Math 121. Practice Problems from Chapter 3 Fall 2016

1 Remainder and Factor Theorem

1. Use long division to find

(2x4 − 4x3 + 2x + 5)÷ (x2 + 4).

Solution: The long division is as follows

2x2 − 4x − 8x2 + 4 2x4 − 4x3 0 x2 + 2x + 5

−(2x4 + 8x2)− 4x3 − 8x2

−(−4x3 − 16x)− 8x2 + 18x

−(− 8x2 − 32)18x + 37

Therefore,

(2x4 − 4x3 + 2x + 5)÷ (x2 + 4) = 2x2 − 4x− 8 +18x + 37

x2 + 4

Note: to check your answer you can check to see if

quotient × divisor + remainder = dividend.

That is you should check to see if

(2x2 − 4x− 8)(x2 + 4) + 18x + 37 is 2x4 − 4x3 + 2x + 5

2. Use long division to find

(6x3 − 2x2 − 3x + 2)÷ (2x2 + 1).

Solution: The long division is as follows

3x − 12x2 + 1 6x3 − 2x2 −3x + 2

−(6x3 + 3x)− 2x2 − 6x−(−2x2 − 1)

− 6x + 3

Therefore,

(6x3 − 2x2 − 3x + 2)÷ (2x2 + 1) = 3x− 1 +3− 6x

2x2 + 1

Note: to check your answer you can check to see if

quotient × divisor + remainder = dividend.

That is you should check to see if

(3x− 1)(2x2 + 1) + 3− 6x is (6x3 − 2x2 − 3x + 2)

3. Use synthetic division to find (x4 − 4x2 − 6x + 1)÷ (x− 3)

Solution: The synthetic division with c = 3 is as follows

3 1 0 −4 −6 13 9 15 27

1 3 5 9 28

Therefore, the answer is:

x3 + 3x2 + 5x + 9 +28

x− 3.

4. Use synthetic division to find (2x5 + 9x4 − 5x2 − 5x + 1)÷ (x + 2)

Solution: The synthetic division with c = −2 is as follows

−2 2 9 0 −5 −5 1−4 −10 20 −30 70

2 5 −10 15 −35 71

Therefore,

(2x5 + 9x4 − 5x2 − 5x + 1)÷ (x + 2) = 2x4 + 5x3 − 10x2 + 15x− 35 +71

x + 2

5. Consider the polynomials P and Q defined as follows

P (x) = x12 + 5x− 4101 and Q(x) = x12 + 7x− 4082

(a) Use a calculator to find P (−2), P (2), Q(−2) and Q(2).

(b) Use your answers from (a) and the remainder and factor theorems to answer thefollowing questions.

Page 2

• What is the remainder of P (x)÷ (x− 2)? Is (x− 2) a factor of P (x)?

• What is the remainder of P (x)÷ (x + 2)? Is (x + 2) a factor of P (x)?

• What is the remainder of Q(x)÷ (x− 2)? Is (x− 2) a factor of Q(x)?

• What is the remainder of Q(x)÷ (x + 2)? Is (x + 2) a factor of Q(x)?

Solution: (a) The values are as follows:

P (−2) = (−2)12 + 5(−2)− 4101 = −15,

P (2) = (2)12 + 5(2)− 4101 = 5,

Q(−2) = (−2)12 + 7(−2)− 4082 = 0, and

Q(2) = (2)12 + 7(2)− 4082 = 28

(b) Recall that the remainder of P (x)÷ (x− c) is P (c), and then (x− c) is a factorof P (x) when P (c) = 0. Therefore, using the answers from (a) we have

• The remainder of P (x) ÷ (x − 2) is P (2) = 5. Since 5 6= 0, (x − 2) is not afactor of P (x).

• The remainder of P (x) ÷ (x + 2) is P (−2) = −15. Since −15 6= 0, (x + 2) isnot a factor of P (x).

