San Diego State University Quadratic Functions and Models Solved Practice

Show full work.  HW 3.1-3.4 and 3.5-3.7

Math 3—College Algebra
HW 3.1—3.4
Name: ______________________
Show all your work for full credit:
3.1 Quadratic Functions and Models
Q1. The graph of 𝑓(𝑥) = 3(𝑥 − 2)2 − 6 is a parabola that opens _________, with its vertex at
( _____, _____) , and 𝑓(2) = ____________ is the ( minimum/ maximum) ___________ value
of 𝑓.
Q2—Q4. A quadratic function 𝑓 is given. (a) Express 𝑓 in standard form. (b) Find its vertex, 𝑥 −
𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(𝑠) and 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 of 𝑓. (c) sketch the graph of 𝑓 (d) Find the domain and range
of 𝑓.
Q2. 𝑓(𝑥) = 2𝑥 2 + 4𝑥 + 3
Q3. 𝑓(𝑥) = 2𝑥 2 − 20𝑥 + 57
Q4. 𝑓(𝑥) = −4𝑥 2 − 12𝑥 + 1
Q5—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(𝑥) = 3𝑥 2 − 6𝑥 + 1
Q6. 𝑓(𝑥) = −𝑥 2 − 3𝑥 + 3
Q7. 𝑓(𝑥) = 3𝑥 2 − 12𝑥 + 13
Q8—Q9. Find the maximum and minimum value of the function.
Q8. 𝑓(𝑡) = −3 + 80𝑡 − 20𝑡 2
1
Q9. ℎ(𝑥) = 2 𝑥 2 + 2𝑥 − 6
Q10. If a ball is thrown directly upward with a velocity of 40 𝑓𝑡/𝑠, its height ( in feet) after 𝑡
seconds is given by 𝑦 = 40𝑡 − 16𝑡 2 . What is the maximum height attained by the ball?
1
3.2 Polynomial Functions and their graphs
Q1—Q4. Sketch the graph of the polynomial function. Make sure your graphs shows all
intercepts and exhibits the proper end behavior.
Q1. 𝑃(𝑥) = −𝑥( 𝑥 − 3)(𝑥 + 2)
Q2. 𝑃(𝑥) = (𝑥 + 2)(𝑥 + 1)(𝑥 − 2)(𝑥 − 3)
Q3. 𝑃(𝑥) = −2𝑥( 𝑥 − 2)2
Q4. 𝑃(𝑥) = 𝑥 3 (𝑥 + 2)(𝑥 − 3)2
Q5—Q9. Factor the polynomial and use the factored form to find the zeros. Then sketch the
graph.
Q5. 𝑃(𝑥) = 𝑥 3 − 𝑥 2 − 6𝑥
Q6. 𝑃(𝑥) = 𝑥 4 − 3𝑥 3 + 2𝑥 2
Q7. 𝑃(𝑥) = 2𝑥 3 − 𝑥 2 − 18𝑥 + 9
Q8. 𝑃(𝑥) = 𝑥 4 − 2𝑥 3 − 8𝑥 + 16
Q9. 𝑃(𝑥) = −𝑥 3 + 𝑥 2 + 12𝑥
2
3.3 Dividing Polynomials
Q1. Two polynomials P and D are given. Use synthetic and long division to divide 𝑃(𝑥) by
𝑃(𝑥)
𝑅(𝑥)
𝐷(𝑥), and express the quotient 𝑃(𝑥)/𝐷(𝑥) in the form
=
𝑄(𝑥)
+
𝐷(𝑥)
𝐷(𝑥)
Q1. 𝑃(𝑥) = 2𝑥 2 − 5𝑥 − 7
𝐷(𝑥) = 𝑥 − 2
Q2—Q4. Find the quotient and remainder using synthetic division.
Q2.
2𝑥 2 −5𝑥+3
𝑥−3
Q3.
𝑥 3 −8𝑥+2
𝑥+3
Q4.
