A Hitchhiker s Guide to Precalculus

Transcription

1 A Hitchhiker s Guide to Precalculus Trent Dean Glen Allen HS, Virginia

2 Contents Trigonometry Unit Circle... Inverse Trig... 4 Trig Identities... 5 Trig Equations... 6 Graphs of Trigonometric Functions... 7 Algebra Rational Functions Even/Odd Functions... 9 Piecewise Functions Compositions of Functions Exponents & Logarithms Inverses... 1 Binomial Expansion (Pascal s Triangle)... 1 Factoring... 1 Systems of Equations... 1 Graphing & Transformations Algebraic Atrocities Conics Slope & Linear Equations Formulas Calculator Techniques (TI-84) Notes & Forward The purpose of A Hitchhiker s Guide to Precalculus is to help students review important concepts from their previous Precalculus course as they prepare for AP Calculus AB. Though not extensively long, this primer is merely a quick refresher for what students should have already seen in great detail. The list of topics and examples is comprehensive, yet I have left out several key topics for the sake of brevity. The following were left out because they are more integral for a BC Calculus course, and will not have much impact in AB Calculus: Sequences & Series Parametric & Polar Equations Vectors Matrices Students should be exposed to these concepts in Math Analysis or Precalculus in order to ensure success in future Calculus courses (BC/Calculus and Calculus ). An argument might be made to have included Series because of its relation to Riemann Sums, but I usually don t discuss Sequences and Series with much depth in AB Calculus when setting up integrals. There are other concepts from Precalculus that I have left out for the sake of brevity and significance: Synthetic Division/Rational p s & q s Complex Numbers Limits & Tangent Lines In all my years of teaching AP Calculus, I have not seen Synthetic Division on any version of the AP exam, and am still baffled by Precalculus teachers who spend more than a class period on it. I never learned synthetic division until I was a math teacher, and I did not need to! Complex numbers also have no bearing in AB Calculus since we only work with the real number line. It is good to learn it in Algebra, but there are more important concepts to spend time on in Precalculus. Limits and Tangents are vital to our understanding of Calculus, and for that reason I start the school year with Limits; I left it out of here to keep this shorter even though most students cover the beginning of Derivatives in their Precalculus classes. Calculus is such an awesome subject because it is a new horizon in math that will open up new ways for you to relate to the world around you which is very powerful! Everything a student has done from middle school through Precalculus has been a subset of Algebra simplifying expressions, solving equations, and graphing functions. Calculus now takes that, puts it in motion, and opens you up to a whole new type of math that deals with change. Students typically struggle with two things in Calculus: the new type of thinking, and adequately remembering and applying previously learned material. Hopefully, this primer will help you with the latter in your journey through the discovery of Calculus. Trent Dean tgdean@henrico.k1.va.us

3 Trigonometry Unit Circle Purpose: Use to quickly recite sine and cosine values. Use to also find tangent, cosecant, secant, and cotangent values. A unit circle is simply a circle with a radius = 1. Therefore if we draw corresponding reference angles, the hypotenuse of each resulting triangle has a length of 1. To use: Memorize the position of all angle measures in radians (use patterns). The sine values (opposite/hypotenuse) pertain to the corresponding y-values sin ( π 6 ) = opp hyp = 1/ 1 = 1 The cosine values (adjacent/hypotenuse) pertain to the corresponding x-values cos ( π ) = adj = / = 6 hyp 1 Use trig identities to find tangent, cosecant, secant, and cotangent values. Examples: sin ( π ) = cos ( 5π 6 ) = sin ( 7π 6 ) = 1 cos ( 5π ) = 1 tan ( π 6 ) = sin(π/6) cos(π/6) = 1 sin ( π ) = 1 = 1 csc ( π 4 ) = cos (7π 4 ) = 1 sin(π/4) = 1 =