• The remainder of Q(x) ÷ (x − 2) is Q(2) = 28. Since 28 6= 0, (x − 2) is a notfactor of Q(x).

• The remainder of Q(x)÷ (x + 2) is Q(−2) = 0. Since 0 = 0, (x + 2) is a factorof Q(x).

6. Ken was trying to factor a polynomial, so he programmed the formula for P (x) in hiscalculator and he correctly found that

P (−8) = −20, P (−7) = 0, P (−6) = 0,P (0) = −14, P (3) = 0, and P (8) = 15.

(a) Even with this information Ken was still puzzled, so he asked his brilliant girlfriendJoAnn, and she gave him a hint by telling him that there are three obvious factors fromthe given information. Is JoAnn correct? If so, what are the factors?

(b) Based on the values of P given above. Answer the following questions.

(i) What is the remainder of P (x)÷ (x + 8)?

(ii) What is the remainder of P (x)÷ (x + 7)?

(iii) What is the remainder of P (x)÷ x?

(iv) What is the remainder of P (x)÷ (x− 8)?

Page 3

Solution: (a) JoAnn is correct because (x − c) is a factor of P (x) when P (c) = 0.Therefore,

• (x + 7) is a factor of P (x) because P (−7) = 0.

• (x + 6) is a factor of P (x) because P (−6) = 0.

• (x− 3) is a factor of P (x) because P (3) = 0.

(b) Using the values Ken found,we know

(i) The remainder of P (x)÷ (x + 8) is P (−8) = −20.

(ii) The remainder of P (x)÷ (x + 7) is P (−7) = 0 (i.e. (x + 7) is a factor of P ).

(iii) The remainder of P (x)÷ x is P (0) = −14

(iv) The remainder of P (x)÷ (x− 8) is P (8) = 15.

7. Use synthetic division to show that (x + 7) is a factor of the polynomial P given by

P (x) = x3 + 12x2 + 36x + 7

Then write P as the product of (x + 7) and a quadratic factor.

Solution: We divide (x + 7) out by synthetic division:

−7 1 12 36 7−7 −35 −7

1 5 1 0

Because the remainder is 0, the factor theorem says (x + 7) is a factor of P , and

P (x) = (x + 7)(x2 + 5x + 1).

Page 4

2 Polynomials of Higher Degree

1. Determine the far right and far left behavior of the polynomial

P (x) = −2x7 + 800x6 − 6x3 + 7x− 6.

(i) What is the leading coefficient of P?

(ii) What is the degree of P? Is it even or odd?

(iii) Because the degree of P is (even or odd) and because the leadingcoefficient of P is (positive or negative) we know the long-term behaviorof P is (choose best response)

(a) Up to far left and up to far right

(b) Up to far left and down to far right

(c) Down to far left and down to far right

(d) Down to far left and up to far right

(e) None of the above.

Solution: (i) The leading coefficient of P is −2 which is negative .

(ii) The degree of P is its highest power is 7 which is odd .

(iii) The leading coefficient of P is −2 which is negative , and the highest power ofP is 7 which is odd and so the graph of P goes up to the far left and down to thefar right, i.e., the answer is (b).

2. Determine the far right and far left behavior of the four polynomials given below

P (x) = −5x11 + 200x10 − 4x3 + 9x− 6.

Q(x) = 5x11 + 200x10 − 4x3 + 9x− 6.

R(x) = −5x12 + 200x10 − 4x3 + 9x− 6.

S(x) = 5x12 + 200x10 − 4x3 + 9x− 6.

For each of the polynomials above, state whether its leading coefficient is negative orpositive, and state whether its degree is even or odd, and then choose one of the followingoptions.

(a) Up to far left and up to far right

(b) Up to far left and down to far right

(c) Down to far left and down to far right

(d) Down to far left and up to far right

(e) None of the above.

Page 5

Solution:

• The leading coefficient of P is −5 which is negative, and the highest power ofP is 11 which is odd and so the graph of P goes up to the far left and down tothe far right, i.e., the answer is (b) for P .