2𝑥 3 +3𝑥 2 −2𝑥+1
𝑥−
1
2
Q5—Q7. Use synthetic division and the Remainder Theorem to evaluate 𝑃(𝑐).
Q5. 𝑃(𝑥) = 4𝑥 2 + 12𝑥 + 5, 𝑐 = −1
Q6. 𝑃(𝑥) = 𝑥 7 − 3𝑥 2 − 1, 𝑐 = 3
Q7. 𝑃(𝑥) = 𝑥 3 + 2𝑥 2 − 3𝑥 − 8, 𝑐 = 0.1
Q8—Q9. Use the Factor Theorem to show that 𝑥 − 𝑐 is a factor of 𝑃(𝑥) for the given value(s) of
c.
Q8. 𝑃(𝑥) = 𝑥 3 − 3𝑥 2 + 3𝑥 − 1, 𝑐 = 1
1
Q9. 𝑃(𝑥) = 2𝑥 3 + 7𝑥 2 + 6𝑥 − 5, 𝑐 = 2
Q10—Q11. Show that the given value(s) of c are zeros of 𝑃(𝑥), and find all other zeros of 𝑃(𝑥).
Q10. 𝑃(𝑥) = 𝑥 3 + 2𝑥 2 − 9𝑥 − 18, 𝑐 = −2
Q11. 𝑃(𝑥) = 3𝑥 4 − 8𝑥 3 − 14𝑥 2 + 31𝑥 + 6, 𝑐 = −2, 3
Q12—Q13. Find a polynomial of the specified degree that has the given zeros.
Q12. Degree 3; zeros −1, 1, 3
Q13. Degree 4, zeros −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(𝑥) = 2𝑥 5 + 3𝑥 3 + 4𝑥 2 − 8
Q2. 𝑇(𝑥) = 4𝑥 4 − 2𝑥 2 − 7
Q3—Q6. Find all the real zeros of the polynomial and write the polynomial in factored form.
Q3. 𝑃(𝑥) = 𝑥 4 − 5𝑥 2 + 4
Q4. 𝑃(𝑥) = 4𝑥 3 + 4𝑥 2 − 𝑥 − 1
Q5. 𝑃(𝑥) = 4𝑥 3 − 6𝑥 2 + 1
Q6. 𝑃(𝑥) = 2𝑥 4 + 15𝑥 3 + 17𝑥 2 + 3𝑥 − 1
Q7—Q8. A polynomial P is given, (a) Find all the real zeros of P (b) sketch a graph of P.
Q7. 𝑃(𝑥) = 𝑥 3 − 3𝑥 2 − 4𝑥 + 12
Q8. 𝑃(𝑥) = 𝑥 4 − 5𝑥 3 + 6𝑥 2 + 4𝑥 − 8
Q9—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(𝑥) = 2𝑥 6 + 5𝑥 4 − 𝑥 3 − 5𝑥 − 1
Q10. 𝑃(𝑥) = 𝑥 5 + 4𝑥 3 − 𝑥 2 + 6𝑥
Q11—Q12. Show that the given values for a and b are lower and upper bounds for the real zeros
of the polynomial.
Q11. 𝑃(𝑥) = 2𝑥 3 + 5𝑥 2 + 𝑥 − 2
𝑎 = −3, 𝑏 = 1
Q12. 𝑃(𝑥) = 𝑥 4 + 2𝑥 3 + 3𝑥 2 + 5𝑥 − 1
𝑎 = −2, 𝑏 = 1
Q13—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(𝑥) = 2𝑥 4 + 3𝑥 3 − 4𝑥 2 − 3𝑥 + 2
Q14. 𝑃(𝑥) = 4𝑥 4 − 21𝑥 2 + 5
4
Math 3—College Algebra
HW 3.5—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(𝑥) = 𝑥 4 + 4𝑥 2
Q2. 𝑃(𝑥) = 𝑥 3 − 2𝑥 2 + 2𝑥
Q3. 𝑃(𝑥) = 𝑥 3 + 8
Q4—Q6. Factor the polynomial completely and find all its zeros. State the multiplicity of each
zero.
