex: Factor (1) 15x 2 7x 2 (2) 12x 2 5x 2 (3) 6x 2 + 19x 7 (4) 2x 2 7xy + 3y 2 (5) 2x 2 x 6 (6) 8x 2 22x + 5 5.5 : Factoring special forms Recall: a 2 b 2 = (a + b)(a b) Ex: Factor (1) x 2 9 (2) 9x 2 100 (3) 36y 6 49x 4 What about...? x 2 5, x 7 25, 1 x 6 y 4, 3y 3x 6 y 5 Recall: a 2 + 2ab + b 2 = (a + b) 2 and a 2 2ab + b 2 = (a b) 2 Ex: Factor (1) x 2 + 14x + 49 (2) 4x 2 + 12xy + 9y 2 (3) 9y 4 12y 2 + 4 What about...? x 2 8x + 16 y 2, a 2 b 2 + 10b 25 (factor twice!) Recall: a 3 + b 3 = (a + b)(a 2 ab + b 2 ) and a 3 b 3 = (a b)(a 2 + ab + b 2 ) Ex: Factor (1) x 3 + 125 (2) x 6 64y 3 (2) 27 y 3 5.6 : A general factoring strategy (1) Factor out GCF from polynomial. (2) Is the polynomial only two terms? Try a special form. (3) Is the polynomial three terms? (a) Try a special form, then trial/error using the list of factors. 14
(b) If the coefficient on the x 2 term is not 1, factor by grouping i.e. the ac method (4) Is the polynomial four terms? Factor by grouping. (5) Make sure it is factored completely. (6) Check your answer by multiplying it out. Ex: Factor (1) 5x 3 20x (2) 7x 4 7 (3) x 3 + 2x 2 9x 18 (4) 6x 2 + 6x + 12 (5) 8x 5 2x 3 (6) y 2 + 8y 16 prime (7) 12x 3 y 12xy 2 (8) x 3 xy 2 + x 2 y y 3 (9) 16x 4 y y 5 5.7 : Polynomial equations and applications Ex: 2x 2 + x 1 = 0 In order to solve polynomial equations, we use the zero product principle: if ab = 0 then a = 0 or b = 0 How to... solve polynomial equations by factoring (1) Re-write the equation in standard form ax 2 + bx + c = 0. (2) Factor completely (3) Apply sero product principle i.e. set each factor containing a variable to zero (4) Solve the equations in step (3) (5) Check your solutions in original equation Ex: Solve (1) (3x 2)(x+2) = 0 (2) 2x 2 +7x 4 = 0 (3) 2x 2 5x = 12 (4) x 2 = 6x + 4 (5) 5x 2 = 20x (6) (x 7)(x + 5) = 20 (7) x 3 + x 2 = 4x + 4 Modeling motion You Throw a ball straight up on the roof of a building, the ball misses the roof and falls all the way down to the ground. If the function h(t) = 16t 2 + 32t + 384 describes the hight of the ball after t second, how long will it take for the ball to hit the ground? Landscape question 15
A rectangular garden measures 80ft by 60ft. You want to add a large path along both shorter sides and one of the longer sides of the garden. By adding the path, you want to double the area of the garden. How wide should the path be? (call width of path x and draw diagram) 6.1 : Rational Expressions and Functions intro; multiplying; dividing 13x Ex: x 1, x + 2 x 2 + 2x 1, x + 1 1 2y Domain of rational functions Ex: Find domain (1) f(x) = 4 x 2 (2) f(x) = 2x + 1 x 2 x 2 (3) f(x) = 2 x 2 + 1 Graphs of rational functions Ex: Graph (1) y = 1 x (2) y = 1 x 2 (3) y = 1 x Simplifying rational expressions How to... (1) Factor numerator and denominator completely. (2) Divide any common factors Ex: Simplify (1) 5x + 35 20x (4) x2 + 4x + 3 x + 1 Notice: Graph f(x) = x2 + 4x + 3 x + 1 Watch out!... Simplify (5 2x) = 5 + ( 2x) = 2x + 5 = ( 1)(2x 5) (2) x3 + x 2 x + 1 (5) x2 7x 18 2x 2 + 3x 2 (2x + 5)(2x 5) 3(5 2x) (3) x2 + 6x5 x 2 25 and g(x) = x + 3 on different graphs. (6) 3x2 + 12xy + 15y 2 6x 3 6xy 2 Multiply rational expressions 16
Ex: x 2 y + 3 x + 5 y 7 = x2 (x + 5) (y + 3)(y 7) How to... (1) Factor all numerators and denominators completely (2) Divide all common factors (3) Multiply remaining factors Ex: Multiply (1) 7 x + 3 x 2 5 (2) x + 3 x 4 x2 2x 8 x 2 9 (3) 4x + 8 6x 3x 2 3x2 4x 4 9x 2 4 Dividing rational expressions P Recall: Q R S = P Q S R = P S QR Why...? x Recall that x 7 6 y is just another way to write... 7 6 y Ex: Divide (1) x 7 6 y (3) x2 + 3x 10 2x x2 5x + 6 x 2 3x (2) (4x 2 + 3x 10) 2x + 5 14 6.2 : Add / subtract rational equations Same denominators Recall: 2 9 + 5 9 = 2 + 5 9 = 7 9 Ex: Add / Subtract (1) 2x 1 3 + x + 4 3 (2) (4) x2 + 2x 2 5x + 12 x 2 + + 3x 10 x 2 + 3x 10 x 2 x 2 9 + 9 6x x 2 9 (5) x 2 x 5 +4x + 5 5 x (3) 2x + 3 x + 1 x + 2 x + 1 (6) 3y3 5x 3 6x 3 x 2 y 2 4y3 x 2 y 2 Different denominators?... Find the Least Common Denominator (LCD) (1) Factor each denominator completely 17