• The leading coefficient of Q is 5 which is positive, and the highest power of Pis 11 which is odd and so the graph of Q goes down to the far left and up tothe far right, i.e. the answer is (d) for Q.

• The leading coefficient of R is −5 which is negative, and the highest power ofR is 12 which is even and so the graph of R goes down to the far left and downto the far right, i.e. the answer is (c) for R.

• The leading coefficient of S is 5 which is positive, and the highest power of Sis 12 which is even and so the graph of S goes up to the far left and up to thefar right, i.e. the answer is (a) for S.

3. Use the Intermediate Value Theorem to determine whether P has a zero between a andb.

P (x) = 3x3 + 2x2 − 3x− 4; a = 1, b = 4

First, find P (1) = and P (4) = . Then choose the bestresponse from the following:

(a) Because P (1) and P (4) have opposite signs, we know that P has at least one realzero between 1 and 4.

(b) Because P (1) and P (4) have opposite signs, we do not know if P has at least onereal zero between 1 and 4.

(c) Because P (1) and P (4) have the same sign, we know that P has at least one realzero between 1 and 4.

(d) Because P (1) and P (4) have the same sign, we do not know if P has at least onereal zero between 1 and 4.

Solution: FirstP (1) = 3(1)3 + 2(1)2 − 3(1)− 4 = −2

andP (4) = 3(4)3 + 2(4)2 − 3(4)− 4 = 208

Because P (1) < 0 and P (4) > 0, i.e. they have opposite signs, the IntermediateValue Theorem ensures that P has at least one zero between 1 and 4, so (a) is theappropriate choice.

4. Find the real zeros of the polynomial function by factoring; for this polynomial tryfactoring by grouping.

P (x) = x3 − 3x2 − 16x + 48

Page 6

Solution: First, factoring by grouping we find

P (x) = x2(x− 3)− 16(x− 3) = (x2 − 16)(x− 3) = (x + 4)(x− 4)(x− 3)

Therefore, the zeros of P are found by solving

(x + 4)(x− 4)(x− 3) = 0

and so the zeros of P are x = −4, x = 4 and x = 3.

5. Determine the x-intercepts for the graph of P . For each x-intercept, use the Even andOdd Powers of (x− c) Theorem to determine whether the graph of P crosses the x-axisor intersects but does not cross the x-axis at that intercept.

P (x) = (x− 3)2(x + 6)5(x− 7)7(x + 3)10

Solution: The zeros of P from smallest to biggest are x = −6,−3, 3, 7.

At the x-intercept (−6, 0) the graph of P crosses the x-axis because the power of(x + 6) is 5 which is is odd.

At the x-intercept (−3, 0) the graph of P intersects but does not cross the x-axisbecause the power of (x + 3) is 10 which is even.

At the x-intercept (3, 0) the graph of P intersects but does not cross the x-axisbecause the power of (x− 3) is 2 which is even.

At the x-intercept (7, 0) the graph of P crosses the x-axis because the power of (x−7)is 7 which is odd.

6. Use the polynomial P given below in both standard and factored form to answer thefollowing questions.

P (x) = x3 − 5x2 + 3x + 9 = (x− 3)2(x + 1)

(a) Determine the far right and far left behavior of P .

(b) List the x-intercepts, and at each intercept determine whether the graph of P crossesor intersects but does not cross the x-axis.

(c) Find the y-intercepts.

(d) Use the above information to sketch a rough graph of P .

Solution: (a) The degree of P is odd (i.e. it is 3), and its leading coefficient is 1which is positive, and so the graph goes down to the far left and up to the far right.

(b) One x-intercept is (−1, 0) and the graph of P crosses at this intercept becausethe power of the (x + 1) factor is 1 which is odd.

Page 7

The other x-intercept is (3, 0) and the graph of P intersects but does not cross thex-axis at this intercept because the power of (x− 3) is 2 which is even.

