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 𝑓(π‘₯) = 3(π‘₯ βˆ’ 2)2 βˆ’ 6 is a parabola that opens _________, with its vertex at ( _____, _____) , and 𝑓(2) = ____________ is the ( minimum/ maximum) ___________ value of 𝑓. Q2β€”Q4. A quadratic function 𝑓 is given. (a) Express 𝑓 in standard form. (b) Find its vertex, π‘₯ βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘(𝑠) and 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ of 𝑓. (c) sketch the graph of 𝑓 (d) Find the domain and range of 𝑓. Q2. 𝑓(π‘₯) = 2π‘₯ 2 + 4π‘₯ + 3 Q3. 𝑓(π‘₯) = 2π‘₯ 2 βˆ’ 20π‘₯ + 57 Q4. 𝑓(π‘₯) = βˆ’4π‘₯ 2 βˆ’ 12π‘₯ + 1 Q5β€”Q7. A quadratic function 𝑓 is given. (a) Express 𝑓 in standard form. (b) sketch a graph of 𝑓. (c) Find the maximum or minimum value of 𝑓. Q5. 𝑓(π‘₯) = 3π‘₯ 2 βˆ’ 6π‘₯ + 1 Q6. 𝑓(π‘₯) = βˆ’π‘₯ 2 βˆ’ 3π‘₯ + 3 Q7. 𝑓(π‘₯) = 3π‘₯ 2 βˆ’ 12π‘₯ + 13 Q8β€”Q9. Find the maximum and minimum value of the function. Q8. 𝑓(𝑑) = βˆ’3 + 80𝑑 βˆ’ 20𝑑 2 1 Q9. β„Ž(π‘₯) = 2 π‘₯ 2 + 2π‘₯ βˆ’ 6 Q10. If a ball is thrown directly upward with a velocity of 40 𝑓𝑑/𝑠, its height ( in feet) after 𝑑 seconds is given by 𝑦 = 40𝑑 βˆ’ 16𝑑 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. 𝑃(π‘₯) = βˆ’π‘₯( π‘₯ βˆ’ 3)(π‘₯ + 2) Q2. 𝑃(π‘₯) = (π‘₯ + 2)(π‘₯ + 1)(π‘₯ βˆ’ 2)(π‘₯ βˆ’ 3) Q3. 𝑃(π‘₯) = βˆ’2π‘₯( π‘₯ βˆ’ 2)2 Q4. 𝑃(π‘₯) = π‘₯ 3 (π‘₯ + 2)(π‘₯ βˆ’ 3)2 Q5β€”Q9. Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. Q5. 𝑃(π‘₯) = π‘₯ 3 βˆ’ π‘₯ 2 βˆ’ 6π‘₯ Q6. 𝑃(π‘₯) = π‘₯ 4 βˆ’ 3π‘₯ 3 + 2π‘₯ 2 Q7. 𝑃(π‘₯) = 2π‘₯ 3 βˆ’ π‘₯ 2 βˆ’ 18π‘₯ + 9 Q8. 𝑃(π‘₯) = π‘₯ 4 βˆ’ 2π‘₯ 3 βˆ’ 8π‘₯ + 16 Q9. 𝑃(π‘₯) = βˆ’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 𝑃(π‘₯) by 𝑃(π‘₯) 𝑅(π‘₯) 𝐷(π‘₯), and express the quotient 𝑃(π‘₯)/𝐷(π‘₯) in the form = 𝑄(π‘₯) + 𝐷(π‘₯) 𝐷(π‘₯) Q1. 