(2) List all the factors in 1 st denominator (3) Put any other factors from the 2 st denominator on the list (4) Multiply all factors on the list, this is the LCD. Ex: Find LCD 3 (1) 10x 2, 7 15x (2) 9 7x 2 + 28x, 11 x 2 + 8x + 16 Ex: Add / subtract 3 (1) 10x 2 + 7 15x (3) x x + 3 + x 1 x + 3 (2) (4) 9 7x 2 + 28x + 11 x 2 + 8x + 16 x 1 x 2 + x 6 x 2 x 2 + 4x + 3 6.6 : Rational Equations Ex: Solve (1) x + 4 2x + x + 20 = 3 multiply by the LCD to clear fractions 3x Notice: x cannot be zero, but the answer is 4 0 so we are okay! (2) x + 1 x + 10 = x 2 x + 4 (3) x x 3 = 3 x 3 + 9 (4) x 3 + 9 x = 4 (5) 2x x 3 + 6 x + 3 = 6.7 : Formulas and applications 28 x 2 9 Solve for a variable (1) Solve s c 1 r for r (2) Solve 1 p + 1 q = 1 f for p Questions involving average cost Ex: A company has to pay $5000 rent each month, and it costs $100 to produce each whatchama-call-it. (1) Write an equation for the cost of producing x what-chama-call-its. (2) Write an equation for the average cost of producing x what-chama-call-its. (3) How many what-chama-call-its must be produced each month to keep the average cost down to $200 per unit. Questions using motion 18
Ex: You commute to work 40miles and return home the same route. Your average rate driving home is 30 mph faster than your average rate driving to work. If the round trip takes 2 hours, what is your average rate driving to work? Distance = (rate)(time), also, T ime = distance rate Since we want to find rate, this will be our variable we have to solve for. Table: Distance / Rate / Time=D/R To work: 40 x 40/x From work: 40 x + 30 40/(x + 30) Then, 40/x + 40/(x + 30) = 2 since the round trip takes 2 hours. Questions using work You can design a webpage in 15 hours. A friend can design a webpage in 10 hours. How long would it take to design a webpage working together? Table: Rate / Time / Work done You: 1/15 x x/15 Friend: 1/10 x x/10 Then x/15 + x/10 = 1 where 1 is the whole webpage you are trying to make. 6.3 : Simplifying complex rational expressions How to... (1) Find the LCD of smaller fractions (2) Multiply numerator and denominator of larger fraction by LCD (3) After simplifying you should have only one fraction (4) Factor and simplify more if possible Ex: Simplify (1) 1 x + y x 2 1 y + x y 2 (2) 6.4 : Division of polynomials x y 1 x 2 y 2 1 Dividing a monomial by a monomial How to... (1) Divide coefficients (2) Cancel factors (3) 1 x + h 1 x h Ex: Divide 19
(1) 25x12 5x 4 (2) 12x8 4x 2 (3) 2x3 8x 3 (4) 15x5 y 4 3x 2 y Dividing a polynomial by a monomial How to... Divide each term in the numerator by the denominator Ex: Divide (1) (15x 3 5x 2 + x) (5x) (2) 10x8 + 15x 6 5x 3 (3) 16x5 9x 4 + 8x 3 2x 3 Dividing a polynomial by a binomial How to... This is exactly the same as high school long division Ex: Divide (1) (x 2 14x + 24) (x 2) (2) (4 5x x 2 + 6x 3 ) (3x 2) (3) 8x3 1 2x 1 Ex: Simplify (1) 25x12 5x 4 (4) (x 3 1) (x 1) (2) (15x 3 5x 2 + x + 5) (5x) (3) (8x 4 y 5 10x 4 y 3 + 12x 2 y 3 ) (4x 3 y 2 ) (4) (x 2 14x + 24) (x 2) (5) (4 5x x 2 + 6x 3 ) (3x 2) (6) (6x 4 + 5x 3 + 3x 5) (3x 2 2x) 7.1 : Radical expressions and functions Notice: 5 2 = 25 = ( 5) 2, by 25 we mean the positive value i.e. 25 = 5 2 = 5 Ex: Evaluate (1) 81 (2) 9 (3) The square root function 4 49 (a) Evaluate: Find f(3) given f(x) = 2x 2 (4) 36 + 64 (5) 36 + 64 (b) Graph: f(x) = x and compare to f(x) = x 2 Domain of a square root function Ex: f(x) = 3x + 12 this has to be positive so set up inequality! Square roots and cube roots 20
