# Math 3βCollege Algebra HW

Math 3βCollege Algebra HW 3.1β3.4 Name: ______________________ Show all your work for full credit: 3.1 Quadratic Functions and Models Q1. The graph of π(π₯) = 3(π₯ β 2)2 β 6 is a parabola that opens _________, with its vertex at ( _____, _____) , and π(2) = ____________ is the ( minimum/ maximum) ___________ value of π. Q2βQ4. A quadratic function π is given. (a) Express π in standard form. (b) Find its vertex, π₯ β πππ‘ππππππ‘(π ) and π¦ β πππ‘ππππππ‘ of π. (c) sketch the graph of π (d) Find the domain and range of π. Q2. π(π₯) = 2π₯ 2 + 4π₯ + 3 Q3. π(π₯) = 2π₯ 2 β 20π₯ + 57 Q4. π(π₯) = β4π₯ 2 β 12π₯ + 1 Q5βQ7. A quadratic function π is given. (a) Express π in standard form. (b) sketch a graph of π. (c) Find the maximum or minimum value of π. Q5. π(π₯) = 3π₯ 2 β 6π₯ + 1 Q6. π(π₯) = βπ₯ 2 β 3π₯ + 3 Q7. π(π₯) = 3π₯ 2 β 12π₯ + 13 Q8βQ9. Find the maximum and minimum value of the function. Q8. π(π‘) = β3 + 80π‘ β 20π‘ 2 1 Q9. β(π₯) = 2 π₯ 2 + 2π₯ β 6 Q10. If a ball is thrown directly upward with a velocity of 40 ππ‘/π , its height ( in feet) after π‘ seconds is given by π¦ = 40π‘ β 16π‘ 2 . What is the maximum height attained by the ball? 1 3.2 Polynomial Functions and their graphs Q1βQ4. Sketch the graph of the polynomial function. Make sure your graphs shows all intercepts and exhibits the proper end behavior. Q1. π(π₯) = βπ₯( π₯ β 3)(π₯ + 2) Q2. π(π₯) = (π₯ + 2)(π₯ + 1)(π₯ β 2)(π₯ β 3) Q3. π(π₯) = β2π₯( π₯ β 2)2 Q4. π(π₯) = π₯ 3 (π₯ + 2)(π₯ β 3)2 Q5βQ9. Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. Q5. π(π₯) = π₯ 3 β π₯ 2 β 6π₯ Q6. π(π₯) = π₯ 4 β 3π₯ 3 + 2π₯ 2 Q7. π(π₯) = 2π₯ 3 β π₯ 2 β 18π₯ + 9 Q8. π(π₯) = π₯ 4 β 2π₯ 3 β 8π₯ + 16 Q9. π(π₯) = βπ₯ 3 + π₯ 2 + 12π₯ 2 3.3 Dividing Polynomials Q1. Two polynomials P and D are given. Use synthetic and long division to divide π(π₯) by π(π₯) π(π₯) π·(π₯), and express the quotient π(π₯)/π·(π₯) in the form = π(π₯) + π·(π₯) π·(π₯) Q1. π(π₯) = 2π₯ 2 β 5π₯ β 7 π·(π₯) = π₯ β 2 Q2βQ4. Find the quotient and remainder using synthetic division. Q2. 2π₯ 2 β5π₯+3 π₯β3 Q3. π₯ 3 β8π₯+2 π₯+3 Q4. 2π₯ 3 +3π₯ 2 β2π₯+1 π₯β 1 2 Q5βQ7. Use synthetic division and the Remainder Theorem to evaluate π(π). Q5. π(π₯) = 4π₯ 2 + 12π₯ + 5, π = β1 Q6. π(π₯) = π₯ 7 β 3π₯ 2 β 1, π = 3 Q7. π(π₯) = π₯ 3 + 2π₯ 2 β 3π₯ β 8, π = 0.1 Q8βQ9. Use the Factor Theorem to show that π₯ β π is a factor of π(π₯) for the given value(s) of c. Q8. π(π₯) = π₯ 3 β 3π₯ 2 + 3π₯ β 1, π = 1 1 Q9. π(π₯) = 2π₯ 3 + 7π₯ 2 + 6π₯ β 5, π = 2 Q10βQ11. Show that the given value(s) of c are zeros of π(π₯), and find