4 Inverse Trig Values Inverse Sine Inverse Cosine Inverse Tangent sin 1 x or arcsinx cos 1 x or arccosx tan 1 x or arctanx Domain: [ 1,1] Range: [ π, π ] Domain: [ 1,1] Range: [0, π] Domain: (, ) Range: ( π, π ) When you evaluate an inverse trig function, you get back an angle (in radians). There is only one correct output, and make sure that output correctly corresponds to the range of that function. Pay careful attention to the range of each function. Examples Solution Explanation 1. arcsin ( ) π Since sin π =, therefore arcsin ( ) = π. Clarification: Even though sin ( π ) also equals, a function may only have one output. The range of inverse sine is [ π, π ]. π is outside this range. Therefore the (only) answer is π.. arccos ( 1 ) π. tan 1 ( ) π Both cos π & cos 4π = 1, but π is in the range of inverse cosine. Thus arccos ( 1 ) = π For negative inputs in inverse tangent, look at quadrant IV. tan ( π ) =, thus tan 1 ( ) = π. More examples (without explanations): arcsin ( 1 ) = π 6 arcsin ( ) = π sin 1 ( 1) = π arccos ( 1 ) = π cos 1 ( ) = 5π 6 arccos( 1) = π tan 1 ( ) = π arctan(1) = π 4 tan 1 ( ) = π 6 Example (Composite Trig Inverse): Find sin 1 (sin 7π 6 ) Solution: π 6 Explanation: Be very careful! Work from the inside-out, and mind the range of inverse sine. Do not just cancel the inverses out! Sin 7π 6 = 1, but sin 1 ( 1 ) must be π 6 since π 6 it is between π and π. 4

5 Trigonometric Identities Reciprocal Identities Quotient Identities Pythagorean Identities Even/Odd Identities sin θ = 1 cscθ csc θ = 1 sinθ tan θ = sin θ cosθ sin θ + cos θ = 1 sin( θ) = sinθ cos θ = 1 secθ tan θ = 1 cotθ sec θ = 1 cosθ cot θ = 1 tanθ cot θ = cos θ sinθ 1 + cot θ = csc θ tan θ + 1 = sec θ cos( θ) = cosθ tan( θ) = tan θ Sum and Difference Formulas sin(x ± y) = sin x cos y ± cos x sin y cos(x ± y) = cos x cos y sin x sin y Double-Angle Formulas sin(θ) = sin θ cos θ cos(θ) = cos θ sin θ = cos θ 1 = 1 sin θ Power-Reducing Formulas sin θ = cos θ = 1 cos θ 1 + cos θ Example 1: Verify tan x + cot x = sec x csc x tan x + cot x = sin x cos x + cos x sin x = sin x + cos x cos x sin x = 1 cos x sin x = 1 cos x 1 sin x Start with left side, quotient identities Add fractions by getting common denominators Pythagorean identity Product of fractions Example : Verify tanθ 1 cos θ = = sin θ cos θ sin θ tanθ 1 cos θ sin θ cos θ 1 sin θ = 1 cos θ 1 sin θ = sec θ csc θ = secθ cscθ Quotient identity & Pythagorean identity Dividing fractions rewritten sin θ s cancel Reciprocal identities = sec x csc x Reciprocal identities Example : Derive first two Double-Angle Formulas for cos(θ) cos(θ) = cos(θ + θ) = cos θ cos θ sin θ sin θ = cos θ sin θ cos θ sin θ = cos θ (1 cos θ) = cos θ 1 + cos θ = cos θ 1 Start by recognizing that θ is the sum of θ + θ. Sum identity for cosine Products > Identity #1 Start with Identity #1; variant of Pythagorean identity Distribute Simplify > Identity # Guidelines for Verifying Trig Identities: 1) Work with each side of the equation independently of the other side. Start with the more complicated side and transform it in a stepby-step fashion until it looks exactly like the other side. ) Analyze the identity and look for opportunities to apply the fundamental identities. Rewriting the more complicated side of the equation in terms of sines and cosines is often helpful. ) If sums or differences of fractions appear on one side, use the least common denominator and combine the fractions 4) Don t be afraid to stop and start over again if you are not getting anywhere. Creative puzzle solvers know that strategies leading to dead ends often provide good problem-solving ideas. 5