Square root: (a 2 ) = a look at graphs of f(x) = (x 2 ) and f(x) = x (1) ( 6) 2 (2) (x + 5) 2 (3) 25x 6 (4) x 2 4x + 4 Cube root: 3 (a 3 ) = a look at graphs of f(x) = 3 (x 3 ) and f(x) = x (1) 3 64 (2) 3 27y 3 (3) 3 8x 6 7.2 : Rational exponents ( ) Ex: 7 1 3 1 3 = 7 3 3 = 7 1 = 7 Definition: n a = a 1 n Ex: Compute these are two different ways of writing the same thing (1) 64 1 2 = 64 = 8 (2) ( 125) 1 3 = 3 125 = 5 From the definition, ( n a) m = ( ) a 1 m 1 n = a n m = a m n and so ( n a ) m = a m n also different ways of writing the same thing Also from the definition, a m n = a m 1 n = (a m ) 1 n = n a m and so a m n = n a m also different ways of writing the same thing How to... simplify rational exponents and radicals (1) Write out your number as a prime factor to some power (2) Multiply the exponents using the appropriate rule of exponents i.e. (b m ) n = b m n Ex: Compute (1) 1000 2 3 (2) 16 3 2 (3) 32 3 5 (4) 27 2 3 All of the rules of exponents can also be used on fractional exponents ( ) ( ) Recall: b m b n = b m+n Now: a k l a m n = a k l + m n Recall: a n = 1 a n Now: a m n = 1 a m n Also (b m ) n = b m n, Ex: Simplify b m b n = bm n, etc. 21
(1) 10 x 5 = ( x 5) 1 10 = x 5 10 = x 1 2 (2) 3 27x 15 (3) 4 x 6 y 2 (4) x 3 x (5) 3 x 7.3 : Multiplying and simplifying radical expressions Ex: ( 4)( 9) = 2 3 = 6 and also ( 4)( 9) = 36 = 6 Yes, this always works! Why...?? I m glad you asked ( 4)( 9) = 4 1 2 9 1 2 = (4 9) 1 2 = (36) 1 2 = 6 The product rule for radicals a b = a b Ex: Multiply (1) 3 7 (2) x + 7 x 7 (3) 3 4 3 3 Simplifying radical expressions How to... (1) re-write as product of factors (2) Take the root of appropriate factors Ex: Simplify (1) 75 (2) 3 54 (3) 5 64 (4) 500xy 2 Simplifying a radical function Ex: f(x) = 2x 2 + 4x + 2 and g(x) = 2 x + 1, look at graphs This is one way to see why it is necessary to take the absolute value. Notice: The book says Assume that all variables in a radicand represent positive real numbers, however, we are going to learn the correct way to do this Radicals rule! n a n = a if n is odd and n a n = a if n is even Note: a 2 = a 2 and similarly a n = a n so long as n is even i.e. 4 x 8 = 4 (x 2 ) 4 = x 2 = x 2 Ex: Simplify (1) 3 x 3 (2) 4 x 8 (3) 4x 7 (4) 9x 9 22
(5) x 5 y 1 3z 7 (2) 3 32x 8 y 1 6 (3) 5 64x 3 y 7 z 2 9 Multiply then simplify Ex: (1) 15 3 (2) 7 3 4 6 3 6 (3) 4 8x 3 y 2 4 8x 5 y 3 7.4 : Adding, subtracting, and dividing radical expressions Add / Subtract Like radicals should be thought of as like terms Ex: Add/Sub (1) 7 2 + 8 2 (2) 3 5 5 3 5 2 3 5 (3) 7 + 2x 7 2 7 (4) 8 6 5x 5 6 5x + 4 3 5x Add/sub radicals after simplification Ex: Add/Sub (1) 7 18 + 5 8 (2) 4 27x 8 12x (3) 7 3 2 5 (4) 2 3 16 4 3 54 5 3 xy 2 + 3 8x 4 y 5 7.5 : Multiplying with more than one term and rationalizing denominators Ex: Multiply (1) 7(x + 2) (2) 3 x( 3 6 3 x 2 ) (3) (5 2 + 2 3)(5 2 2 3) (4) ( 3 + 7) 2 Rationalizing denominators with one term Ex: Write without a radical in the denominator 5 7 3 2 (1) (2) 3 (3) (4) 3 6 25 7 9 Rationalizing denominators with two terms Ex: Rationalize 8 (1) 3 2 + 4 (4) (2) 15 6 + 1 (5) 2 + 5 6 3 (3) 11 5 11 + 5 h x + h x (6) 3 x + y y 3 x 7.6 : Radical Equations Ex: x = 9 Solving Radical Equations 23