Q4. 𝑃(𝑥) = 𝑥 4 + 2𝑥 2 + 1
Q5. 𝑃(𝑥) = 𝑥 3 + 𝑥 2 + 9𝑥 + 9
Q6. 𝑃(𝑥) = 𝑥 5 + 6𝑥 3 + 9𝑥
Q7—Q10. Find a polynomial with integer coefficients that satisfies the given conditions.
Q7. 𝑃 ℎ𝑎𝑠 𝑑𝑒𝑔𝑟𝑒𝑒 2 𝑎𝑛𝑑 𝑧𝑒𝑟𝑜𝑠 1 + 𝑖 𝑎𝑛𝑑 1 − 𝑖
Q8. P has degree 3 and zeros 2 and 𝑖.
Q9. R has degree 4 and zeros 1 − 2𝑖 and 1, with 1 a zero of multiplicity 2.
Q10. T has degree 4, zeros 𝑖 and 1 + 𝑖, and constant term 12.
Q11—Q13. Find all zeros of the polynomial.
Q11. 𝑃(𝑥) = 𝑥 3 + 2𝑥 2 + 4𝑥 + 8
Q12. 𝑃(𝑥) = 2𝑥 3 + 7𝑥 2 + 12𝑥 + 9
Q13. 𝑃(𝑥) = 𝑥 4 + 𝑥 3 + 7𝑥 2 + 9𝑥 − 18
1
3.6 Rational Functions
Q1—Q10. Graph the rational function. Show clearly all 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(𝑠) and 𝑦 −
𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(𝑠), asymptotes, and state the domain and range of 𝑟.
5
Q1. 𝑟(𝑥) = 𝑥−2
Q3. 𝑟(𝑥) =
4𝑥−4
𝑥+2
2𝑥−4
Q5. 𝑟(𝑥) = 𝑥 2 +𝑥−2
Q7. 𝑟(𝑥) =
Q9. 𝑟(𝑥) =
𝑥 3 +27
𝑥+4
𝑥 2 +4𝑥−5
𝑥 2 +𝑥−2
Q2. 𝑟(𝑥) =
Q4. 𝑟(𝑥) =
2𝑥−3
𝑥−2
3𝑥 2 −12𝑥+13
𝑥 2 −4𝑥+4
(𝑥−1)(𝑥+2
Q6. 𝑟(𝑥) = (𝑥+1)(𝑥−3)
Q8. 𝑟(𝑥) =
Q10. 𝑟(𝑥) =
2
𝑥 2 +2
𝑥−1
𝑥 2 −2𝑥−3
𝑥+1
3.7 polynomial and rational inequalities
Q1—Q10. Solve the inequality.