6 Trigonometric Equations To solve a trigonometric equation, use standard algebraic techniques such as collecting like terms and factoring. Your preliminary goal in solving a trigonometric equation is to isolate the trigonometric function in the equation. Example 1: sin x 1 = 0 sin x = 1 sin x = 1 To solve for x, note that the equation sin x = 1 has solutions x = π/6 and x = 5π/6 in the interval [0, π). Moreover, because sin x has a period of π, there are infinitely many other solutions, which can be written as x = π 6 + nπ and x = 5π 6 + nπ where n is an integer, as shown in the graph below. Example : tan x 1 = 0 tan x = 1 tan x = 1 tan x = ± 1 ± Here we isolated tan x to one side of the equation. Because tan x has a period of π, first find all solutions in the interval [0, π). These are x = π/6 and x = 5π/6. Finally, add nπ to each of these solutions to get the general form x = π 6 + nπ and x = 5π 6 + nπ where n is an integer. Another way to see that the equation sin x = 1 has infinitely many solutions is indicated in the figure below. Any angles that are coterminal with π/6 or 5π/6 are also solutions of the equation. Example : Find all solutions of sin x sin x 1 = 0 in the interval [0, π). 6

7 Graphs of Trigonometric Functions Parent Graphs (x-axis in intervals of π/, y-axis intervals of 1): y = sin x y = cos x y = tan x period: π period: π period: π y = Asin[B(x C)] + D y = Acos[B(x C)] + D Transformations to Sine and Cosine graphs A Amplitude half the distance between the max and min values of the function. Negative values of A flip the graph. Examples (parent graph appears thinner) y = sin x y = 1 cos x y = sin x y = cos ( π 4 x) Period shift. period = π B B The period is the distance of one cycle. Period: π Period: 8 y = sin (x π ) y = cos(x + π) C Horizontal Shift Shifted π units to the right y = sin x Shifted π units to the left y = cos x + 1 D Vertical Shift Shifted down units Shifted up 1 unit 7

8 Rational Functions Algebra 1. Find all horizontal (or oblique) asymptotes. Let n be degree of numerator, d be degree of denominator xn + x d + asymptotes Example function Asymptote n < d Horizontal: y = 0 f(x) = 1 x + 1 y = 0 n = d Horizontal: y = the ratio of leading coefficients f(x) = x + 7 x + x + 1 y = n is one greater than d Oblique/Slant: y = the quotient, in the long division of the fraction f(x) = x + x + 1 x y = x + 1. Find vertical asymptotes or holes If factor in denominator Does not cancel out with a factor in the numerator Cancels out with a factor in numerator You have Vertical asymptote where x-value(s) make denominator equal 0. Hole where x-value is x-coordinate of the hole. y = 1 x + Vertical asymptote is x = (x + 1)(x + ) y = x + Hole at x = Use simplified form to get coordinate: (, ). Use key points to determine shape of the graph and location of curves y-intercept x-intercepts Try points to the left and right of asymptotes & holes. Plug in zero for x Using simplified form, Set numerator equal to zero, solve for x. This ensures that your curve is correctly above or below horizontal asymptotes. Example 1: f(x) = x x 1 x x 8 1) Horizontal Asymptote: y = 1 Factor: f(x) = (x+)(x 4) (x 4)(x+) ) Vertical Asymptote: x = ) Hole: x =4; 4) y-intercept: (0, ) 5) x-intercept: (, 0) 6) Additional points: x f(x) 4/ 1/ (x+)(x 4) (x 4)(x+) x+ x = 7 6. The hole is at (4, 7 6 ) 8

9 Example : f(x) = x x 4x+ 1) Horizontal Asymptote: y = 0 Factor: f(x) = 1(x ) (x )(x 1) ) Vertical Asymptote: x = 1 ) Hole: x =; 4) y-intercept: (0,1) 5) x-intercept: none 6) Additional points: x 4 1 f(x) 1 1/ 1/ 1(x ) 1 1 = 1. (x )(x 1) x 1 1 The hole is at (, 1 ) Example : find the slant asymptote of f(x) = x 5x+5 x Solution: Use long division, and ignore the remainder. Slant asymptote is the line y = x 1. Even & Odd Functions Symmetric to y-axis Even function Symmetric to origin Odd function Symmetric to x-axis Not a function A function f is even if, for each x in the domain of f, f( x) = f(x). A function f is odd if, for each x in the domain of f, f( x) = f(x). 9