Ex: Solve (1) 2x + 3 = 5 (2) x 3+6 = 5 (3) x+ 26 11x = 4 (4) 3x + 1 x + 4 = 1 (5) (3x 1) 1 3 + 4 = 0 7.7 : Complex numbers Imaginary numbers! Definition: i = 1 and so i 2 = 1 Ex: 25 = 25( 1) = 25 1 = 5i Notice: To take the square root of a negative number, b = b( 1) = b 1 = ( b)(i) General written convention: 2i 5 and 4i and i 7 Ex: Write as an imaginary number using i (1) 9 (2) 3 (3) 80 Complex numbers Complex numbers can be written in the form a + bi where a and b are real numbers We call a the real part of the complex number We call b the imaginary part of the complex number Notice: The Complex numbers contain the Real numbers Adding and subtracting complex numbers How to... Treat imaginary numbers as like terms, add the coefficients of i i.e. (a + bi) + (c + di) = (a) + (c) + (b)i + (d)i = (a + c) + (b + d)i and (a + bi) (c + di) = (a) + ( c) + (b)i + ( d)i = (a c) + (b d)i Ex: Simplify (1) (5 11i) + (7 + 4i) (2) ( 5 + 7i) ( 11 6i) (3) (5 2i) + (3 + 3i) (4) (2 + 6i) (12 4i) 24
Multiply complex numbers Ex: Multiply (1) 4i(3 5i) (2) (7 3i)( 2 5i) (3) 7i(2 9i) (4) (5 + 4i)(6 7i) Conjugates Ex: (4 + 7i)(4 7i) = 65 also recall (3 2)(3 + 2) = 9 2 = 7 Just like multiplying square root conjugates gets rid of the square root, multiplying complex conjugates gets rid of the imaginary part of a complex number. Ex: Divide (1) 7 + 4i 2 5i (2) 5i 4 3i (3) 6 + 2i 4 3i (4) 3 2i 4i Simplifying powers of i How to... (1) Write as i 2 to some power i.e. i 7 = (i 6 )(i) = (i 2 ) 3 (i) (2) Then simplify using i 2 = 1 i.e. i 7 = (i 6 )(i) = (i 2 ) 3 (i) = ( 1) 3 (i) = ( 1)(i) = i Ex: Simplify (1) i 12 (2) i 39 (4) i 50 8.1 : Square root property and completing the square The square root property is if u 2 = d, then u = d or u = d i.e. u = ± d Ex: Solve (1) x 2 = 9 (2) 3x 2 = 18 (3) 2x 2 7 = 0 (4) 9x 2 + 25 = 0 (5) (x 1) 2 = 5 Completing the square This method ALWAYS works to solve quadratic equations. We use completing the square to derive the quadratic formula (after which we can just use the quadratic formula). Given a quadratic equation with leading coefficient one x 2 + bx + c = 0 25
We complete the square by adding and subtracting ( b 2 )2 like so x 2 + bx + ( ) b 2 2 ( ) b 2 + c = 0 2 After which the first three terms factor as a perfect square like so We then solve for the perfect square ( x + b 2) 2 ( x + b 2) 2 = ( ) b 2 + c = 0 2 ( ) b 2 c 2 and finally use the square root property to finish solving. Ex: Solve by completing the square (1) x 2 6x+4 = 0 (2) 9x 2 6x 4 = 0 (3) 2x 2 x+6 = 0 (4) x 2 +4x 1 = 0 (5) 2x 2 +3x 4 = 0 (6) 3x 2 9x+8 = 0 8.2 : The Quadratic Formula We start with a quadratic equation ax 2 + bx + c = 0 and solve for x to get the quadratic formula x = b ± b 2 4ac 2a This method ALWAYS works to solve quadratic equations. Ex: Solve (1) x 2 +8x+12 = 0 (2) 2x 2 = 4x+5 (3) 2x x 10 = 0 (4) (x 2 5) 2 +3(x 2 5) = 0 (5) 10x 2 +7x 1 +1 = 0 (6) 5x 2 3 +11x 1 3 +2 = 0 8.4 : Equations quadratic in form The essential step for these questions is to identify the correct substitution to make. Ex: Solve (1) x 4 8x 2 9 = 0 (2) x 4 5x 2 +6 = 0 (3) 2x x 10 = 0 (4) (x 2 5) 2 +3(x 2 5) = 0 (5) 10x 2 +7x 1 +1 = 0 (6) 5x 2 3 +11x 1 3 +2 = 0 26
8.3 : Graphing quadratic functions (and the discriminant from section 8.2) Mini Lesson on the Discriminant Given a quadratic function f(x) = ax 2 + bx + c, the discriminant is b 2 4ac (part of the quadratic formula) (1) If b 2 4ac > 0, then the function has two real solutions (draw graph). (2) If b 2 4ac = 0, then the function has one real solution (draw graph). (3)b 2 4ac < 0, then the function has two imaginary solutions (draw graph). Ex: Compute the discriminant, determine number and type of solutions. (1) f(x) = 3x 2 + 4x 5 (2) f(x) = 3x 2 8x + 7 (3) f(x) = 9x 2 6x + 1 The graph of a quadratic function f(x) = ax 2 + bx + c = 0, (a 0) is called a parabola. Properties of Quadratic Functions (1) Draw parabola with a < 0 and another with a > 0, label axis of symmetry