Q1. 2𝑥 2 ≥ 𝑥 + 3
Q2. (𝑥 − 3)(𝑥 + 5)(2𝑥 + 5) < 0 Q3. 𝑥 3 + 4𝑥 2 ≥ 4𝑥 + 16 Q4. 2𝑥 3 − 𝑥 2 < 9 − 18𝑥 Q5. 𝑥 4 − 7𝑥 2 − 18 < 0 𝑥−1 Q6. 𝑥−10 < 0 2𝑥+5 Q7. 𝑥 2 +2𝑥−35 ≥ 0 𝑥−3 Q8. 2𝑥+5 ≥ 1 1 3 Q9. 2 + 1−𝑥 ≤ 𝑥 3 2.1 Functions Q1. 𝑓(𝑥) = 𝑥 2 − 6 𝑓(−3) = (−3)2 − 6 = 9 − 6 = 𝟑 𝑓(3) = 32 − 6 = 9 − 6 = 𝟑 𝑓(0) = 02 − 6 = 0 − 6 = −𝟔 1 1 2 1 𝟐𝟑 𝑓( ) = ( ) −6 = −6 = − 2 2 4 𝟒 1 − 2𝑥 3 1 − 2(2) = −𝟏 3 1 − 2(−2) 𝟓 𝑓(−2) = = 3 𝟑 1 1 − 2 (2) 1 𝑓( ) = =𝟎 2 3 𝟏 − 𝟐𝒂 𝑓(𝑎) = 𝟑 1 − 2(−𝑎) 𝟏 + 𝟐𝒂 𝑓(−𝑎) = = 3 𝟑 1 − 2(𝑎 − 1) 𝟑 − 𝟐𝒂 𝑓(𝑎 − 1) = = 3 𝟑 𝑓(𝑥) = 𝑥 2 + 2𝑥 𝑓(0) = 02 + 2(0) = 𝟎 𝑓(3) = 32 + 2(3) = 𝟏𝟓 𝑓(−3) = (−3)2 + 2(−3) = 𝟑 𝑓(𝑎) = 𝒂𝟐 + 𝟐𝒂 𝑓(−𝑥) = (−𝑥)2 + 2(−𝑥) = 𝒙𝟐 − 𝟐𝒙 1 1 2 1 𝟏 + 𝟐𝒂 𝑓( ) = ( ) + 2( ) = 𝑎 𝑎 𝑎 𝒂𝟐 Q2. 𝑓(𝑥) = 𝑓(2) = Q3. Q4. 𝑓(𝑥) = { 𝑥2, 𝑥 + 1, 𝑖𝑓 𝑥 < 0 𝑖𝑓 𝑥 ≥ 0 𝑓(−2) = (−2)2 = 𝟒 𝑓(−1) = (−1)2 = 𝟏 𝑓(0) = 0 + 1 = 𝟏 𝑓(1) = 1 + 1 = 𝟐 𝑓(2) = 1 + 1 = 𝟑 Q5. 𝑥 2 − 2𝑥, 𝑖𝑓 𝑥 ≤ −1 𝑓(𝑥) = { 𝑥, 𝑖𝑓 − 1 < 𝑥 ≤ 1 −1, 𝑖𝑓 𝑥 ≥ 1 𝑓(−4) = (−4)2 − 2(−4) = 𝟐𝟒 3 3 2 3 𝟐𝟏 𝑓 (− ) = (− ) − 2 ( ) = 2 2 2 𝟒 𝑓(−1) = (−1)2 − 2(−1) = 𝟑 𝑓(0) = 𝟎 𝑓(25) = −𝟏 Q6. 𝑓(𝑥) = 1 𝑥−3 The domain of the function is the values of x that will not make the denominator zero. 𝑥−3≠0 𝐷𝑜𝑚𝑎𝑖𝑛: 𝒙 ≠ 𝟑 Q7. The domain of the function is the values of x that will not make the radicand negative. 𝑓(𝑡) = √𝑡 + 1 𝑡+1 ≥ 0 𝐷𝑜𝑚𝑎𝑖𝑛: 𝒕 ≥ −𝟏 Q8. 𝑔(𝑥) = √1 − 2𝑥 The domain of the function is the values of x that will not make the radicand negative. 1 − 2𝑥 ≥ 0 1 ≥ 2𝑥 𝐷𝑜𝑚𝑎𝑖𝑛: 𝒙 ≤ 𝟏 𝟐 Q9. A linear function’s domain is the set of real numbers. 𝐷𝑜𝑚𝑎𝑖𝑛: 𝒙𝝐𝑹 𝑓(𝑥) = 3𝑥 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(𝑥) = −𝑥 + 3, −3 ≤ 𝑥 ≤ 3 𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥) -3 6 1 2 -2 5 2 1 -1 4 3 0 0 3 Q2. 𝑓(𝑥) = −𝑥 2 𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥) -3 -9 1 -1 -2 -4 2 -4 -1 -1 3 -9 0 0 Q3. 𝑓(𝑥) = 1 + √𝑥 𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥) 0 1 2 3 1 4 3 2 25 4 7 2 1 2 3 4 9 4 5 2 Q4. 𝑓(𝑥) = |2𝑥| 𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥) -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(𝑥) = { 3, 𝑥 − 1, 𝑖𝑓 𝑥 < 2 𝑖𝑓 𝑥 ≥ 2 𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥) -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(𝑥) = { 𝑥, 𝑥 + 1, 𝑖𝑓 𝑥 ≤ 0 𝑖𝑓 𝑥 > 0
𝑥
𝑓(𝑥)
𝑥
𝑓(𝑥)
-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 𝑥 < 2 𝑓(𝑥) = {𝑥 2 , 𝑖𝑓 − 2 ≤ 𝑥 ≤ 2 −𝑥 + 6, 𝑖𝑓 𝑥 > 2
𝑥
𝑓(𝑥)
𝑥
𝑓(𝑥)
-3
4
1
1
-2
4
2
4
-1
1
3
3
0
0
4
2
Q8.
