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# ALGEBRA & TRIG FOR CALCULUS MA 110

UK

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This 72 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 110 at University of Kentucky taught by Jakayla Robbins in Fall. Since its upload, it has received 11 views. For similar materials see /class/228143/ma-110-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15

Table 39o39f contents Cartesian Coordinate System Distance Q Midpoints Graphs g Circles 6 Lines El 539 39 E QQCV Axes Quadrantes Coordinates Your personal review 0 Locate points on a Cartesian Coordinate System 2 Discuss the four quadrants of a Cartesian Coordinate System J Identify the X 7 axis the y 7 axis and the origin on a Cartesian Coordinate System Quick Review Every point one the X axis has A an X coordinate that equals 0 B a y coordinate that equals 0 C Both A and B QQCV Quick Review Every point one the X axis has A an X coordinate that equals 0 B a y coordinate that equals 0 C Both A and B ANSWER A Distance Example Use the Pythagorean Theorem to find the distance between the points 275 and 7372 QQCV A Divsit ce Example Use the Pythagorean Theorem to find the distance between the points 275 and 7372 Cr qo A Distance Example Use the Pythagorean Theorem to find the distance between the points 25 and 7372 A Distance Example Use the Pythagorean Theorem to find the distance between the points 25 and 7372 A Distance Example Use the Pythagorean Theorem to find the distance between the points 25 and 7372 ANSWER v52 72 The Distance F rlnula Use the Pythagorean Theorem to find thedistance between the points x1y1 and X27y2 QQCV The Distance F rlnula Use the Pythagorean Theorem to find thedistance between the points x1y1 and X27y2 X27 2 QQCV The Distance Formula Use the Pythagorean Theorem to find the distance between the points thl and X27y2 X27 YQ X17 Y1 The Distance Formula Use the Pythagorean Theorem to find the distance between the points thl and X27y2 4 I I y YZ4 Y1 The Distance Formula Use the Pythagorean Theorem to find the distance between the points thl and X27y2 X27 YQ D 1 4 gt12 X1 12 X17Y1 The Distance Formula Use the Pythagorean Theorem to find the distance between the points thl and X27y2 X27 YQ D 1 4 gt12 X1 12 X17Y1 ANSWERZ X1 XZ Z Y1 MP X1 X22 Y1 n The Distance Formula Theorem The Distance Formula The distance between the points X17 yl and x27y2 is X1 7 X22 n 7 n QQCV Mic lpo in39ts 72 0 to 05 Is the distance from 27 0 to 07 5 equal to the distance from QQCV Midpoints Is the distance from 27 0 to 07 5 equal to the distance from 727 0 to 05 ANSWER Midpoints Is the distance from 27 0 to 07 5 equal to the distance from 727 0 to 05 ANSWER Is 07 5 the midpoint of the line segment from 20 to 727 O Midpoints Is the distance from 27 0 to 07 5 equal to the distance from 727 0 to 05 ANSWER Is 07 5 the midpoint of the line segment from 20 to 727 O ANSWER Mic lpo in ts To prove that 97 f is the midpoint39of the line segment between 37 b and c7 d we need to show QQCV Midpoin39ts39 To prove that e7 f is the midpoint of the line segment between 37 b and c7 d we need to show 1 The distance between 37 b and e7 1 equals the distance between c d and e7 f AN QQCV M idin n39ts To prove that e7 f is the midpoint of the line segment between 37 b and 27 d we need to show 1 The distance between 37 b and e7 1 equals the distance between 67d and sf AN 2 e f is on the line segment between 37 b and 67 d QQCV llvj liicdlipminits gaming lg l jil To prove that e7 f is the midpoint of the line segment between 37 b and c7 d we need to show imam 1 The distance between 37 b and e7 f equals the distance between c7 d and e7f AND 2 e7 f is on the line segment between 37 b and c7 d We can show the e7 f is on the line segement between 37 b and c7 d by showing that the distance between 37 b and e7 f plus the distance between c7 d and e7 f equals the distance between 37 b and c7 d Think about triangles Midpoint Formula Theorem The Midpoint Of the line segment from X17Y11 to XQJZ is 272 QQO Proof of the Midpoint Formula What is the distance from X17y1 to my What is the distance from an to Xliz xg What is the distance from thl to X27y239 How are these distances related El 539 39 E QQCV l39Sgolu tions of Equations and Graphs The graph of an equation is the Set of all equation solutions of the Is 2 3 on the graph of y 3X l 5 s 18 on the graph of y 3X l 57 0 30 l39Sgolu tions of Equations and Graphs The graph of an equation is the Set of all solutions of the equation Is 2 3 on the graph of y 3X