and vertex (2) Graph y = x 2 and y = 2x 2 and y = 1 2 x2 and y = x 2 + 3 and y = x 2 3 Graphing quadratic equations of the form f(x) = a(x h) 2 + k Given f(x) = a(x h) 2 + k f(x) has a vertex at (h, k) and axis of symmetry at x = h. If a > 0 opens upward, if a < 0 opens downward How to... graph, given f(x) = a(x h) 2 + k form (1) Plot vertex at the point (h, k) note: f(h) = k, so (h, k) is really (h, f(h)) (2) Determine if parabola opens upward or downward (3) If parabola crosses x-axis, find x-intercept(s) by solving f(x) = 0 (4) Compute f(0) to find y-intercept (5) Draw a nice arcing curve from vertex Ex: Graph (1) f(x) = 2(x 3) 2 + 8 (2) f(x) = (x + 3) 2 + 1 Graphing of the form f(x) = ax 2 + bx + c Start with f(x) = ax 2 + bx + c = a(x 2 + b a x) + c 27
and complete the square to get... f(x) = a(x + b 2a )2 + c b2 4a Similar to f(x) = a(x h) 2 + k?? Yes!! h = b and k = c b2 2a 4a Since h is the x-value of the vertex, the vertex is the point ( b 2a, f( b 2a )) How to... graph, given f(x) = ax 2 + bx + c form (1) Find vertex at ( b 2a, f( b 2a )) (2) Determine if parabola opens upward or downward (3) If parabola crosses x-axis, find x-intercept(s) by solving f(x) = 0 (4) Compute f(0) to find y-intercept (5) Draw a nice arcing curve from vertex Ex: Graph (1) f(x) = x 2 2x + 1 (2) f(x) = x 2 3x + 2 (3) f(x) = x 2 x + 3 8.5 : Polynomial and rational inequalities Ex: x 2 1 > 0 or x 2 1 < 0, draw graphs and show solutions Solving polynomial inequalities How to... (1) Express the inequality in the form f(x) > 0 or f(x) < 0 (2) Solve the equation f(x) = 0. The real solutions are boundary points. (3) Plot the boundary points on a number line, dividing the number line into intervals. (4) Choose a test value from each interval (a) If the number is positive, f(x) > 0 for the entire interval (b) If the number is negative, f(x) < 0 for the entire interval (5) Write the solution set including intervals that satisfy the given inequality Ex: Solve (1) x 2 7x + 10 (2) 2x 2 + x > 15 (3) 4x 2 1 2x (use quadratic eqtn) (4) x 3 + x 2 4x + 4 (5) x 3 + 3x 2 < x + 3 Solving rational inequalities 1 Ex: x > 0 or 1 < 0, draw graphs and show solutions x 28
How to... Exactly the same as solving polynomial inequalities, except we change step (2)... (2) Since f(x) is a rational function it can be written f(x) = p(x) q(x) (a) Solve p(x) = 0, solutions are boundary points (b) Solve q(x) = 0, these are also Ex: Solve (1) x + 3 x 7 < 0 (2) x 5 x + 2 < 0 (3) x + 1 x + 3 2 (4) 2x x + 1 1 9.2 : Function composition and inverse functions Function Composition Ex: f(x) = x 2 + 1 and g(x) = 3x + 5 then f(g(x)) is... The composition of a function f with another function g is written as (f g)(x) where Ex: Find (f g)(x) and (g f)(x) for (f g)(x) = f(g(x)) (a) f(x) = 2x 7 and g(x) = x + 3 (b) f(x) = x and g(x) = x + 3 (c) f(x) = 2x 2 x+1 and g(x) = x+2 (d) f(x) = x + 1 2 Inverse functions Ex: f(x) = 2x and g(x) = 1 2x are inverse functions since and g(x) = x+5 f(g(x)) = f( 1 2 x) = 2( 1 2 x) = x and g(f(x)) = g(2x) = 1 2 (2x) = x Definition: Given a function f(x) and a function g(x) that satisfies f(g(x)) = x and g(f(x)) = x for all values of x we call g(x) the inverse function of f(x). We use the notation f 1 (x) to denote the inverse of the function f(x). Using this notation, f(f 1 (x)) = x and f 1 (f(x)) = x Important! f 1 (x) does NOT mean 1 f(x) as in negative exponents. Ex: Verify that the following functions are inverse functions 29