𝑃(π‘₯) = 2π‘₯ 2 βˆ’ 5π‘₯ βˆ’ 7 𝐷(π‘₯) = π‘₯ βˆ’ 2 Q2β€”Q4. Find the quotient and remainder using synthetic division. Q2. 2π‘₯ 2 βˆ’5π‘₯+3 π‘₯βˆ’3 Q3. π‘₯ 3 βˆ’8π‘₯+2 π‘₯+3 Q4. 2π‘₯ 3 +3π‘₯ 2 βˆ’2π‘₯+1 π‘₯βˆ’ 1 2 Q5β€”Q7. Use synthetic division and the Remainder Theorem to evaluate 𝑃(𝑐). Q5. 𝑃(π‘₯) = 4π‘₯ 2 + 12π‘₯ + 5, 𝑐 = βˆ’1 Q6. 𝑃(π‘₯) = π‘₯ 7 βˆ’ 3π‘₯ 2 βˆ’ 1, 𝑐 = 3 Q7. 𝑃(π‘₯) = π‘₯ 3 + 2π‘₯ 2 βˆ’ 3π‘₯ βˆ’ 8, 𝑐 = 0.1 Q8β€”Q9. Use the Factor Theorem to show that π‘₯ βˆ’ 𝑐 is a factor of 𝑃(π‘₯) for the given value(s) of c. Q8. 𝑃(π‘₯) = π‘₯ 3 βˆ’ 3π‘₯ 2 + 3π‘₯ βˆ’ 1, 𝑐 = 1 1 Q9. 𝑃(π‘₯) = 2π‘₯ 3 + 7π‘₯ 2 + 6π‘₯ βˆ’ 5, 𝑐 = 2 Q10β€”Q11. Show that the given value(s) of c are zeros of 𝑃(π‘₯), and find all other zeros of 𝑃(π‘₯). Q10. 𝑃(π‘₯) = π‘₯ 3 + 2π‘₯ 2 βˆ’ 9π‘₯ βˆ’ 18, 𝑐 = βˆ’2 Q11. 𝑃(π‘₯) = 3π‘₯ 4 βˆ’ 8π‘₯ 3 βˆ’ 14π‘₯ 2 + 31π‘₯ + 6, 𝑐 = βˆ’2, 3 Q12—Q13. Find a polynomial of the specified degree that has the given zeros. Q12. Degree 3; zeros βˆ’1, 1, 3 Q13. Degree 4, zeros βˆ’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 don’t check to see which actually are zeros). Q1. 𝑅(π‘₯) = 2π‘₯ 5 + 3π‘₯ 3 + 4π‘₯ 2 βˆ’ 8 Q2. 𝑇(π‘₯) = 4π‘₯ 4 βˆ’ 2π‘₯ 2 βˆ’ 7 Q3β€”Q6. Find all the real zeros of the polynomial and write the polynomial in factored form. Q3. 𝑃(π‘₯) = π‘₯ 4 βˆ’ 5π‘₯ 2 + 4 Q4. 𝑃(π‘₯) = 4π‘₯ 3 + 4π‘₯ 2 βˆ’ π‘₯ βˆ’ 1 Q5. 𝑃(π‘₯) = 4π‘₯ 3 βˆ’ 6π‘₯ 2 + 1 Q6. 𝑃(π‘₯) = 2π‘₯ 4 + 15π‘₯ 3 + 17π‘₯ 2 + 3π‘₯ βˆ’ 1 Q7β€”Q8. A polynomial P is given, (a) Find all the real zeros of P (b) sketch a graph of P. Q7. 𝑃(π‘₯) = π‘₯ 3 βˆ’ 3π‘₯ 2 βˆ’ 4π‘₯ + 12 Q8. 𝑃(π‘₯) = π‘₯ 4 βˆ’ 5π‘₯ 3 + 6π‘₯ 2 + 4π‘₯ βˆ’ 8 Q9β€”Q10. Use Descartes’ 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. 𝑃(π‘₯) = 2π‘₯ 6 + 5π‘₯ 4 βˆ’ π‘₯ 3 βˆ’ 5π‘₯ βˆ’ 1 Q10. 𝑃(π‘₯) = π‘₯ 5 + 4π‘₯ 3 βˆ’ 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. 