A BRIEF REVIEW OF ALGEBRA AND TRIGONOMETRY Some Key Concepts: 1. The slope and the equation of a straight line 2. Functions and functional notation 3. The average rate of change of a function and the DIFFERENCE- QUOTIENT OF A FUNCTION 4. Domains of some common types of functions 5. Graphs of functions and piecewise functions 6. Expanding, factoring, and simplifying expressions 7. Solving algebraic equations 8. Basic trigonometry 9. Solving trigonometric equations 1 1. The slope and the equation of a straight line: (a) Slope of a straight line Y y2 Q x2,y2 Dy=y2-y1 P x1,y1 y1 Dx=x2-x1 X x1 x2 rise change of y y y y Slope of the line PQ m 2 1 run change of x2 x1 x Four different kinds of lines and their slopes: Y Negative Positive Zero Undefined X (b) Equation of a straight line The equation of a straight line can be written in any one of the following three ways: (i) The Point-Slope Form The equation of a straight line that passes through the point (x1 , y1 ) and having slope m is given by 2 y y1 m(x x1) (ii) The Slope-Intercept Form The equation of a line with slope m and having the y-intercept (0, b) is given by y mx b (iii) The Standard Form In this form, you move all x and y terms of the equation of the given line, to one side, and the constant term to the other side. It is normally written as Ax By C Example: 1. Find the equation of the line that passes through the points (1, 2) and (2, 9) . Provide the exact answer in the standard form with integral coefficients. Solution: Make sure to draw your own sketch (not included here). Let P(x1 , y1 ) be (1, 2) and Q(x2 , y2 ) be (2, 9) . Then the slope m of the line PQ is given by m (y2 y1 ) /(x2 x1 ) (9 2) /(2 1) 7 / 3 The equation of this line PQ can be found either by using the point-slope form or by the slope-intercept form. For example, using the point-slope form y y1 m(x x1 ) , its equation is given by, 7 y 2 (x 1) 3 Multiply both sides by 3 to get: 3(y 2) 7(x 1) 3y 6 7x 7 In the standard form, the equation is: 7x 3y 13 (c) Parallel and perpendicular lines: If two non-vertical lines are parallel, then they have equal slopes. In other words, if m1 and m2 are the slopes of the parallel lines, then m1 m2 . On the other hand, if two-non vertical lines are perpendicular, then the slope of one line is equal to the negative reciprocal of the slope of the other. In other words, if m1 and m2 are the slopes of the perpendicular lines, then m1 1/ m2 . 3 Example: 1. Find the equation of the line that is perpendicular to the line 2x 3y 4 and which passes through the point (1,3) . Provide the exact answer in the slope-intercept form. Solution: Make sure to make a brief sketch – coordinate axes are not necessary (not included). First, find the slope of the given line 2x 3y 4. To do this, arrange this equation in the form y mx b (i.e. solve for y) and look for the coefficient of x: 2 4 2x 4 3y; 3y 2x 4; y x 3 3 Therefore, the slope of the given line is 2/3. So, the slope of a line perpendicular to the original line is the negative reciprocal of 2/3, i.e. 1/(2 / 3) or 3/ 2. Now, to find the equation of the required perpendicular line, use the point-slope form with m 3/ 2 and (x1 , y1 ) (1, 3) : 3 y 3 (x 1) 2 To put in the slope-intercept form, solve the above equation for y: 3 3 y x 2 2 2. Functions and the functional notation: Quite often, a function can be given by an equation, such as y 2x 2 3x 4 . The same function can be re-written as f (x) 2x 2 3x 4 , which is called the functional notation. Thus, f (x) is just another name for “y”. Thus, as a specific example,f (3 ) means the y-value of the function when x is equal to 3. So, this can be found by substituting each x-variable in the expression 2x 2 3x 4 by “3”. Therefore, we have f (3) 2(3) 2 3(3) 4 18 9 4 13 f (3) 13 Note that f (3) 13 just means that, when x 3 , the y-value of the function f is equal to 13. 