10 Piecewise Functions A piecewise function is pieces of different functions on different intervals. They do not have to connect. x +, x 1 Example 1: f(x) = { x + 4, x > 1 This piecewise-defined function is composed of two linear functions. To the left of x = 1, the graph is the line given by y = x +. To the right of x = 1, the graph is the line given by y = x + 4. Notice that the point (1, 5) is a solid dot and the point (1,) is an open dot. This is because f(1) = 5. 5, x > 4 Example : f(x) = {(x 1) 4, 1 < x 4 4x + 8, x 1 The first piece, which is graphed to the right of x =4, is the horizontal line y = 5. The second piece, which is graphed on the interval ( 1, 4], is a quadratic function in vertex form. From the parent graph of y = x, this parabola contains a horizontal shift 1 unit to the right, and a vertical shift 4 units down. Notice at x = 4, the two pieces meet at the same y-value. Therefore, simply connect the two pieces. To the left of x = 1, the graph is the line given by y = 4x + 8. Compositions of Functions Example: Let f(x) = x + x and g(x) = x 1. Find a) f(g(x)), and b) g(f(4)). a) f(g(x)) = f(x 1) = (x 1) + (x 1) = x x x = x 1 b) work from the inside out f(4) = 4 + (4) = 4 g(f(4)) = g(4) = 4 1 = 10

11 Exponents and Logarithms Properties of Exponents e a e b = e a+b e a = ea b eb (e a ) b = e a b e a = 1 e a a 0 = 1 (a 0) Properties of Logarithms log b x = a b a = x log b xy = log b x + log x y log b x y = log b x log b y log b (x) y = y log b x Formulas Compound Interest: A = P (1 + r n )nt P = principal, r = rate (decimal), n = # of times interest compounded per year, t = time (years) Continuously Compounded: A = Pe rt Half-Life: A = A 0 ( 1 )t/h Example 1: Use properties to expand log 8 ( a 8b ). Leave no exponents. Solution: log 8 ( a 8b ) = log 8 ( a 8b ) We have to take care of the first. Move it down using the fourth property. = [log 8 a log 8 (8b )] = [log 8 a (log log 8 b )] = [log 8 a (log log 8 b)] = [log 8 a 1 log 8 b] = log 8 a 6 log 8 b Split division using the third property. Split multiplication using nd property. Notice how we can t move the down when given 8b. The exponent must be alone in order to move it to the front; we do that next when we have b. log 8 8 = 1 Distribute. Example : Let R = log 7 4 and S = log 7. Write log 7 Solution: log 7 16 = log 7 log 7 16 = S log 7 (4) = S log 7 4 = S R 16 in terms of R and S. Get the logarithms in terms of just log 7 4 and log 7. Your final answer will not contain any logarithms. Substitute S for log 7. Use 4 = 16 to get in terms of R. Example : Example 4: Example 5: log x + log (x ) = 1 log(x + ) + log (x ) = log(7) e x 5e x + 6 = 0 log x (x ) = 1 1 = x(x ) = x x 0 = x x 0 = (x )(x + 1) x =, 1 x = 1 extraneous b/c it doesn t fit original equation. Can t evaluate logarithm of a negative number. log[(x + )(x )] = log 7 log(x 9) = log 7 If the logarithms with the same base (10) are equal, therefore their arguments are equal. x 9 = 7 x = 16 x = 4, 4 x = 4 is extraneous. (e x ) 5e x + 6 = 0 (e x )(e x ) = 0 e x = 0 e x = x = ln e x = 0 e x = x = ln 11