(c) As with any polynomial, the y-intercept is (0, P (0)) which is (0, 9)

(d) A graph of P is as follows:

−5−4−3−2−1 1 2 3 4 5

−50−40−30−20−10

1020304050

x

y

Page 8

3 Zeros of Polynomial Functions

1. Find the zeros of the polynomial function, and state the multiplicity of each zero.

P (x) = x2(2x− 3)4(x2 − 1)2

Solution: First rewrite P as

P (x) = x2(2x− 3)4[(x− 1)(x + 1)]2 = x2(2x− 3)4(x− 1)2(x + 1)2

and so the zeros of P are

• x = −1, multiplicity 2

• x = 0, mutliplicity 2

• x = 3/2, multiplicity 4

• x = 1, multiplicity 2

2. Use the graph of the polynomial P given below to answer the following questions.

−3 −2 −1 1 2 3 4 5

−5−4−3−2−1

12345

x

y

(a) Use the far right and far left behavior to determine whether the degree of P is evenor odd.

(b) Use the far right and far left behavior to determine whether the leading coefficientof P is positive or negative.

(c) List the zeros of P and for each zero, determine whether it has even multiplicity orodd multiplicity.

Solution: (a) Because the far right and far left behaviors are the same, the degreeof P is even.

(b) Because the graph goes down to the far right, the leading coefficient of P isnegative.

Page 9

(c) The zeros of P are the x-values where the graph crosses the x-axis. Therefore,the zeros of P are −1, 2 and 4.

• The zero x = −1 has even multiplicity because the graph of P does not crossthe x-axis at the intercept (−1, 0).

• The zero x = 2 has odd multiplicity because the graph of P crosses the x-axisat (2, 0).

• The zero x = 4 has odd multiplicity because the graph of P crosses the x-axisat (4, 0).

3. Use the rational zero theorem to determine the possible rational zeros of

P (x) = 21x10 − 2x7 + 7x4 + 3x− 15

Solution: The possible rational zeros of P are

± all factors of 15

all factors of 21.

The factors of 15 are 1, 3, 5, 15 and the factors of 21 are 1, 3, 7, 21. Therefore, thepossible rational zeros of P are:

±1,±3,±5,±15,±1

3,±3

3,±5

3,±15

3,±1

7,±3

7,±5

7,±15

7,± 1

21,± 3

21,± 5

21,±15

21

then simplifying and eliminating repeats, we obtain that the possible rational zerosof P are as follows

±1, ±3, ±5, ±15, ±1

3, ±5

3, ±1

7, ±3

7, ±5

7, ±15

7, ± 1

21, ± 5

21

4. Use Descartes’ Rule of Signs to state the number of possible positive and negative realzeros of the given polynomial functions.

(a) P (x) = 2x5 + 3x4 + 5x3 + x2 + 6x + 5

(b) Q(x) = 6x6 − 7x5 − x4 + 6x3 − 5x2 + 3x + 20

Solution: (a) P (x) has no sign variations, and

P (−x) = −2x5 + 3x4 − 5x3 + x2 − 6x + 5

has five sign variations, so there are no positive real zeros and 5 or 3 or 1 negative real zeros.

Page 10

(b) Q(x) has four sign variations, and

Q(−x) = 6x6 + 7x5 − x4 − 6x3 − 5x2 − 3x + 20

which has two sign variations. Thus there are 4, 2 or 0 positive real zeros and2 or 0 negative real zeros.

5. Find the zeros of the polynomial P by using synthetic division to divide out one rationalzero of P , and then solve the remaining quadratic to find the remaining two zeros

P (x) = x3 + 11x2 + 29x + 7

Solution: Observe that Descarte’s rule of signs implies there can be no positive realzero, so we will check to see if x = −7 is a real zero of P . We divide it out bysynthetic division.

−7 1 11 29 7−7 −28 −7

1 4 1 0

Because the remainder is 0, x = −7 is a zero of P , and so P (x) = (x+7)(x2 +4x+1).To find the remaining two zeros of P , we solve solve

x2 + 4x + 1 = 0.