𝑃(π‘₯) = 2π‘₯ 3 + 5π‘₯ 2 + π‘₯ βˆ’ 2 π‘Ž = βˆ’3, 𝑏 = 1 Q12. 𝑃(π‘₯) = π‘₯ 4 + 2π‘₯ 3 + 3π‘₯ 2 + 5π‘₯ βˆ’ 1 π‘Ž = βˆ’2, 𝑏 = 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, Descartes’ Rule of Signs, the Quadratic Formula, or other factoring techniques. Q13. 𝑃(π‘₯) = 2π‘₯ 4 + 3π‘₯ 3 βˆ’ 4π‘₯ 2 βˆ’ 3π‘₯ + 2 Q14. 𝑃(π‘₯) = 4π‘₯ 4 βˆ’ 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. 𝑃(π‘₯) = π‘₯ 4 + 4π‘₯ 2 Q2. 𝑃(π‘₯) = π‘₯ 3 βˆ’ 2π‘₯ 2 + 2π‘₯ Q3. 𝑃(π‘₯) = π‘₯ 3 + 8 Q4β€”Q6. Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. Q4. 𝑃(π‘₯) = π‘₯ 4 + 2π‘₯ 2 + 1 Q5. 𝑃(π‘₯) = π‘₯ 3 + π‘₯ 2 + 9π‘₯ + 9 Q6. 𝑃(π‘₯) = π‘₯ 5 + 6π‘₯ 3 + 9π‘₯ Q7—Q10. Find a polynomial with integer coefficients that satisfies the given conditions. Q7. 𝑃 β„Žπ‘Žπ‘  π‘‘π‘’π‘”π‘Ÿπ‘’π‘’ 2 π‘Žπ‘›π‘‘ π‘§π‘’π‘Ÿπ‘œπ‘  1 + 𝑖 π‘Žπ‘›π‘‘ 1 βˆ’ 𝑖 Q8. P has degree 3 and zeros 2 and 𝑖. Q9. R has degree 4 and zeros 1 βˆ’ 2𝑖 and 1, with 1 a zero of multiplicity 2. Q10. T has degree 4, zeros 𝑖 and 1 + 𝑖, and constant term 12. Q11β€”Q13. Find all zeros of the polynomial. Q11. 𝑃(π‘₯) = π‘₯ 3 + 2π‘₯ 2 + 4π‘₯ + 8 Q12. 𝑃(π‘₯) = 2π‘₯ 3 + 7π‘₯ 2 + 12π‘₯ + 9 Q13. 𝑃(π‘₯) = π‘₯ 4 + π‘₯ 3 + 7π‘₯ 2 + 9π‘₯ βˆ’ 18 1 3.6 Rational Functions Q1β€”Q10. Graph the rational function. Show clearly all π‘₯ βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘(𝑠) and 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘(𝑠), asymptotes, and state the domain and range of π‘Ÿ. 5 Q1. π‘Ÿ(π‘₯) = π‘₯βˆ’2 Q3. π‘Ÿ(π‘₯) = 4π‘₯βˆ’4 π‘₯+2 2π‘₯βˆ’4 Q5. π‘Ÿ(π‘₯) = π‘₯ 2 +π‘₯βˆ’2 Q7. π‘Ÿ(π‘₯) = Q9. π‘Ÿ(π‘₯) = π‘₯ 3 +27 π‘₯+4 π‘₯ 2 +4π‘₯βˆ’5 π‘₯ 2 +π‘₯βˆ’2 Q2. π‘Ÿ(π‘₯) = Q4. π‘Ÿ(π‘₯) = 2π‘₯βˆ’3 π‘₯βˆ’2 3π‘₯ 2 βˆ’12π‘₯+13 π‘₯ 2 βˆ’4π‘₯+4 (π‘₯βˆ’1)(π‘₯+2 Q6. π‘Ÿ(π‘₯) = (π‘₯+1)(π‘₯βˆ’3) Q8. π‘Ÿ(π‘₯) = Q10. π‘Ÿ(π‘₯) = 2 π‘₯ 2 +2 π‘₯βˆ’1 π‘₯ 2 βˆ’2π‘₯βˆ’3 π‘₯+1 3.7 polynomial and rational inequalities Q1β€”Q10. Solve the inequality. Q1. 