12 Inverses of Functions Let f(x) be a function where its inverse is denoted f 1 (x). If f is a function mapping x to y, then its inverse, f 1 (x), maps y back to x. Graphically, a function and its inverse reflect each other across the line y = x. Two functions, f(x) and g(x), are inverses of each other if and only if f(g(x)) = x and g(f(x)) = x Example 1: Find the inverse of f(x) = 5 x. y = 5 x x = 5 y x = 5 y x 5 = y y = x 5 f 1 (x) = 5 x Replace f(x) with y. Interchange x and y, then solve for y. This expression is equivalent to the one above it. Example : Find the inverse of f(x) = 4x 5 x 1. y = 4x 5 x 1 x = 4y 5 y 1 x(y 1) = 4y 5 xy x = 4y 5 xy 4y = x 5 y(x 4) = x 5 y = x 5 x 4 f 1 (x) = x 5 x 4 Replace f(x) with y. Interchange x and y, then solve for y. Cross multiply Distribute Collect y terms to one side, other terms to other. Factor y Divide by x 4. Example : Verify that f(x) = x 1 and g(x) = x+1 are inverses of each other. Solution: f(g(x)) = f ( x + 1 ) = ( x + 1 ) = ( x+1 ) 1 = x = x 1 g(f(x)) = g(x 1) = (x 1) + 1 = x = x = x Binomial Expansion (Pascal s Triangle) Expand: (x + ) 4 without using FOIL Use the fifth line: (x) 4 () (x) () (x) () + 4 (x) 1 () + 1 (x) 0 () 4 16x x + 16x + 16x

13 Factoring GCF 15x 5 9x + x x (5x x + 1) 8x y + xy 5 xy (4x + y ) Difference of Squares 5x 6y 6 (5x + 6y )(5x 6y ) x 4 81 = (x + 9)(x 9) = (x + 9)(x + )(x ) Trinomials (leading coefficient 1) x + 4x 1 Find the factors of your last term that add up to your middle term. (x + 7)(x ) Super Advanced Mega GCF Factor GCF think smallest exponent (x 1) / x 1/ (x 1) 5/ x 1/ GCF: (x 1) / x 1/ (x 1) / x 1/ [x 1 (x 1) 1 ] Simplify (x 1) / x 1/ [ x + 1] Grouping 4a 8a + 9a 8a (a 1) + (a 1) (a 1)(8a + ) Sum/Diff of Cubes (SOAP) a + b = (a + b)(a ab + b ) a b = (a b)(a + ab + b ) Trinomials (leading coefficient not 1) There are hundreds of different methods. In this one 1. Multiply the coefficients of your first & last terms.. Then, find factors that add up to middle term.. Rewrite middle term as sum of those two factors. 4. Factor by grouping. ex1) 1x 5x ex) x 17x ( ) = 4 (0) = ( 5) 1x 8x + x 4x(x ) + 1(x ) (x )(4x + 1) x 1x 5x + 0 x(x 4) 5(x 4) (x 4)(x 5) Systems of Equations Substitution Elimination Graphing c. { x + y = 16 y = x 6 by graphing c. Solve the first equation for y. y = 16 x y = ± 16 x Use a graphing calculator to graph y 1 = 16 x y = 16 x y = x 6 On the TI-84, Press nd > Trace > 5: Intersection Find each intersecting point one at a time. Move your cursor as close as possible to the intersecting point. Make sure that you use the correct Y functions for first curve and second curve relative to each point of intersection. Solutions: (.9498,.7016) (.9498,.7016) ( ,.7016) (1.5161,.7016) 1

14 Graphing & Transformations Let f(x) be given by the graph: Vertical Shift f(x) + c shifts graph upward c units Horizontal Shift f(x + c) shifts graph left c units *f(x) is the thinner curve, and changes to f(x) are displayed as the thicker curve Shifts: c > 0 Stretch/Compressions: c > 1 The value of c is abstract and may differ from one graph to the next. f(x) c shifts graph downward c units f(x c) shifts graph right c units Vertical Stretch/Compress cf(x) stretches graph vertically by a factor of c Horizontal Stretch/Compress f(cx) compresses graph horizontally by a factor of c Reflections f(x) reflects graph about x-axis 1 c f(x) compresses graph vertically by a factor of c f ( x ) stretches graph horizontally c by a factor of c f( x) reflects graph about y-axis f(x) Absolute Value f( x ) Values below x-axis are reflected across x-axis. Values to the right of y-axis are reflected across y-axis. Values to the left of y-axis are replaced accordingly. 14

