1. Find the domain of the following functions. Write your answer using interval notation. (9 pts.)
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1 MATH- Sample Eam Spring 7. Find the domain of the following functions. Write your answer using interval notation. (9 pts.) a. 9 f ( ) b. g ( ) Write the equation of the circle in standard form given the following information. a. Center: (, ) ; Radius: 5 (5 pts.) b. The endpoints of a diameter are (, ) and ( 5, ). (8 pts.) c. The center is (, ) and the point (, ) is a point on the circle. ( pts.). Find the center and radius of the circle whose equation is given by y 8 y. (8 pts.). If f ( ) 5, then find the average rate of change of the function f on the interval [, h], where h. (8 pts.) 5. If f ( ) 8 and g ( ) 7, then find ( f g )( ). (5 pts.). Sketch the graph of h ( ). (5 pts.) 7. For the function f ( ) ( ), sketch the graph and identify the following. a. horizontal shift ( pts.) b. vertical shift ( pts.) c. range of the function ( pts.) d. interval(s) on which the function is increasing ( pts.) e. interval(s) on which the function is decreasing ( pts.)
2 f. value of relative (local) maimum(s) and location(s) ( pts.) g. value of relative (local) minimum(s) and location(s) ( pts.) h. -intercept(s) (5 pts.) 8. Find the zeros (roots) and their multiplicities of g ( ) ( )( ). Determine the sign of the infinity that the polynomial values approaches as approaches positive infinity and negative infinity. Sketch a graph of the polynomial. ( pts.) 9. If h ( ) 5, then use the Remainder Theorem to find h (7). ( pts.). Identify the possible rational zeros (roots) of the polynomial f ( ) 9. Then find the zeros (roots). Write a factorization for 9. ( pts.). Given the rational function g ( ), then find the following. a. vertical asymptote(s) ( pts.) b. horizontal asymptote(s) ( pts.) c. if the function has a horizontal asymptote, determine if the graph crosses the asymptote ( pts.) d. sketch the graph of the function ( pts.). Solve. Write your answer using interval notation. (8 pts.) SOLUTIONS: a. f ( ) 9 8 8
3 Want (Need): ( ) ( )( ) and and Answer:,, (, ) Back to Problem. b. g ( ) 9 Want (Need): Answer:, 9 Back to Problem. a. Center: (, ) ; Radius: 5 The standard form of the equation for a circle: ( h) ( y k ) r, where the point ( h, k ) is the center of the circle and r is the radius. h, k, r 5 ( ) ( y ) 5
4 Answer: ( ) ( y ) 5 Back to Problem. b. The endpoints of a diameter are (, ) and ( 5, ). NOTE: The diameter of a circle is a line segment passing through the center of the circle and whose endpoints lie on the circle. The midpoint of the diameter is the center of the circle and the length of the diameter is twice the radius of the circle. Use the midpoint formula to find the center of the circle: 5,, Use the distance formula to find the length of the diameter in order to find the radius: d ( 5) ( ) ( ) ( 5) 5 Since the diameter is, then the radius is. Center:, ; Radius: The standard form of the equation for a circle: ( h) ( y k ) r, where the point ( h, k ) is the center of the circle and r is the radius.
5 h, k, r y NOTE: Answer: y Back to Problem. c. The center is (, ) and the point (, ) is a point on the circle. The standard form of the equation for a circle: ( h) ( y k ) r, where the point ( h, k ) is the center of the circle and r is the radius. h, k, r? ( ) ( y ) r ( ) y r In order for the point (, ) to be on the circle, it must satisfy the equation for the circle. Thus, ( ) r r ( 7) 9 85 r 85 ( ) y 85 Answer: ( ) y 85 Back to Problem.. y 8 y
6 We will need to complete the squares for both and y in order to put the equation in standard form. y 8 y 5 y 8 y Half 9 Half Square 5 Square 8 ( 5) ( y 9) 8 Center: ( 5, 9) Radius: 8 Back to Problem.. f ( ) 5 m sec y f ( h) h f () 8 9h h h 8 9h h h h(9 h h) 9 h f ( h) ( h) 5( h) ( 8h h ) 5h 8 h h 5h 8 9h h
7 f ( ) 8 8 Answer: 9 h Back to Problem. 5. f ( ) 8 and g ( ) 7 ( f g )( ) f ( g ( )) f ( 7) 8 ( 7) 8 9 Answer: 9 Back to Problem 5.. h ( ) The graph of y is the graph of y shifted units to the left. y Back to Problem.