l 5 ANSWER NO ls 18 on the graph of y 3X l 5 E OQCV Solutions of Equations and Graphs The graph of an equation is the Set of all solutions of the equation Is 27 3 on the graph of y 3X l 5 ANSWER NO ls 18 on the graph of y 3X l 5 ANSWER YESl E OQO Graphs of Circl es Find an equation for the circle with center 725 and radius 4 QQCV Graphs of Circl es Find an equation for the circle with center 725 and radius 4 ANSWER X 22 i y 7 52 16 QQCV Graphs of Circl es Find an equation for the circle with center X17y1 and radius r QQCV Graphs of Circl es Find an equation for the circle with center X17y1 and radius r ANSWER X 7 X12 i y 7y12 r2 QQCV Circles through the point 477639 Find an equation for the circle39with center 271 that passes QQCV Ci rcles Find an equation for the circle39with center 271 that passes through the point 476 ANSWER lt223912 53 eqo Circles an equation for the circle A diameter of a circle has endpoints 25 and 471 Find QQCV Ci rcles A diameter of a circle has endpoints 27 5 and 471 Find an equation for the circle ANSWER x 7 32 y 7 22 10 QQCV Ci rcles its center and radius Is the graph of X2 10xy27 6y 32 O a circle If so find ANSWER YESL Center753 Radius QQCV Circles its center and radius Is the graph of X2 10xy27 6y l 34 O a circle If so find ANSWER This graph is the single point 753 Sometimes people call this a degenerate circle since the radius equals zero QQCV Circles its center and radius satisfy this equation is the graph of X2 10xy27 6y i 40 O a circle If so find ANSWER This graph is empty No real values of X and y QQCV Mountain Climbing Consider the mountain below Is it equally steep everywhere QQCV Mountain Climbing What kind of mountain would be equally steep everywhere QQCV Mountain Climbing What kind of mountain would be equally steep everywhere QQCV The steepness of a line is determined by comparing the change in the vertical distance to the change in the horizontal distance The ratio of the change in vertical distance to the change in horizontal distance is constant for a line For a line this ratio is called the slope of the line If X1 X2 then the le pe of the line through the points X17y1 and X27y2 is E n Y1 AX X2 7X1 El 539 39 3 ORG The slope of a line is a rate of change If the line is in terms of the variables x and y then the slope of the line is the rate of change of y with respect to X This means that the slope is the ratio of the change in y to the change in X The word per is often associated with rates of change Think about speed Speed can be measured in miles per hour Speed is a rate of change of distance with respect to time El 539 39 E OCH R39atge orchange A particle is traveling along a straight line Its position 5 at time 1 seconds is given by s 601 where s is measured in feet 1 Is this a linear equation 2 What is the slope of this equation 3 What are the units on the slope of the equation 4 Express the slope of the equation as a rate ofchange QQCV R atge atChange A particle is traveling along a straight line Its position 5 at ANSWER Yes time 1 seconds is given by s 601 where s is measured in feet 1 Is this a linear equation 2 What is the slope of this equation 3 What are the units on the slope of the equation 4 Express the slope of the equation as a rate ofchange QQCV R atge atChange A particle is traveling along a straight line Its position 5 at time 1 seconds is given by s 601 where s is measured in feet 1 M A 4gt Is this a linear equation ANSWER Yes What is the slope of this equation ANSWER 6O What are the units on the slope of the equation Express the slope of the equation as a rate ofchange QQCV R39a39tge Oi Ch39ahge A particle is traveling along a straight line Its position 5 at time 1 seconds is given by s 601 where s is measured in feet 1 Is this a linear equation ANSWER Yes What is the slope of this equation ANSWER 6O 3 What are the units on the slope of the equation ANSWER feetsecond Express the slope of the equation as a rate ofchange M 4gt n 5 E ciao A particle is traveling along a straight line Its position 5 at time 1 seconds is given by s 601 where s is measured in feet 1 Is this a linear equation ANSWER Yes 2 What is the slope of this equation ANSWER 60 lm 3 What are the units on the slope of the equation ANSWER feetsecond 4 Express the slope of the equation as a rate of change ANSWER The rate of change of the position with respect to time is 60 feetsec R39atge o fuChange If the radius r of a circle is measured in meters then the circumference C of the circle in meters is given by C 27rr 1 Is this a linear equation 2 What is