(a) f(x) = 5x and g(x) = x 5 (b) f(x) = 3x + 2 and g(x) = x 2 3 How to... (find the inverse of a given function) (1) Replace f(x) with y (2) Interchange x and y (3) Solve for y (4) Replace y with f 1 (x) Ex: Find the inverse of (a) f(x) = 7x 5 (b) f(x) = 2x + 7 One-to-one functions and the horizontal line test The horizontal line test is a way to tell if a function is one-to-one. A function must be one-to-one in order to have an associated inverse function f 1. Why?? If a function isn t one-to-one then the inverse will not be a function! Ex: Which functions have inverses? (a) f(x) = x + 1 (b) f(x) = x 2 Graphs of inverse functions Ex: Graph function and inverse (a) f(x) = 7x 5 (b) f(x) = 2x + 7 9.1 : Exponential functions An exponential function f(x), with base b is f(x) = b x or y = b x where b > 0 and b 1 NOT exponential functions: f(x) = x 2, f(x) = 1 x, f(x) = ( 1) x, f(x) = x x Graphing exponential functions Ex: Graph using table (a) f(x) = 2 x (b) f(x) = ( 1 2 )x = 2 x then also draw graphs for f(x) = 7 x and f(x) = ( 3 4 )x then also draw graphs for f(x) = 3 x and f(x) = 3 x+1 (shifted to left one unit) 30
Compound interest Formula for compound interest: After t years, the balance A of an account with principle P and annual interest rate r (in decimal form) (1) For n compounding periods per year: A = P (2) For continuously compounding: A = P e rt Ex: You want to invest $8000 for 6 years, which choice is better? (a) One choice pays 7% per year, compounded monthly (b) The other pays 6.85% per year, compounded continuously 9.3 : Logarithmic functions Logarithms are the inverse functions for exponential functions! i.e. If f(x) = log b x and g(x) = b x then (f g)(x) = log b b x = x ( 1 + r ) nt n and (g f)(x) = b log b x = x Definition of logarithmic functions In the equation y = log b (x) the value y is the answer to the question To what power must b be raised, in order to get x? i.e. What value of y must satisfy the equation In these two equations we need b > 0, b 1. b y = x Ex: Change between logarithmic and exponential notation (1) 2 = log 5 x means (2) 3 = log b 64 means (3) log 3 7 = y means (4) 12 2 = x means (5) b 3 = 8 means (6) e y = 9 means Ex: Evaluate (1) log 2 16 (2) log 3 9 (3) log 25 5 Some properties of logs (1) log b b = 1 Why?? (b 1 = b) (2) log b 1 = 0 Why?? (b 0 = 1) 31
Graphing logs Ex: Graph f(x) = 2 x and g(x) = log 2 x using tables. How?? Do a table for f(x) and switch input/output in table for g(x). Properties of logarithmic functions (1) The domain of f(x) = log b x is (0, ), and the range of f(x) = log b x is (, ) (2) Notice for f(x) = log b x we have f(1) = log b 1 = 0, so x-intercept is always 1. There is never a y-intercept. (3) If b > 1 then f(x) = log b x is increasing function (4) If 0 < b < 1 then f(x) = log b x is decreasing function (5) The graph as a vertical asymptote at the y-axis Domain of logs The domain of f(x) = log b x is (0, ), why?? (look at exponential equation) Ex: Find domain (1) f(x) = log 4 (x + 3) (2) f(x) = log 2 (x 5) Note: If a log does not have a base then it is assumed to have base 10 i.e. log(2x) is log 10 (2x) Natural logarithms A logarithm with a base of e is a natural logarithm and is denoted f(x) = ln x i.e. f(x) = log e x = ln x Ex: Evaluate (1) ln 1 (2) ln e 7 (3) eln 15 9.4 : Properties of logarithms The properties of exponents correspond to the properties of logarithms as follows... The product rule for exponents: b m b n = b m+n corresponds to The product rule for logs: Ex: Expand each logarithmic expression log b (mn) = log b m + log b n (1) log 4 (7 5) (2) log(10x) (= 1 + log x) 32