all other zeros of π(π₯). Q10. π(π₯) = π₯ 3 + 2π₯ 2 β 9π₯ β 18, π = β2 Q11. π(π₯) = 3π₯ 4 β 8π₯ 3 β 14π₯ 2 + 31π₯ + 6, π = β2, 3 Q12—Q13. Find a polynomial of the specified degree that has the given zeros. Q12. Degree 3; zeros β1, 1, 3 Q13. Degree 4, zeros β1, 1, 3, 5 3 3.4 Real zeros of polynomials Q1—Q2. List all possible rational zeros given by the Rational zeros Theorem (but donβt check to see which actually are zeros). Q1. π(π₯) = 2π₯ 5 + 3π₯ 3 + 4π₯ 2 β 8 Q2. π(π₯) = 4π₯ 4 β 2π₯ 2 β 7 Q3βQ6. Find all the real zeros of the polynomial and write the polynomial in factored form. Q3. π(π₯) = π₯ 4 β 5π₯ 2 + 4 Q4. π(π₯) = 4π₯ 3 + 4π₯ 2 β π₯ β 1 Q5. π(π₯) = 4π₯ 3 β 6π₯ 2 + 1 Q6. π(π₯) = 2π₯ 4 + 15π₯ 3 + 17π₯ 2 + 3π₯ β 1 Q7βQ8. A polynomial P is given, (a) Find all the real zeros of P (b) sketch a graph of P. Q7. π(π₯) = π₯ 3 β 3π₯ 2 β 4π₯ + 12 Q8. π(π₯) = π₯ 4 β 5π₯ 3 + 6π₯ 2 + 4π₯ β 8 Q9βQ10. Use Descartesβ Rule of Signs to determine how many positive and how many negative real zeros of the polynomial can have. Then determine the possible total number of real zeros. Q9. π(π₯) = 2π₯ 6 + 5π₯ 4 β π₯ 3 β 5π₯ β 1 Q10. π(π₯) = π₯ 5 + 4π₯ 3 β π₯ 2 + 6π₯ Q11βQ12. Show that the given values for a and b are lower and upper bounds for the real zeros of the polynomial. Q11. π(π₯) = 2π₯ 3 + 5π₯ 2 + π₯ β 2 π = β3, π = 1 Q12. π(π₯) = π₯ 4 + 2π₯ 3 + 3π₯ 2 + 5π₯ β 1 π = β2, π = 1 Q13βQ14. Find all rational zeros of the polynomial, and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartesβ Rule of Signs, the Quadratic Formula, or other factoring techniques. Q13. π(π₯) = 2π₯ 4 + 3π₯ 3 β 4π₯ 2 β 3π₯ + 2 Q14. π(π₯) = 4π₯ 4 β 21π₯ 2 + 5 4 Math 3βCollege Algebra HW 3.5β3.7 Name: ______________________ 3.5 Complex zeros and the Fundamental Theorem of Algebra Q1—Q3. A polynomial P is given. (a) Find all zeros of P, real and complex (b) Factor P completely. Q1. π(π₯) = π₯ 4 + 4π₯ 2 Q2. π(π₯) = π₯ 3 β 2π₯ 2 + 2π₯ Q3. π(π₯) = π₯ 3 + 8 Q4βQ6. Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. Q4. π(π₯) = π₯ 4 + 2π₯ 2 + 1 Q5. π(π₯) = π₯ 3 + π₯ 2 + 9π₯ + 9 Q6. π(π₯) = π₯ 5 + 6π₯ 3 + 9π₯ Q7—Q10. Find a polynomial with integer coefficients that satisfies the given conditions. Q7. π βππ  ππππππ 2 πππ π§ππππ  1 + π πππ 1 β π Q8. P has degree 3 and zeros 2 and π. Q9. R has degree 4 and zeros 1 β 2π and 1, with 1 a zero of multiplicity 2. Q10. T has degree 4, zeros π and 1 + π, and constant term 12. Q11βQ13. Find all zeros of the polynomial. Q11. π(π₯) = π₯ 3 + 2π₯ 2 + 4π₯ + 8 Q12. π(π₯) = 2π₯ 3 + 7π₯ 2 + 12π₯ + 9 Q13. π(π₯) = π₯ 4 + π₯ 3 + 7π₯ 2 + 9π₯ β 