8 7. f ( ) ( ) Graph the equation y ( ) The graph of y ( ) is the graph of to the right and units downward. y shifted units y a. horizontal shift: units to the right b. vertical shift: units downward c. range of the function: [, ) d. interval(s) on which the function is increasing: (, ) e. interval(s) on which the function is decreasing: (, ) f. value of relative (local) maimum(s) and location(s): None g. value of relative (local) minimum(s) and location(s): at h. 8 -intercept(s):, Set y : ( ) ( ) 8 Back to Problem 7.
9 8. g ( ) ( )( ) Zeros (Roots) of g : q ( ) ( )( ) ( )( ), Since the factor produces the zero (root) of, its multiplicity is one. Since the factor ( ) produces the zero (root) of, its multiplicity is two. Implication for the Graph Zero (Root) Multiplicity of the Polynomial Touches the -ais at (, ) Crosses the -ais at (, ) For infinitely large values of, g ( ) ( ) As As,. Thus, since, then q ( ).,. Thus, since, then f ( ). y Back to Problem 8.
10 9. h ( ) 5 Find h (7) Answer: 55 Back to Problem 9.. f ( ) 9 Back to Problem. Factors of :,,, 5,,, 5, Factors of :,, Using a denominator of :,,, 5,,, 5, Using a denominator of :,, 5, 5 Using a denominator of :,, 5, 5 Trying : of 9 9 Coeff 9 9
11 Thus, f ( ) is not a factor of f and is not a zero (root) of f. Trying : of 9 9 Coeff Thus, f ( ) 8 is not a factor of f and is not a zero (root) of f. Trying : of 9 9 Coeff Thus, f ( ) 75 is not a factor of f and is not a zero (root) of f. NOTE: By the Bound Theorem, is an upper bound for the positive zeros (roots) of f since all the numbers are positive in the third row of the 5 5 synthetic division. Thus,, 5,,,, 5, and can t be rational zeros (roots) of f. Trying : of 9 9 Coeff 8 8 Thus, f ( ) is not a factor of f and is not a zero (root) of f.
12 Trying : of 9 9 Coeff 8 5 Thus, f ( ) is factor of f and is a zero (root) of f. The third row in the synthetic division gives us the coefficients of the other factor starting with. Thus, the other factor is 5. Thus, we have that 9 = ( )( 5). Now, we can try to find a factorization for the epression 5. However, it does not factor. Thus, we have that 9 = ( )( 5). Thus, 9 ( )( 5), 5 We will need to use the Quadratic Formula to solve 5. Thus, b b ac = a 8 ( )5 = [ 8 ()5] = = 8 = i 8 = i
13 Answer: Zeros (Roots):, i, i Factorization: 9 = ) 5 )( ( Back to Problem.. ) )( ( ) ( g a. vertical asymptotes:, ) )( ( Answer:, b. horizontal asymptotes: y ) ( g As, ) ( g c. ) ( g Answer:,
14 d. y Back to Problem.. Back to Problem. NOTE: This is a two part problem. One part of the problem is to solve the nonlinear inequality. The other part of the problem is to solve the equation. We will use the three step method to solve the nonlinear inequality : Step :
15 undefined Step : Plot all the numbers found in Step on the real number line. + Sign of Step : Use the real number line to identify the open intervals determined by the plotted numbers. Pick a test value for each open interval. Interval Test Value Sign of, ( ) ( ), ( ) ( ) 7 (, ) 7 ( ) ( ) Thus, the solution for the nonlinear inequality is the set of real numbers given by, (, ). The solution for was found in Step above. Thus, the solution for is the set
16 . Putting these two solutions together, we have that the solution for is the set of real numbers, [, ). Answer:, [, )
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