The quotient rule for exponents: The quotient rule for logs: b m b n = bm n corresponds to ( m ) log b = log n b m log b n Ex: Expand each logarithmic expression ( ) ( ) 19 e 3 (1) log 7 (2) ln x 7 The power rule for exponents: (b m ) n = b mn corresponds to The power rule for logs: log b (m p ) = p log b m Ex: Expand each logarithmic expression (1) log 5 (7 4 ) (2) ln x (3) log(10x) ( (4) log b (x 2 3 ) x y) (5) log 6 36y 4 Ex: Condense each logarithmic expression (1) log 4 2 + log 4 32 (2) log(4x 3) log x (Note that coefficient out front must be 1 before you can proceed) (3) 1 2 log x + 4 log(x 1) (4) 3 ln(x + 7) ln x (3) 4 log b x 2 log b 6 1 2 log b y Change of base property: log b m = log a m log a b Here we are changing from a base of b to a base of a. Ex: (i) First change to a base of 10 (ii) then change to a natural logarithm (base e) (1) log 5 140 (2) log 6 17 3.1 : Systems of linear equations in two variables Solving systems of linear equations by graphing 33
Ex: Is (3, 4) a solution to this system? { x + y = 7 x y = 1 Ex: Is ( 4, 3) a solution to this system? { x + 2y = 2 x 2y = 6 Solving by graphing How to... (1) Graph em both. (2) If they intersect, find the point! (3) Check the point in both equations. Ex: Solve (1) (2) { 3x + 2y = 6 2x + y = 2 { y = 3x + 2 y = 5x 6 Systems with ly many and zero solutions Ex: One solution 34
f 4. 3. 2. 1. 5. 4. 3. 2. 1. 0 1. 2. 3. 4. 1. 2. 3. 4. Ex: Infinitely many solutions 5. 4. 3. 2. 1. 5. 4. 3. 2. 1. 0 1. 2. 3. 4. 1. 2. 3. 4. 5. 35
Ex: No solution 4. 3. 2. 1. 5. 4. 3. 2. 1. 0 1. 2. 3. 4. 1. 2. 3. 4. A system with no solution: 5. A system with infinitely many solutions: Solving systems using substitution How to... { y = 2x 1 y = 2x + 3 { 2x + y = 3 4x + 2y = 6 (1) Solve either equation for one of the variables. (2) Substitute the equation found in (1) into the other equation, in the end the equation only has one variable. (3) Solve for the variable, this is half your answer. (4) Substitute the result from (3) into equation from step (1). (5) Check the solution in both equations. Ex: Solve using the substitution method 36
(1) (2) { y = x 1 4x 3y = 24 { 5x 4y = 9 x 2y = 3 Recognizing (1) the no solution situation and (2) the infinitely many solutions situation. Ex: Same slope? Then no solution or infinitely many solutions. (1) Same slope but different line { y 5x = 4 y = 5x 1 (2) These are the same line { y = 2x + 3 4x + 2y = 6 Solving systems using addition method How to... (1) Re-write the equations in Ax + By = C format. (2) Multiply by a constant (if necessary) so that the sum of the x-coefficients or y-coefficients is zero. (3) Add equations from step (2), the sum should have only one variable. (4) Solve for the variable. (5) Substitute the result from (4) back into one of the original equations to find the other variable. Ex: Solve using the addition method (1) { y = x 1 4x 3y = 24 (2) { 5x 4y = 9 x 2y = 3 37
3.2 : Application questions using systems of equations How to... (1) What are you being asked to find? Label your variables. (2) Figure out how your variables are related and write two equations. (3) Solve for one variable, substitute back to find the other variable. (4) Make sure you answered the question, and check your answer. Ex: Fencing a waterfront lot Perimeter of rectangular lot is 1000ft. Front decorative fencing costs $25 / foot, fencing for two sides costs $5 / foot, no rear fencing along water. If the total cost of the fencing was $9500, what are the dimensions of the lot? Table: Cost per foot number of feet = total cost { 25x + 2(5y) = 9500 2x + 2y = 1000 Ex: A chemist needs to mix an 18% acid solution with a 45% acid solution to obtain 12 liters of a 36% acid solution. Table: Acid per liter number of liters = total acid { 18%x + 45%y = (36%)12 x + y = 12 Ex: Flying with the wind a dragon can travel 450 miles in 3 hours. When the dragon turns around and goes against the wind she can travel the same distance in 5 hours. Find the average speed of the dragon in still air, and the average speed of the wind. Table: Rate time = distance { 3(x + y) = 450 5(x y) = 450 Ex: You have two options to install as heating systems. The solar system costs $29,700 to install and costs $150 / year. The electric system costs $5,000 to install and costs $1100 / year. After how many years will the total cost of each system be the same. Table: { y = R(x) = 29, 000 + 150x y = S(x) = 5000 + 1100x 4.4 : Linear inequalities in two variables Ex: 2x 3y 6 Graphing using test points 38