18 1 3.6 Rational Functions Q1βQ10. Graph the rational function. Show clearly all π₯ β πππ‘ππππππ‘(π ) and π¦ β πππ‘ππππππ‘(π ), asymptotes, and state the domain and range of π. 5 Q1. π(π₯) = π₯β2 Q3. π(π₯) = 4π₯β4 π₯+2 2π₯β4 Q5. π(π₯) = π₯ 2 +π₯β2 Q7. π(π₯) = Q9. π(π₯) = π₯ 3 +27 π₯+4 π₯ 2 +4π₯β5 π₯ 2 +π₯β2 Q2. π(π₯) = Q4. π(π₯) = 2π₯β3 π₯β2 3π₯ 2 β12π₯+13 π₯ 2 β4π₯+4 (π₯β1)(π₯+2 Q6. π(π₯) = (π₯+1)(π₯β3) Q8. π(π₯) = Q10. π(π₯) = 2 π₯ 2 +2 π₯β1 π₯ 2 β2π₯β3 π₯+1 3.7 polynomial and rational inequalities Q1βQ10. Solve the inequality. Q1. 2π₯ 2 β₯ π₯ + 3 Q2. (π₯ β 3)(π₯ + 5)(2π₯ + 5) < 0 Q3. π₯ 3 + 4π₯ 2 β₯ 4π₯ + 16 Q4. 2π₯ 3 β π₯ 2 < 9 β 18π₯ Q5. π₯ 4 β 7π₯ 2 β 18 < 0 π₯β1 Q6. π₯β10 < 0 2π₯+5 Q7. π₯ 2 +2π₯β35 β₯ 0 π₯β3 Q8. 2π₯+5 β₯ 1 1 3 Q9. 2 + 1βπ₯ β€ π₯ 3 2.1 Functions Q1. π(π₯) = π₯ 2 β 6 π(β3) = (β3)2 β 6 = 9 β 6 = π π(3) = 32 β 6 = 9 β 6 = π π(0) = 02 β 6 = 0 β 6 = βπ 1 1 2 1 ππ π( ) = ( ) β6 = β6 = β 2 2 4 π 1 β 2π₯ 3 1 β 2(2) = βπ 3 1 β 2(β2) π π(β2) = = 3 π 1 1 β 2 (2) 1 π( ) = =π 2 3 π β ππ π(π) = π 1 β 2(βπ) π + ππ π(βπ) = = 3 π 1 β 2(π β 1) π β ππ π(π β 1) = = 3 π π(π₯) = π₯ 2 + 2π₯ π(0) = 02 + 2(0) = π π(3) = 32 + 2(3) = ππ π(β3) = (β3)2 + 2(β3) = π π(π) = ππ + ππ π(βπ₯) = (βπ₯)2 + 2(βπ₯) = ππ β ππ 1 1 2 1 π + ππ π( ) = ( ) + 2( ) = π π π ππ Q2. π(π₯) = π(2) = Q3. Q4. π(π₯) = { π₯2, π₯ + 1, ππ π₯ < 0 ππ π₯ β₯ 0 π(β2) = (β2)2 = π π(β1) = (β1)2 = π π(0) = 0 + 1 = π π(1) = 1 + 1 = π π(2) = 1 + 1 = π Q5. π₯ 2 β 2π₯, ππ π₯ β€ β1 π(π₯) = { π₯, ππ β 1 < π₯ β€ 1 β1, ππ π₯ β₯ 1 π(β4) = (β4)2 β 2(β4) = ππ 3 3 2 3 ππ π (β ) = (β ) β 2 ( ) = 2 2 2 π π(β1) = (β1)2 β 2(β1) = π π(0) = π π(25) = βπ Q6. π(π₯) = 1 π₯β3 The domain of the function is the values of x that will not make the denominator zero. π₯β3β 0 π·πππππ: π β  π Q7. The domain of the function is the values of x that will not make the radicand negative. π(π‘) = βπ‘ + 1 π‘+1 β₯ 0 π·πππππ: π β₯ βπ Q8. π(π₯) = β1 β 2π₯ The domain of the function is the values of x that will not make the radicand negative. 1 β 2π₯ β₯ 0 1 β₯ 2π₯ π·πππππ: π β€ π π Q9. A linear functionβs domain is the set of real numbers. π·πππππ: πππΉ π(π₯) = 3π₯ 2.2 Graphs of a Function Q1. Since the domain is from -3 to 3, find points between -3 and 3 and connect them to draw the function. Draw a line from -3 to 3 only! Do not extend the line! π(π₯) = βπ₯ + 3, β3 β€ π₯ β€ 3 π₯ π(π₯) π₯ π(π₯) -3 6 1 2 -2 5 2 1 -1 4 3 0 0 3 Q2. π(π₯) = βπ₯ 2 π₯ π(π₯) π₯ π(π₯) -3 -9 1 -1 -2 -4 2 -4 -1 -1 3 -9 0 0 Q3. π(π₯) = 1 + βπ₯ π₯ π(π₯) π₯ π(π₯) 0 1 2 3 1 4 3 2 25 4 7 2 1 2 3 4 9 4 5 2 Q4. π(π₯) = |2π₯| π₯ π(π₯) π₯ π(π₯) -3 6 1 2 -2 4 2 4 -1 2 3 6 0 0 Q5. Take note of the hollow dot at (2,3), because the value of the function at x=2 is at y=2 and not at y=3. π(π₯) = { 3, π₯ β 1, ππ π₯ < 2 ππ π₯ β₯ 2 π₯ π(π₯) π₯ π(π₯) -1 3 2 1 0 3 3 2 1 3 4 3 Q6. Take note of the hollow dot at (0,1), because the value of the function at x=0 is at y=0 and not at y=1. π(π₯) = { π₯, π₯ + 1, ππ π₯ β€ 0 ππ π₯ > 0 π₯ π(π₯) π₯ π(π₯) -2 -2 1 2 -1 -1 2 3 0 0 3 4 Q7. Note that there are no holes in this piecewise function. 4, ππ π₯ < 2 π(π₯) = {π₯ 2 , ππ β 2 β€ π₯ β€ 2 βπ₯ + 6, ππ π₯ > 2 π₯ π(π₯) π₯ π(π₯) -3 4 1 1 -2 4 2 4 -1 1 3 3 0 0 4 2 Q8. 3π₯ β 5π¦ = 7 β5π¦ = β3π₯ + 7 3 7 π¦= π₯β 5 5 The equation is a linear function in the form π(π₯) = ππ₯ + π. Thus, y is defined as a function of x. Q9. 2π₯ β 4π¦ 2 = 4 π₯ β 2π¦ 2 = 2 β2π¦ 2 = βπ₯ + 2 π₯ π¦2 = β 1 2 π₯ π¦ = Β±β β 1 2 π₯ π₯ The equation for y is defined by two functions: β2 β 1 and ββ2 β 1. This makes the relation one-tomany. Therefore, y is NOT defined as a function of x. Q10. 2|π₯| + π¦ = 0 π¦ = β2|π₯| The equation is an absolute value function in the form π(π₯) = π|π₯|. Thus, y is defined as a function of x. 2.3 Getting information from the graph of a function. Q1. a. β(β2) = 1, β(0) = β1, β(2) = 3, β(3) = 4 b. The leftmost point of the function is at x=-3. The rightmost point of the function is at x=4. The domain is β3 β€ π₯ β€ 4. The lowest point of the function is at y=-1. The highest point of the function is at y=4. The range is β1 β€ π¦ β€ 4. c. Intersect the line y=3 to the function. The values of x are where the line and function intersect. The values are x=-3, 2, 4. d. Intersect the line y=3 to the function. The values of x are where the function is below or at the line. The values are β3 β€ π₯ β€ 2. e. Net change is π(3) β π(β3) = 4 β 3 = 1. Q2. a. π(0) = 3, π(0) = 0.5. The larger value is π(0). b. π(β3) = β1, π(β3) = 2. The larger value is π(β3). c. Find the values of x where the two functions intersect. The two functions intersect at x=-2, 2. d. Find the values of x where the red function f(x) is below or at the blue function g(x). The values of x are β4 β€ π₯ β€ β2 βͺ 2 β€ π₯ β€ 4. e. Find the values of x where the red function f(x) is above the blue function g(x). The values of x are β2 < π₯ < 2. Q3. Domain: The domain of linear functions is the set of real numbers. π·πππππ: ππ πΉ Range: The range of linear functions is the set of real numbers. π·πππππ: ππ πΉ π(π₯) = 2π₯ + 3 Q4. Domain: The domain is given from the condition stated by the function. π·πππππ: βπ β€ π β€ π Range: The lowest point of