3𝑥 − 5𝑦 = 7
−5𝑦 = −3𝑥 + 7
3
7
𝑦= 𝑥−
5
5
The equation is a linear function in the form 𝑓(𝑥) = 𝑚𝑥 + 𝑏. Thus, y is defined as a function of x.
Q9.
2𝑥 − 4𝑦 2 = 4
𝑥 − 2𝑦 2 = 2
−2𝑦 2 = −𝑥 + 2
𝑥
𝑦2 = − 1
2
𝑥
𝑦 = ±√ − 1
2
𝑥
𝑥
The equation for y is defined by two functions: √2 − 1 and −√2 − 1. This makes the relation one-tomany. Therefore, y is NOT defined as a function of x.
Q10.
2|𝑥| + 𝑦 = 0
𝑦 = −2|𝑥|
The equation is an absolute value function in the form 𝑓(𝑥) = 𝑎|𝑥|. Thus, y is defined as a function of x.
2.3 Getting information from the graph of a function.
Q1.
a. ℎ(−2) = 1, ℎ(0) = −1, ℎ(2) = 3, ℎ(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 −3 ≤ 𝑥 ≤ 4.
The lowest point of the function is at y=-1. The
highest point of the function is at y=4. The range
is −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 −3 ≤ 𝑥 ≤ 2.
e.
Net change is 𝑓(3) − 𝑓(−3) = 4 − 3 = 1.
Q2.
a. 𝑓(0) = 3, 𝑔(0) = 0.5. The larger value is 𝑓(0).
b. 𝑓(−3) = −1, 𝑔(−3) = 2. The larger value is
𝑔(−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 −4 ≤ 𝑥 ≤ −2 ∪ 2 ≤ 𝑥 ≤ 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 −2 < 𝑥 < 2. Q3. Domain: The domain of linear functions is the set of real numbers. 𝐷𝑜𝑚𝑎𝑖𝑛: 𝒙𝝐 𝑹 Range: The range of linear functions is the set of real numbers. 𝐷𝑜𝑚𝑎𝑖𝑛: 𝒚𝝐 𝑹 𝑓(𝑥) = 2𝑥 + 3 Q4. Domain: The domain is given from the condition stated by the function. 𝐷𝑜𝑚𝑎𝑖𝑛: −𝟐 ≤ 𝒙 ≤ 𝟓 Range: The lowest point of the function is at y=-4 and the highest point of the function is at y=3. 𝐷𝑜𝑚𝑎𝑖𝑛: −𝟒 ≤ 𝒚 ≤ 𝟑 𝑓(𝑥) = 𝑥 − 2, −2 ≤ 𝑥 ≤ 5 Q5. Domain: The domain is given from the condition stated by the function. 𝐷𝑜𝑚𝑎𝑖𝑛: −𝟑 ≤ 𝒙 ≤ 𝟑 Range: The lowest point of the function is at y=-1 and the highest point of the function is at y=8. 𝐷𝑜𝑚𝑎𝑖𝑛: −𝟏 ≤ 𝒚 ≤ 𝟖 𝑓(𝑥) = 𝑥 2 − 1, −3 ≤ 𝑥 ≤ 3 Q6. a. Domain: The leftmost point is at x=-1 and the rightmost point is at x=4. The domain is −1 ≤ 𝑥 ≤ 4. Range: The lowest point is at y=-1 and the highest point is at y=3. The domain is −1 ≤ 𝑦 ≤ 3. b. Increasing: The function is increasing at −1 ≤ 𝑥 < 1 ∪ 2 < 𝑥 ≤ 4. Decreasing: The function is decreasing at 1 < 𝑥 < 2. Q7. a. Domain: The leftmost point is at x=-3 and the rightmost point is at x=3. The domain is −3 ≤ 𝑥 ≤ 3. Range: The lowest point is at y=-2 and the highest point is at y=2. The domain is −2 ≤ 𝑦 ≤ 2. b. Increasing: The function is increasing at −2 < 𝑥 < −1 ∪ 1 < 𝑥 < 2. Decreasing: The function is decreasing at −3 ≤ 𝑥 < −2 ∪ −1 < 𝑥 < 1 ∪ 2 < 𝑥 ≤ 3. Q8. a. The graph of the function is shown. b. Domain: The domain of quadratic functions is the set of real numbers. 𝐷𝑜𝑚𝑎𝑖𝑛: 𝒙𝝐 𝑹 Range: Find the vertex of the function, which is the extreme point of the function. 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑏 −5 ℎ=− →ℎ= = −2.5 2𝑎 2 𝑓(−2.5) = (−2.5)2 − 5(−2.5) = −6.25 𝑓(𝑥) = 𝑥 2 − 5𝑥 Since the function is an upward parabola, the range is: 𝑅𝑎𝑛𝑔𝑒: 𝒚 ≥ −𝟔. 𝟐𝟓 c. From the vertex and axis of symmetry in the graph, the function is increasing at 𝑥 > 2.5 and the function is decreasing at
𝑥 < 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(𝑥) = 3𝑥 − 2, 𝑥 = 2, 𝑥 = 3 3 Q2. ℎ(𝑡) = −𝑡 + 2 , 𝑡 = −4, 𝑡 = 1 Q3. ℎ(𝑡) = 2𝑡 2 − 𝑡, 𝑡 = 3, 𝑡 = 6 Q4. 𝑓(𝑥) = 𝑥 3 − 4𝑥 2 , 𝑥 = 0, 𝑥 = 10 1 𝑥 Q5. 𝑔(𝑥) = , 𝑥 = 1, 𝑥 = 𝑎 𝑓(2) = 3(2) − 2 = 4 𝑓(3) = 3(3) − 2 = 7 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝑓(3) − 𝑓(2) = 3 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑎𝑡𝑒 = =3 3−2 3 ℎ(−4) = 4 + = 5.5 2 3 ℎ(1) = −1 + = 0.5 2 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 = ℎ(1) − ℎ(−4) = −5 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑅𝑎𝑡𝑒 = = −1 1 − (−4) ℎ(3) = 2(3)2 − 3 = 15 ℎ(6) = 2(6)2 − 6 = 66 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 = ℎ(6) − ℎ(3) = 51 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑅𝑎𝑡𝑒 = = 17 6−3 3 2 𝑓(0) = 0 − 4(0) = 0 𝑓(10) = 103 − 4(10)2 = 600 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝑓(10) − 𝑓(0) = 600 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑅𝑎𝑡𝑒 = = 60 10 − 0 1 𝑓(1) = = 1 1 1 𝑓(𝑎) = 𝑎 1 1−𝑎 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝑓(𝑎) − 𝑓(1) = − 1 = 𝑎 𝑎 1−𝑎 𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 1 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑎𝑡𝑒 = = 𝑎 =− 𝑎−1 𝑎−1 𝑎