4 Examples: 1. Given f (x) 2x Sin(x) , find the exact value of f ( / 3) . Solution: 2 3 4 3 3 f 2 Sin 3 3 3 3 2 6 4 3 3 f 3 6 2. Given f (x) 3x 2 4x 1, find x if f (x) 21. Solution: DO NOT PLUG IN x 21 in the given equation, as that would be a major mistake! Rather set f (x) 21, and solve the resulting equation for x, as follows: 3x 2 4x 1 21 3x 2 4x 20 0 The above is a quadratic equation. BE VERY FAMILIAR with the methods of solving quadratic equations and other equations. For example, one can use the Factoring Method or the Quadratic Formula x (b b 2 4ac) /(2a) . Let us use the Factoring Method, since the right-hand side of the last equation can be found to be factorable: (3x 10)(x 2) 0 Now use the zero-product property: 3x 10 0 or x 2 0 x 10 / 3 or x 2 Thus, there are two x-values for which f (x) 21. 3. VERY IMPORTANT EXAMPLE: Given f (x) 3x 2 4x 1, find f (x h) . Expand and simplify your answer. Solution: This problem illustrates one of the subtle points of functional notation: DO NOT simply add “h” to the right-hand side of the equation to obtain 3x 2 4x 1 h as the answer for f (x h) . In fact, 3x 2 4x 1 h is equal to f (x) h , which is quite different from the desired f (x h) . In order to find f (x h) , replace all the x-terms on the right hand-side of the given equation by x h : f (x h) 3(x h) 2 4(x h) 1 5 Now use rules of algebra such as the Foil Method or special squaring formulas to expand the right-hand side: f (x h) 3(x 2 2xh h 2 ) 4x 4h 1 3x 2 6xh 3h 2 4x 4h 1 f (x h) 3x 2 6xh 3h 2 4x 4h 1 Just to summarize, given a function f (x) , the two quantities f (x h) and f (x) h are totally different! For example, for the function f (x) 1 x , we can find these two quantities as follows: Given : f (x) 1 x f (x h) 1 (x h) 1 x h But f (x) h 1 x h In general, f (x h) f (x) h 3. Average rate of change of a function and the difference-quotient: (a) The average rate of change of a function Y y=f x Q f x2 Dy=f x2 -f x1 P f x1 Dx=x2-x1 X x1 x2 In the above diagram, when the x-value is changed from x1 to x2 , the y-value of the function f (x) changes from y1 f (x1 ) to y2 f (x2 ) . Then, the change of x, or the increment of x, is given by x2 x1 , and is denoted by x . Similarly, the change of y, or the increment of y, is given by y2 y1 f (x2 ) f (x1 ) , and is denoted by y . 6 We define the average rate of change of f (x) from x x1 to x x2 as the ratio y / x . It is very important to observe that, in the above diagram y / x is equal to the slope of the secant line PQ. A secant line of a curve is a line joining any two points of the curve. Thus we have: The average rate of change of f (x) from x x1 to x x2 is eqaul to any of y y y f (x ) f (x ) 2 1 2 1 slope of the secant line PQ x x2 x1 x2 x1 Examples: 1. Find the average rate of change of the function f (x) Cos 2 (2x) from x 0 to x / 3 . Solution First make sure to draw your own diagram to see the corresponding secant line (not included). The graphical view point is very essential for a successful study of especially calculus. Let x1 0 and x2 / 3. Then the average rate of change of f (x) from x x1 to x x2 is given by: f (x ) f (x ) f ( / 3) f (0) Cos 2 (2 / 3) Cos 2 (0) (1/ 2) 2 (1) 2 (3/ 4) 9 2 1 x2 x1 ( / 3) 0 ( / 3) ( / 3) ( / 3) 4 (b) The difference-quotient of a function In part (a), we just defined the average rate of change of a function f (x) from x x1 to x x2 . The difference quotient of a function is just a special case of the average rate of change – in other words, the difference-quotient is just the average rate of change of f (x) from x to x h .