15 Algebraic Atrocities Below is a list of the most common (and egregious) mathematical errors used when simplifying expressions. Each error contains an explanation, and a fix or counter example involving numbers. Variables represent an arbitrary number, and mathematically correct expressions must hold true for all numbers (denominator 0). Error Explanation Debunk (by counter-example) or Fix (a + b) = a + b Binomial Expansion or FOIL must be applied. Fix: (a + b) = a + a b + ab + b You cannot split sums or Debunk: Let a =, b =. a + b = a + b differences under a radical a + b = + = = sign. a + b = + = 5 x + y + z Must split sums over common x + y + z = x + y Fix: = x z denominator. z z + y z + z z = x z + y z + 1 x + 1 When canceling factors, must (x + 1)(x + ) = x + x + 1 have placeholder of 1 in Fix: (x + 1)(x + ) = 1 x + numerator ax + by y x x + y = 1 y = ax + b x a + b = x a + x b sin(a + B) = sin A + sin B sin(a) = sin A sin(a) A Must divide both terms by common denominator Cannot cancel x s with that sum in the denominator. Cannot split binomial denominator You can t split up the evaluation of an angle this way. Use Sum formula (p. 5). A is an angle. You cannot move a constant or variable to the outside. ax + by Fix: = ax y y + by y = ax y + b Debunk: Let x = 1, y = x x + y = = 1 ; 1 y = 1 Debunk: Let x = 1, a =, b = x a + b = 1 + = 1 5 x a + x b = = = 5 6 Debunk: Let A = 0, B = 60 sin(a + B) = sin( ) = sin 90 = 1 sin A + sin B = sin 0 + sin 60 = 1 + = 1 + Fix: Double Angle Identity sin A = sin A cos A = sin Cannot cancel an expression with an angle being evaluated. log(a + B) = log A + log B log x = log y log log x = y x(x ) = 4 x = 4, a 1 = 1 a 1.05() x =.1 x Dividing each side by log You can t split arguments this way. Fix: Property of Logarithms log A + log B = log(a B) This is not a mathematical operation; it is a poor use of notation. Either take e^ to both sides, or simply remove the log (s). Quadratic equations must be set = 0 in order to solve by factoring or quadratic formula. Negative exponents are not negative numbers, they are a division of powers. Fix: x(x ) = 4 x x = 4 x x 4 = 0 (x 6)(x + 4) = 0 x = 6, 4 Fix: a 1 = 1 a = a Order of operations call for exponents to be evaluated first. Also, the exponent of 1.05 is 1, so you can t combine that with an expression that has an exponent of x. 15

16 Conics Slope & Linear Equations m = rise run = y x = y y 1 x x 1 Point-Slope Form: y y 1 = m(x x 1 ) Slope-Intercept form: y = mx + b Standard form: Ax + By = C Parallel Lines: same slope Perpendicular/Normal/Orthogonal: Opposite reciprocals (m 1 m ) Example 1: Graph the line y = 4 x using its slope and y-intercept. Start on the y-axis at. Go up units (rise), then right 4 units (run) in order to get to another point. You can also go down units and left 4 units. Example : Write the equation of the line that is perpendicular to y = 4x 7 and goes through the point (8, ). Solution: Start with point-slope form. Slope will be 1 ; plug in point. 4 y y 1 = m(x x 1 ) y = 1 (x 8) 4 y = 1 4 x + y = 1 4 x + 5 Positive Slope Negative Slope Zero Slope Undefined Slope Rises from left to right Falls from left to right Horizontal Vertical Formulas Area of Triangle Heron s Formula (all sides known): Area = s(s a)(s b)(s c) Law of Sines: sin A a sin B = b = sin C c Law of Cosines: a = b + c bc cos A b = a + c ac cos B c = a + b ab cos C where s = 1 (a + b + c) Equilateral Triangles Area = 4 a SAS Triangles Area = 1 ab sin C Distance Formula (x x 1 ) + (y y 1 ) Midpoint Formula ( x 1 + x, y 1 + y ) Pythagorean Theorem a + b = c 16