2π‘₯ 2 β‰₯ π‘₯ + 3 Q2. (π‘₯ βˆ’ 3)(π‘₯ + 5)(2π‘₯ + 5) < 0 Q3. π‘₯ 3 + 4π‘₯ 2 β‰₯ 4π‘₯ + 16 Q4. 2π‘₯ 3 βˆ’ π‘₯ 2 < 9 βˆ’ 18π‘₯ Q5. π‘₯ 4 βˆ’ 7π‘₯ 2 βˆ’ 18 < 0 π‘₯βˆ’1 Q6. π‘₯βˆ’10 < 0 2π‘₯+5 Q7. π‘₯ 2 +2π‘₯βˆ’35 β‰₯ 0 π‘₯βˆ’3 Q8. 2π‘₯+5 β‰₯ 1 1 3 Q9. 2 + 1βˆ’π‘₯ ≀ π‘₯ 3 2.1 Functions Q1. 𝑓(π‘₯) = π‘₯ 2 βˆ’ 6 𝑓(βˆ’3) = (βˆ’3)2 βˆ’ 6 = 9 βˆ’ 6 = πŸ‘ 𝑓(3) = 32 βˆ’ 6 = 9 βˆ’ 6 = πŸ‘ 𝑓(0) = 02 βˆ’ 6 = 0 βˆ’ 6 = βˆ’πŸ” 1 1 2 1 πŸπŸ‘ 𝑓( ) = ( ) βˆ’6 = βˆ’6 = βˆ’ 2 2 4 πŸ’ 1 βˆ’ 2π‘₯ 3 1 βˆ’ 2(2) = βˆ’πŸ 3 1 βˆ’ 2(βˆ’2) πŸ“ 𝑓(βˆ’2) = = 3 πŸ‘ 1 1 βˆ’ 2 (2) 1 𝑓( ) = =𝟎 2 3 𝟏 βˆ’ πŸπ’‚ 𝑓(π‘Ž) = πŸ‘ 1 βˆ’ 2(βˆ’π‘Ž) 𝟏 + πŸπ’‚ 𝑓(βˆ’π‘Ž) = = 3 πŸ‘ 1 βˆ’ 2(π‘Ž βˆ’ 1) πŸ‘ βˆ’ πŸπ’‚ 𝑓(π‘Ž βˆ’ 1) = = 3 πŸ‘ 𝑓(π‘₯) = π‘₯ 2 + 2π‘₯ 𝑓(0) = 02 + 2(0) = 𝟎 𝑓(3) = 32 + 2(3) = πŸπŸ“ 𝑓(βˆ’3) = (βˆ’3)2 + 2(βˆ’3) = πŸ‘ 𝑓(π‘Ž) = π’‚πŸ + πŸπ’‚ 𝑓(βˆ’π‘₯) = (βˆ’π‘₯)2 + 2(βˆ’π‘₯) = π’™πŸ βˆ’ πŸπ’™ 1 1 2 1 𝟏 + πŸπ’‚ 𝑓( ) = ( ) + 2( ) = π‘Ž π‘Ž π‘Ž π’‚πŸ Q2. 𝑓(π‘₯) = 𝑓(2) = Q3. Q4. 𝑓(π‘₯) = { π‘₯2, π‘₯ + 1, 𝑖𝑓 π‘₯ < 0 𝑖𝑓 π‘₯ β‰₯ 0 𝑓(βˆ’2) = (βˆ’2)2 = πŸ’ 𝑓(βˆ’1) = (βˆ’1)2 = 𝟏 𝑓(0) = 0 + 1 = 𝟏 𝑓(1) = 1 + 1 = 𝟐 𝑓(2) = 1 + 1 = πŸ‘ Q5. π‘₯ 2 βˆ’ 2π‘₯, 𝑖𝑓 π‘₯ ≀ βˆ’1 𝑓(π‘₯) = { π‘₯, 𝑖𝑓 βˆ’ 1 < π‘₯ ≀ 1 βˆ’1, 𝑖𝑓 π‘₯ β‰₯ 1 𝑓(βˆ’4) = (βˆ’4)2 βˆ’ 2(βˆ’4) = πŸπŸ’ 3 3 2 3 𝟐𝟏 𝑓 (βˆ’ ) = (βˆ’ ) βˆ’ 2 ( ) = 2 2 2 πŸ’ 𝑓(βˆ’1) = (βˆ’1)2 βˆ’ 2(βˆ’1) = πŸ‘ 𝑓(0) = 𝟎 𝑓(25) = βˆ’πŸ Q6. 𝑓(π‘₯) = 1 π‘₯βˆ’3 The domain of the function is the values of x that will not make the denominator zero. π‘₯βˆ’3β‰ 0 π·π‘œπ‘šπ‘Žπ‘–π‘›: 𝒙 β‰  πŸ‘ Q7. The domain of the function is the values of x that will not make the radicand negative. 𝑓(𝑑) = βˆšπ‘‘ + 1 𝑑+1 β‰₯ 0 π·π‘œπ‘šπ‘Žπ‘–π‘›: 𝒕 β‰₯ βˆ’πŸ Q8. 𝑔(π‘₯) = √1 βˆ’ 2π‘₯ The domain of the function is the values of x that will not make the radicand negative. 