the function is at y=-4 and the highest point of the function is at y=3. π·πππππ: βπ β€ π β€ π π(π₯) = π₯ β 2, β2 β€ π₯ β€ 5 Q5. Domain: The domain is given from the condition stated by the function. π·πππππ: βπ β€ π β€ π Range: The lowest point of the function is at y=-1 and the highest point of the function is at y=8. π·πππππ: βπ β€ π β€ π π(π₯) = π₯ 2 β 1, β3 β€ π₯ β€ 3 Q6. a. Domain: The leftmost point is at x=-1 and the rightmost point is at x=4. The domain is β1 β€ π₯ β€ 4. Range: The lowest point is at y=-1 and the highest point is at y=3. The domain is β1 β€ π¦ β€ 3. b. Increasing: The function is increasing at β1 β€ π₯ < 1 βͺ 2 < π₯ β€ 4. Decreasing: The function is decreasing at 1 < π₯ < 2. Q7. a. Domain: The leftmost point is at x=-3 and the rightmost point is at x=3. The domain is β3 β€ π₯ β€ 3. Range: The lowest point is at y=-2 and the highest point is at y=2. The domain is β2 β€ π¦ β€ 2. b. Increasing: The function is increasing at β2 < π₯ < β1 βͺ 1 < π₯ < 2. Decreasing: The function is decreasing at β3 β€ π₯ < β2 βͺ β1 < π₯ < 1 βͺ 2 < π₯ β€ 3. Q8. a. The graph of the function is shown. b. Domain: The domain of quadratic functions is the set of real numbers. π·πππππ: ππ πΉ Range: Find the vertex of the function, which is the extreme point of the function. π(π₯) = ππ₯ 2 + ππ₯ + π π β5 β=β ββ= = β2.5 2π 2 π(β2.5) = (β2.5)2 β 5(β2.5) = β6.25 π(π₯) = π₯ 2 β 5π₯ Since the function is an upward parabola, the range is: πππππ: π β₯ βπ. ππ c. From the vertex and axis of symmetry in the graph, the function is increasing at π₯ > 2.5 and the function is decreasing at π₯ < 2.5. d. From the graph, the local minimum is at (2.5, -6.25), and there is no local maximum. 2.4 Average rate of change of a function. Q1. π(π₯) = 3π₯ β 2, π₯ = 2, π₯ = 3 3 Q2. β(π‘) = βπ‘ + 2 , π‘ = β4, π‘ = 1 Q3. β(π‘) = 2π‘ 2 β π‘, π‘ = 3, π‘ = 6 Q4. π(π₯) = π₯ 3 β 4π₯ 2 , π₯ = 0, π₯ = 10 1 π₯ Q5. π(π₯) = , π₯ = 1, π₯ = π π(2) = 3(2) β 2 = 4 π(3) = 3(3) β 2 = 7 πππ‘ πβππππ = π(3) β π(2) = 3 πππ‘ πβππππ π΄π£πππππ πππ‘π = =3 3β2 3 β(β4) = 4 + = 5.5 2 3 β(1) = β1 + = 0.5 2 πππ‘ πβππππ = β(1) β β(β4) = β5 πππ‘ πβππππ π΄π£πππππ πππ‘π = = β1 1 β (β4) β(3) = 2(3)2 β 3 = 15 β(6) = 2(6)2 β 6 = 66 πππ‘ πβππππ = β(6) β β(3) = 51 πππ‘ πβππππ π΄π£πππππ πππ‘π = = 17 6β3 3 2 π(0) = 0 β 4(0) = 0 π(10) = 103 β 4(10)2 = 600 πππ‘ πβππππ = π(10) β π(0) = 600 πππ‘ πβππππ π΄π£πππππ πππ‘π = = 60 10 β 0 1 π(1) = = 1 1 1 π(π) = π 1 1βπ πππ‘ πβππππ = π(π) β π(1) = β 1 = π π 1βπ πππ‘ πβππππ 1 π΄π£πππππ πππ‘π = = π =β πβ1 πβ1 π
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