17 Popular Geometry Formulas Trapezoid Circle Cylinder Sphere Cone Triangluar Prism S. A. = πr + πrh S. A. = πr + πrl V = 1 bhl Graphing Calculator Techniques (TI-84) 1) Storing a number for a variable This is very important when working with multi-step problems, which you will do a lot of in Calculus! Take your most recent answer and press (located above ON). Then using the key, press your desired letter. For example, pressing the key will bring up the letter H after pressing ALPHA. Press Enter. Now, any time you use the letter H, the calculator will read that value as accurately. You can choose any letter you like. I suggest you stay away from using X and Y, as the stored values will change if you enter a graph (see the end of example 5). Now, with the number you have stored for H, you will preserve accuracy. Even though the calculator seems to only display ten digits, it will hold the long string of decimal places that follow. If you type H, you will get an accurate answer. If you try to square a rounded version, for example or even , you lose accuracy in the final answer. Storing preserves accuracy. Below, accuracy is lost when not storing: It is important to get in the habit of storing numbers in for variables in order to preserve accuracy. Rounded answers typed in along the way of the problem will cause your final answer to be inaccurate. The rest of these examples are going to make use of this problem: The function P(t) = 460t 675 t4 models the number of people entering an auditorium for a rock concert t hours from the door opening. At time t = 0, doors open to let people into the auditorium and at time t = doors are closed for the concert to begin. 4 ) Graphing functions in an arbitrary viewing window Graph P(x) using. Type in the function, and pressing will bring up the following graph (provided your viewing window is set to standard i.e. ). Pressing brings up the viewing window of the graph. Changing the values of Xmin, Xmax, Ymin, and Ymax will change the viewing window of your graph. Changing window values is better and more user friendly than zooming in/out. Changing Xscl and Yscl will change your tick mark intervals on the axes. Since P(t) operates from t = 0 to t =, adjust your Xmin to 1, and Xmax to. Adjust Y-max to This window gives a much better picture of what is happening in the function. (The reason we didn t make the x-min 0 and x-max is because you always want to leave a little bit of room on the edges.) 17

18 We seem to have extra room on the left and right side that can be cropped off more. Change the Xmin to 0.. Change the Xmax to.. Change Yscl to 100 to improve tick mark readability on the y-axis. ) Finding function values How many people were in the auditorium 1 hour after the doors opened? Go to the graph. Press. You can either move along the curve using the left and right arrows, or even better, type the x-value you want. Type 1 and press enter. It will give you the function value at x = 1. There are approximately 91 people in the auditorium one hour in. 4) Solving equations At what time t, 0 t, were there 400 people in the auditorium? This problem is asking you to solve the equation: 400 = 460t t4. To do that, go back to and type 400 in for Y. The graph will draw a horizontal line, and your goal is to find the x-value where they intersect. Now press to bring up the CALCULATE menu. Select option 5: intersect. Move the cursor, using the left and right arrows, as close as you can to the point of intersection. Press ENTER once, twice, thrice. The exact point of intersection appears. There were 400 people in the auditorium at t This is approximately one hour, 8 minutes, and 8 or 9 seconds in. 5) Pulling an answer from the graph into the home screen Let s use the intersecting point from the previous example. After you have found the intersecting point, do not move your cursor at all. Go directly back to the home screen ( ND, MODE). If you type in X and press enter, it will display the x-value you just had. If you type in ALPHA, 1 (which will bring up Y), it will give you the y-value from the graph as well. It is smart to store these values in for different letters right away. If you go back into the graph, and your cursor lands on a different x- or y-value, your calculator will now read X and Y as those new values when you return to the home screen. 18