1 βˆ’ 2π‘₯ β‰₯ 0 1 β‰₯ 2π‘₯ π·π‘œπ‘šπ‘Žπ‘–π‘›: 𝒙 ≀ 𝟏 𝟐 Q9. A linear function’s domain is the set of real numbers. π·π‘œπ‘šπ‘Žπ‘–π‘›: 𝒙𝝐𝑹 𝑓(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! 𝑓(π‘₯) = βˆ’π‘₯ + 3, βˆ’3 ≀ π‘₯ ≀ 3 π‘₯ 𝑓(π‘₯) π‘₯ 𝑓(π‘₯) -3 6 1 2 -2 5 2 1 -1 4 3 0 0 3 Q2. 𝑓(π‘₯) = βˆ’π‘₯ 2 π‘₯ 𝑓(π‘₯) π‘₯ 𝑓(π‘₯) -3 -9 1 -1 -2 -4 2 -4 -1 -1 3 -9 0 0 Q3. 𝑓(π‘₯) = 1 + √π‘₯ π‘₯ 𝑓(π‘₯) π‘₯ 𝑓(π‘₯) 0 1 2 3 1 4 3 2 25 4 7 2 1 2 3 4 9 4 5 2 Q4. 𝑓(π‘₯) = |2π‘₯| π‘₯ 𝑓(π‘₯) π‘₯ 𝑓(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. 𝑓(π‘₯) = { 3, π‘₯ βˆ’ 1, 𝑖𝑓 π‘₯ < 2 𝑖𝑓 π‘₯ β‰₯ 2 π‘₯ 𝑓(π‘₯) π‘₯ 𝑓(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. 𝑓(π‘₯) = { π‘₯, π‘₯ + 1, 𝑖𝑓 π‘₯ ≀ 0 𝑖𝑓 π‘₯ > 0 π‘₯ 𝑓(π‘₯) π‘₯ 𝑓(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, 𝑖𝑓 π‘₯ < 2 𝑓(π‘₯) = {π‘₯ 2 , 𝑖𝑓 βˆ’ 2 ≀ π‘₯ ≀ 2 βˆ’π‘₯ + 6, 𝑖𝑓 π‘₯ > 2 π‘₯ 𝑓(π‘₯) π‘₯ 𝑓(π‘₯) -3 4 1 1 -2 4 2 4 -1 1 3 3 0 0 4 2 Q8. 3π‘₯ βˆ’ 5𝑦 = 7 βˆ’5𝑦 = βˆ’3π‘₯ + 7 3 7 𝑦= π‘₯βˆ’ 5 5 The equation is a linear function in the form 𝑓(π‘₯) = π‘šπ‘₯ + 𝑏. Thus, y is defined as a function of x. Q9. 2π‘₯ βˆ’ 4𝑦 2 = 4 π‘₯ βˆ’ 2𝑦 2 = 2 βˆ’2𝑦 2 = βˆ’π‘₯ + 2 π‘₯ 𝑦2 = βˆ’ 1 2 π‘₯ 𝑦 = ±√ βˆ’ 1 2 π‘₯ π‘₯ The equation for y is defined by two functions: √2 βˆ’ 1 and βˆ’βˆš2 βˆ’ 1. This makes the relation one-tomany. Therefore, y is NOT defined as a function of x. Q10. 2|π‘₯| + 𝑦 = 0 𝑦 = βˆ’2|π‘₯| The equation is an absolute value function in the form 𝑓(π‘₯) = π‘Ž|π‘₯|. Thus, y is defined as a function of x. 2.3 Getting information from the graph of a function. Q1. a. β„Ž(βˆ’2) = 1, β„Ž(0) = βˆ’1, β„Ž(2) = 3, β„Ž(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 βˆ’3 ≀ π‘₯ ≀ 4. The lowest point of the function is at y=-1. The highest point of the function is at y=4. The range is βˆ’1 ≀ 𝑦 ≀ 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 βˆ’3 ≀ π‘₯ ≀ 2. e. Net change is 𝑓(3) βˆ’ 𝑓(βˆ’3) = 4 βˆ’ 3 = 1. Q2. a. 