the slope of this equation 3 What are the units on the slope of the equation 4 Express the slope of the equation as a rate ofchange E ego R39atge o fuChange If the radius r of a circle is measured in meters then the circumference C of the circle in meters is given by C 27rr l N La 4gt Is this a linear equation ANSWER Yes What is the slope of this equation What are the units on the slope of the equation Express the slope of the equation as a rate ofchange QQCV R atge atChange If the radius r of a circle is measured in meters then the circumference C of the circle in meters is given by C 27rr l N La 4gt Is this a linear equation ANSWER Yes What is the slope of this equation ANSWER 27r What are the units on the slope of the equation Express the slope of the equation as a rate ofchange QQCV R39a39tge Oi Ch39ahge If the radius r of a circle is measured in meters then the circumference C of the circle in meters is given by C 27rr l N 9 4gt Is this a linear equation ANSWER Yes What is the slope of this equation ANSWER 27r What are the units on the slope of the equation ANSWER meters of circumferencemeters of radius Express the slope of the equation as a rate ofchange 0 go imam 39 Mae1H4 If the radius r of a circle is measured in meters then the circumference C of the circle in meters is given by C 27rr 1 Is this a linear equation ANSWER Yes 2 What is the slope of this equation ANSWER 27r 3 What are the units on the slope of the equation ANSWER meters of circumferencemeters of radius 4 Express the slope of the equation as a rate of change The rate of change of the circumference of a circle with respect to the radius of the circle is 27139 l was Linear Equations slope 72 Find an equation for the line that passes through 47 3 and has QQCV Linear Equations slope 72 Find an equation for the line that passes through 47 3 and has ANSWER y 7 3 72X 7 4 QQCV Important Things Ab CT the most use form for a j Slopeintercept form is M line Important Things Ab j Slopeintercept form is NOT the most use form for a line if Point slope form is often more useful than slopeintercept form Learn it Important Things Ab j Slopeintercept form is NOT the most use form for a line if Point slope form is often more useful than slopeintercept form Learn it 3 Parallel lines have the same slope Important Things Ab f Slopeintercept form is lllC T the most use form for a line if Point slope form is often more useful than slopeintercept form Learn it 3 Parallel lines have the same slope C1 The product of the slopes of perpendicular lines is 71 Important Things Ab f Slopeintercept form is lllC T the most use form for a line if Point slope form is often more useful than slopeintercept form Learn it 3 Parallel lines have the same slope C1 The product of the slopes of perpendicular lines is 71 I The slope of vertical lines X a is undefined Important Things Ab C Slopeintercept form is lllC T the most use form for a line Point slope form is often more useful than slopeintercept form Learn it 3 Parallel lines have the same slope C1 The product of the slopes of perpendicular lines is 71 I The slope of vertical lines X a is undefined A The slope of horizontal lines y a is zero Important Things Ab if Slopeintercept form is NOT the most use form for a line if Point slope form is often more useful than slopeintercept form Learn it 0 Parallel lines have the same slope C1 The product of the slopes of perpendicular lines is 71 I The slope of vertical lines X a is undefined L3 The slope of horizontal lines y a is zero To find the equation of a line you need the slope and a point Linear Equations Find an equation for the line that is perpendicular to 2X 3y 7 5 O and passes through the point 2771 0 30 Linear Equations Find an equation for the line that is perpendicular to 2x 3y7 5 O and passes through the point 271 3 ANSWER y 1 5X7 2 QQCV NAME SOLVING EQUATIONS ALGEBRAICALLY MA 110 l Principles for Solving Equations Algebraically A Algebraically or Graphically In this class7 we will solve equations in two di erent ways Algebraically and Graphically The only way to guarantee an exact solution is to solve the equation alge braically Solutions that are obtained by graphical means are approxirnations Whether you are solving an equation algebraically or graphically7 you should AL WAYS CHECK potential solutions to ensure that they actually are solutions B Sorne Strategies Two equations are equivalent if they have the same solutions When you are solving an equation7 you want to move from one equation to an equivalent equation that is easier to