Messiah College Calculus 1 Placement Exam Topics and Review: Key

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1 Messiah College Calculus Placement Eam Topics and Review: Key Answers to problems in the tet are listed in the back of the course tetbook: Calculus, 9th edition by Larson, Hostetler, and Edwards. Solutions to additional problems from the review are given below. Graphs and Models. Equations and graphs of parabolas Find the verte of the parabola y = + 3. Where does this parabola intersect the -ais? y-ais? The first coordinate of the verte of a parabola is always b =, when the equation is in standard form y=a +b+c. So the - a coordinate of the verte is =. To find the y-coordinate of the verte, plug =0.5 into the equation of the parabola, to find y=.875. The verte is at (0.5,.875). The parabola opens up, since a>0, and since the verte is the lowest point, we know that the graph has no - intercepts. The y-intercept is at y=3. Find the equation of the parabola shown below. The parabola above intersects the -ais at = - and =, so its equation must have factors (+) and (-). The equation y=(+)(-) is y= +-8 in standard form, which opens up (a>0), not down as required. Try y = -( +-8) = This equation has y-

2 intercept at y=8 and verte at (-,9), which matches the graph. The correct equation is y= Quadratic equations and the quadratic formula. Solve the following quadratic equations: i. + 3 = 0. Use the quadratic formula to find = and = ii. 3 = 3 Subtract 3 from both sides and use the quadratic formula to find = and = Find solution sets for simultaneous linear equations. Solve the following system of equations: y = + 3 y = + 8 Subtract the second equation from the first and solve 5 for to find =. Plug this back in to either equation to get 6 y=. 3 Which of the following graphs shows the system of equations + y = y 5 3 and its solution? y = 5 MAT Placement Eam Review Key

3 The intersection point is (,8), so the second graph is correct. (The second equation is a line with slope ½ and y-intercept at 7.5. This line is drawn correctly only on the second graph.). Find solution sets for simultaneous equations in general. e Solve the following systems of equations: y = y = Substitute for y in the first equation and then factor to find two solutions, at = -6 (y=36) and = (y=). The intersection points are (-6,36) and (,). Find all intersections of the circle y 8y = with the line + y = 8. Sketch the two curves and label the points of intersection. Substitute 8- for y in the first equation, simplify, and factor, to find two solutions at = - (y=0) and = (y=7). Linear Models and Rates of Change. linear equations including perpendicular and parallel lines. Identify the slope and the intercepts for the equation 3 + y =. Sketch the graph of this equation. 3 The slope is. The y-intercept is y=3 and the -intercept is =. Find the equation of the line passing through the point (-,3) and parallel to the line 3 + y =. MAT Placement Eam Review Key 3

4 The line will have slope point-slope form, the equation is ( ) 3 (see above problem). Using the y 3 = +, or y = +. Find the equation of the line passing through the points (,-) and (,000). The line has an undefined slope and is therefore a vertical line, whose equation is =. Functions and Their Graphs (pp. 9-9 Eercises 3,7,9,,5,33,35). Domain of a function: Find the domain of 6 The set of all -values for which the epression is defined is the set for which -6>0. This is the set (-,-) U (, ).. Functional notation and equations as functions: For the function f ( ) =, find: 6 a) f(5) b) f(a) c) f(a+3) Answers: a) f(5)= 3 b) f()= = a a c) f(a+3)= ( ) a a 6 3. Use translations and shifting to describe the graphs of functions. f = to sketch the following graphs: Use the graph of ( ) a) f(+) = (+) b) f()+ = + c) f()-3 = -3 Answers: a) The graph of y=(+) is the same as the graph of y= shifted to the left units. b) The graph of y= + is the same as the graph of y= shifted up units. MAT Placement Eam Review Key

5 c) The graph of y= -3 is the same as the graph of y= stretched vertically by a factor of and shifted down 3 units. Real Numbers and the Real Line. Know the definition of absolute value and solve equalities and inequalities involving absolute value: + 5 = 3 There are two solutions: = - and = -. Shade the portion of the -ais that corresponds to the solution of 3 + > 5, given the graph of y= 3+ below. The solution set of 3+ >5 is (-,-) U ( 3,). Shade these. Solve inequalities in general: Solve + > two portions of the graph above, and you can see that the y-values are greater than 5 on these two regions of the -ais. Subtract from both sides and use the quadratic formula to find that -+=0 at = 3 and + 3. Place these points on a number line and test a point in each interval to find that the, 3 U + 3, inequality is satisfied in the set ( ) ( ) MAT Placement Eam Review Key 5

6 Shade the portion of the -ais that corresponds to the solution of +, given the graph of y = + below. The graph above clearly shows that the function + is always above y= when >0 and below y= when <0 (and undefined when =0). The -ais should be shaded to the right of =0, with an open circle at =0. Solve + 0 The epression on the left factors: (+)(-), which is equal to 0 at = - and =. The solution is the set [-,]. MAT Placement Eam Review Key 6

7 The Cartesian Plane. Distance and midpoint formulas Find the distance between (,3) and (-,). 0 Do the following points form a right triangle: (,0), (,), (-,-5)? Graphing these points and connecting them suggests that the line connecting (,0) and (,) may be perpendicular to the line connecting (,) and (-,-5). The first line has slope 0.5, and the second has slope. So they are perpendicular and yes, the points form a right triangle.. Equation of circles and completing the square: Find the radius and center of the circle + y + y =. Sketch the graph. Complete the square to get the equation in the standard form of a circle: (-) +(y+) =. The circle has radius and center (,-). Find the equation of the circle shown below. The graph shows a center at (,-) and radius. The equation is (-) +(y+) =, as in the problem above. MAT Placement Eam Review Key 7

8 Review of Trigonometric Functions. the amplitude and period of functions of the form y = Asin B or y = A cos B. Sketch the graph of y = 3sin π. What is the amplitude and period of this function. The amplitude is 3 and the period is. Sketch the graph of y = 3+. 5cos. State the amplitude and period of this function. π The amplitude is 0.5 and the period is. 6 π If tanθ = 3 and 0 θ, what is θ in radians? π = radians. 3 MAT Placement Eam Review Key 8

9 What are sin θ, cos θ, and s ecθ? 5 3 sin =, cos =3, sec = 3 5. simple trigonometric equations. Solve the following equations: i. sinθ = cosθ. π = +Βk for integer k Other Problems ii. sinθ cos θ =. 5 cosθ. Factor into (cos )(0.5-sin )=0. The first factor is 0 for π π = +kβ for integer k, and the second factor is 0 for = π k 6 + and = 5 π + πk. 6. Simplify the following: ( 3y )( y ) ( y ) 5 y 3 a b 5a b 3 a b b 0a b MAT Placement Eam Review Key 9

10 ( 5 y )( 5 y ) 0y y y 3. Perform the indicated operation and simplify: a + a + 5a + 0a + a 6 a a ( 3u )( u ) 7u( u ) 0u -6u+ ( a + )( 5a + ) a e y y y + y + y + y 3. Factor polynomials. Factor the following: i. 6 ii ( +)(-)(+) (+)(-3) iii (+)(+3)(-3) MAT Placement Eam Review Key 0