How to... (1) Replace inequality with equality and graph. (2) Choose a test point that is not on the line. (3) If the inequality is true, shade that side of the line, otherwise shade the other side. Ex: Graph (1) 2x 3y 6 (2) x + 2y > 4 Graphing inequalities in the form y < mx + b and y > mx + b Ex: Graph (1) y > 2 x (2) y x + 1 (3) y 3 (4) x < 2 3 10.1 : Distance and midpoint; circles The Distance Formula uses the pythagorean theorem, with d 2 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2, to compute the distance between two points (x 1, y 1 ) and (x 2, y 2 ) Ex: Find the distance between the points d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 (1) (0, 1) and (0, 3) (2) ( 1, 4) and (3, 3) The Midpoint Formula computes the midpoint of a line segment connecting points (x 1, y 1 ) and (x 2, y 2 ) ( x1 + x 2 midpoint is, y ) 1 + y 2 2 2 x 1 + x 2 A good way to see this formula is that is the average of x 1 and x 2. 2 Ex: Find the midpoint of the line segment connecting the points (1) (0, 1) and (0, 3) (2) (1, 6) and ( 8, 4) Circles! Definition: A circle of radius r centered at the point (h, k) is the set of all points a fixed distance r from a single point (h, k). The standard form of the equation of a circle of radius r centered at the point (h, k) (x h) 2 + (x k) 2 = r 2 This is the distance formula used with the center point (h, k) and a point (x, y) on the circle. 39
Ex: Write the equation of a circle of radius r = 3 centered at the point (2, 3) Ex: Graph (1) (x 2) 2 + (y 3) 2 = 9 (2) (x + 1) 2 + (y + 3) 2 = 1 The more general form of the equation of a circle can be turned into the above standard form by completing the square Ex: Graph x 2 + y 2 + Dx + Ey + F = 0 (1) x 2 + y 2 + 4x 6y 23 = 0 (2) x 2 + y 2 + 4x 4y 1 = 0 10.2 : The ellipse! Definition: An ellipse is the set of all points (x, y) the sum of whose distances from two fixed points P 1 and P 2 is constant. These fixed points are called the Foci and between them lies the Center of an ellipse. The standard form of the equation of an ellipse centered at the origin x 2 a 2 + y2 b 2 = 1 with x-intercepts {(a, 0), ( a, 0)} and y-intercepts {(0, b), (0, b)}. Ex: Graph (1) x2 9 + y2 4 = 1 (2) 25x2 + 16y 2 = 400 10.3 : The hyperbola Definition: A Hyperbola is the set of all points (x, y) the difference of whose distances from two fixed points P 1 and P 2 is constant. These fixed points P 1 and P 2 are called the Foci and between them lies the Center of an ellipse. The standard form of the equation of a hyperbola centered at the origin x 2 a 2 y2 b 2 = 1 with x-intercepts {(a, 0), ( a, 0)} and no y-intercepts. Ex: Graph (1) x2 9 + y2 4 = 1 (2) 25x2 + 16y 2 = 400 40
10.4 : The parabola; identify conic sections Definition: A parabola is the set of all points (x, y) that are equidistant from a fixed line (the directix ) and a fixed point (the foci ) 10.5 : Systems of nonlinear equations in two variables What do solutions look like? Ex: A parabola intersecting a circle { y = x 2 x 2 + y 2 = 1 Substitution method Ex: Solve (A line intersecting a parabola) Elimination method Ex: Solve (Intersecting circles) { x 2 = 2y + 10 3x y = 9 { 4x 2 + y 2 = 13 x 2 + y 2 = 10 41