𝑓(0) = 3, 𝑔(0) = 0.5. The larger value is 𝑓(0). b. 𝑓(βˆ’3) = βˆ’1, 𝑔(βˆ’3) = 2. The larger value is 𝑔(βˆ’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 βˆ’4 ≀ π‘₯ ≀ βˆ’2 βˆͺ 2 ≀ π‘₯ ≀ 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 βˆ’2 < π‘₯ < 2. Q3. Domain: The domain of linear functions is the set of real numbers. π·π‘œπ‘šπ‘Žπ‘–π‘›: 𝒙𝝐 𝑹 Range: The range of linear functions is the set of real numbers. π·π‘œπ‘šπ‘Žπ‘–π‘›: π’šπ 𝑹 𝑓(π‘₯) = 2π‘₯ + 3 Q4. Domain: The domain is given from the condition stated by the function. π·π‘œπ‘šπ‘Žπ‘–π‘›: βˆ’πŸ ≀ 𝒙 ≀ πŸ“ Range: The lowest point of the function is at y=-4 and the highest point of the function is at y=3. π·π‘œπ‘šπ‘Žπ‘–π‘›: βˆ’πŸ’ ≀ π’š ≀ πŸ‘ 𝑓(π‘₯) = π‘₯ βˆ’ 2, βˆ’2 ≀ π‘₯ ≀ 5 Q5. Domain: The domain is given from the condition stated by the function. π·π‘œπ‘šπ‘Žπ‘–π‘›: βˆ’πŸ‘ ≀ 𝒙 ≀ πŸ‘ Range: The lowest point of the function is at y=-1 and the highest point of the function is at y=8. π·π‘œπ‘šπ‘Žπ‘–π‘›: βˆ’πŸ ≀ π’š ≀ πŸ– 𝑓(π‘₯) = π‘₯ 2 βˆ’ 1, βˆ’3 ≀ π‘₯ ≀ 3 Q6. a. Domain: The leftmost point is at x=-1 and the rightmost point is at x=4. The domain is βˆ’1 ≀ π‘₯ ≀ 4. Range: The lowest point is at y=-1 and the highest point is at y=3. The domain is βˆ’1 ≀ 𝑦 ≀ 3. b. Increasing: The function is increasing at βˆ’1 ≀ π‘₯ < 1 βˆͺ 2 < π‘₯ ≀ 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 βˆ’3 ≀ π‘₯ ≀ 3. Range: The lowest point is at y=-2 and the highest point is at y=2. The domain is βˆ’2 ≀ 𝑦 ≀ 2. b. Increasing: The function is increasing at βˆ’2 < π‘₯ < βˆ’1 βˆͺ 1 < π‘₯ < 2. Decreasing: The function is decreasing at βˆ’3 ≀ π‘₯ < βˆ’2 βˆͺ βˆ’1 < π‘₯ < 1 βˆͺ 2 < π‘₯ ≀ 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‘œπ‘šπ‘Žπ‘–π‘›: 𝒙𝝐 𝑹 Range: Find the vertex of the function, which is the extreme point of the function. 