solve For exarnple7 2x53z71 iz571 7p76 z6 Notice that z 6 is the only solution for ALL FOUR equations C Equivalent Equations The following operations produce equivalent equations Add or subtract the same number to both sides of the equation Add or subtract the same algebraic expression to both sides of the equation Multiply or divide both sides of the equation by a NONZERO nurnber Add zero to one side of the equation afterglow Multiply one side of the equation by 1 Example 1 TRUE or FALSE Does squaring both sides of an equation always produce an equivalent equation Example 2 TRUE 0r FALSE Does multiplying both sides of an equation by z always produce an equivalent equation ldeally7 we would like to keep equivalent equations as we move from one step to the next in the solution This is not always possible Sometimes you need to square both sides of the equation Sometimes you need to multiply both sides of an equation by an algebraic expression instead of a number These operations can produce extraneous solutions This is why it is important to CHECK YOUR SOLUTIONS Solving Equations with One Variable If there is only one variable in the equation7 you want to unwrap77 the variable until it is by itself on one side of the equation Think about what happens to that variable Unwrap one operation at a time by performing the inverse operation For example7 suppose we want to solve 2x 7 7 7 T 7 8 What is happening to x Divide bv 5 Subtract7 Multil b 2 To solve this equation we should do the following 1 Multiply both sides of the equation by 5 2 Add 7 to both sides of the equation 3 Divide both sides of the equation by 2 Example 3 Solve for s Example 4 Solve for s 2i 6 3 75 5 8 12 E Solving Equations with Variables in the Denominator When an equation has denominators 1 Find a common denominator for all of the fractional expressions in the equa tion Multiply both sides of the equation by the common denominator Solve the new equation Be CAREFUL The new equation may not be equivalent to the original equation Why You may nd some mtmneous solutions when you solve the new equation Check all of your solutions in the original equation Keep only those solutions that are solutions of the original equation Example 5 Solve for y E Solving Power Equations The real solutions of x a isare o z Wifnisodd o zWandx7Eifnisevenanda20 If n is even and a lt 07 then x a does not have any real solutions G What do I do when the variable terms are not the same type What do you do when you have an x and an 2 or a y and a or a p and a When this happens7 there is often no way to nd an equivalent equation which has only one variable We must resort to other techniques Certain special types can still be solved algebraically7 especially those equations that have quadratic type But there are many more equations that cannot be solved algebraically In a few days7 we will learn how to approximate the solutions of these equations geornet rically H Quadratic Equations A quadratic equation in z is any equation that is equivalent to an equation of the form of bx c 0 with a 31 0 2x2 3x 5 0 is a quadratic equation in x 5u2 6U 2 is a quadratic equation in u 4 2 2 7 is a quadratic equation in 2 2x 3 0 is not a quadratic equation 1 7 2 7 2 0 is not a quadratic equation Property 1 Zero Product Property IfAB 0 therz A 0 or B 0 Example 6 Use the Zero Product Property to solve 6x2 17x 5 0 Completing the Square Rewrite 2 bx c in the form x D2 A Example 7 Solve 6x2 17x 5 0 by completing the square Theorem 1 The Quadratic Formula The solutions of azz bx c 0 are i ibi Vb 7 4ac 7 2a gt lt b2 7 4ac is called the discriminant If the discriminant is positive7 the equation has two solutions If it is zero7 the equation has one solution If it is negative7 the equation does not have any real solutions Example 8 Use the Quadratic Formula to solve 6x2 17x 5 0 Quadratic Type Equations Sorne equations have the form auz bu 0 0 where u is an algebraic expression To solve these equations 1 Look for an expression and its square 2 Let u be the expression 3 Substitute u for the expression and u2 for the square of the expression The only variable in the new equation should be u None of the original variables should remain 4 Solve the new equation for u 5 In the solution of the new equation7 substitute the original expression for u This will contain the original variable 6 Solve for the original variable 7 CHECK YOUR SOLUTIONS Example 9 Solve for x z 7 2x2 7 3 0 Example 10 Solve for t 2 i 8 t Example 11 Solve forz

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