𝑓(π‘₯) = π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 𝑏 βˆ’5 β„Ž=βˆ’ β†’β„Ž= = βˆ’2.5 2π‘Ž 2 𝑓(βˆ’2.5) = (βˆ’2.5)2 βˆ’ 5(βˆ’2.5) = βˆ’6.25 𝑓(π‘₯) = π‘₯ 2 βˆ’ 5π‘₯ Since the function is an upward parabola, the range is: π‘…π‘Žπ‘›π‘”π‘’: π’š β‰₯ βˆ’πŸ”. πŸπŸ“ 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. 𝑓(π‘₯) = 3π‘₯ βˆ’ 2, π‘₯ = 2, π‘₯ = 3 3 Q2. β„Ž(𝑑) = βˆ’π‘‘ + 2 , 𝑑 = βˆ’4, 𝑑 = 1 Q3. β„Ž(𝑑) = 2𝑑 2 βˆ’ 𝑑, 𝑑 = 3, 𝑑 = 6 Q4. 𝑓(π‘₯) = π‘₯ 3 βˆ’ 4π‘₯ 2 , π‘₯ = 0, π‘₯ = 10 1 π‘₯ Q5. 𝑔(π‘₯) = , π‘₯ = 1, π‘₯ = π‘Ž 𝑓(2) = 3(2) βˆ’ 2 = 4 𝑓(3) = 3(3) βˆ’ 2 = 7 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ = 𝑓(3) βˆ’ 𝑓(2) = 3 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘Ÿπ‘Žπ‘‘π‘’ = =3 3βˆ’2 3 β„Ž(βˆ’4) = 4 + = 5.5 2 3 β„Ž(1) = βˆ’1 + = 0.5 2 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ = β„Ž(1) βˆ’ β„Ž(βˆ’4) = βˆ’5 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘…π‘Žπ‘‘π‘’ = = βˆ’1 1 βˆ’ (βˆ’4) β„Ž(3) = 2(3)2 βˆ’ 3 = 15 β„Ž(6) = 2(6)2 βˆ’ 6 = 66 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ = β„Ž(6) βˆ’ β„Ž(3) = 51 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘…π‘Žπ‘‘π‘’ = = 17 6βˆ’3 3 2 𝑓(0) = 0 βˆ’ 4(0) = 0 𝑓(10) = 103 βˆ’ 4(10)2 = 600 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ = 𝑓(10) βˆ’ 𝑓(0) = 600 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘…π‘Žπ‘‘π‘’ = = 60 10 βˆ’ 0 1 𝑓(1) = = 1 1 1 𝑓(π‘Ž) = π‘Ž 1 1βˆ’π‘Ž 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ = 𝑓(π‘Ž) βˆ’ 𝑓(1) = βˆ’ 1 = π‘Ž π‘Ž 1βˆ’π‘Ž 𝑁𝑒𝑑 π‘β„Žπ‘Žπ‘›π‘”π‘’ 1 π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘Ÿπ‘Žπ‘‘π‘’ = = π‘Ž =βˆ’ π‘Žβˆ’1 π‘Žβˆ’1 π‘Ž
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