According to the quadratic formula, the solutions are

x =−4±

√(4)2 − 4(1)(1)

2=−4±

√12

2

Thus, the zeros of P are −7,−4−

√12

2and

−4 +√

12

2.

6. Find the zeros of the polynomial P by using synthetic division to successively divide outtwo rational zeros of P and then solve the remaining quadratic to find the remaining twozeros.

P (x) = x4 − 1x3 − 8x2 + 6x + 12

Solution: According to the rational zero theorem, 2 is a possible rational zero, wewill try dividing it out

2 1 −1 −8 6 122 2 −12 −12

1 1 −6 −6 0

Page 11

Thus P (x) = (x− 2)(x3 + 1x2− 6x− 6). According to the rational zero theorem, −1is a possible zero of the factor x3 + 1x2− 6x− 6 which we now try to divide out withsynthetic division.

−1 1 1 −6 −6−1 0 6

1 0 −6 0

Therefore,

P (x) = (x− 2)(x3 + 1x2 − 6x− 6) = (x− 2)(x + 1)(x2 − 6)

the remaining zeros of P satisfy x2 − 6 = 0 and so x = ±√

6. Thus the zeros of Pare −1,−

√6,√

6, 2.

7. (a) Find a polynomial P of degree 3 that has zeros −5, −3 and 5. You may leave youranswer in factored form.

(b) Find a polynomial Q of degree 3 that has zeros −5, −3 and 5 such that Q(0) = −75.You may leave your answer in factored form.

Solution: (a) One such answer is P (x) = (x + 5)(x + 3)(x − 5) and in fact, anyanswer of the form

P (x) = k(x + 5)(x + 3)(x− 5)

where k 6= 0 is valid.

(b) From part (a), we can let Q(x) = k(x + 5)(x + 3)(x− 5) and so

Q(0) = k(0 + 5)(0 + 3)(0− 5) = −75k

and because Q(0) = −75 was given, we know −75k = −75, and so

k =−75

−75= 1.

Therefore the requested polynomial is

Q(x) = (x + 5)(x + 3)(x− 5)

8. (a) Find a polynomial P of degree 3 that has zeros 0, −3 and 3. You may leave youranswer in factored form.

(b) Find a polynomial Q of degree 3 that has zeros 0, −3 and 3 and Q(1) = 16 You mayleave your answer in factored form.

Page 12

Solution: (a) One such polynomial is P (x) = x(x+3)(x−3), and in fact, any answerof the form

P (x) = kx(x + 3)(x− 3)

where k 6= 0 is valid.

(b) From part (a), we let Q(x) = kx(x + 3)(x− 3), and then

Q(1) = k(1)(1 + 3)(1− 3) = −8k = 16.

This implies k = 16−8 = −2 and so

Q(x) = −2x(x + 3)(x− 3)

Page 13

4 The Fundamental Theorem of Algebra

1. Given that 1 + 2i is a zero of Q(x) = x4 − 2x3 + 8x2 − 6x + 15 find the remaining zeros,and write Q(x) as a product of linear factors.

Solution: Because complex (non-real) zeros of polynomials with real coefficientscome in conjugate pairs, we know that both 1 + 2i and 1 − 2i are zeros of Q(x).We now use synthetic division divide the factors with these zeros out of Q(x).

1 + 2i 1 −2 8 −6 151 + 2i −5 3 + 6i −15

1 −1 + 2i 3 −3 + 6i 0

Now use synthetic division again to divide out the factor (x− (1− 2i)).

1− 2i 1 −1 + 2i 3 −3 + 6i1− 2i 0 3− 6i

1 0 3 0

This means Q(x) = (x− (1 + 2i))(x− (1− 2i))(x2 + 3). Now x2 + 3 has zeros ±i√

3.Thus the zeros of Q(x) are 1 + 2i, 1− 2i, i

√3, −i

√3. Therefore,

Q(x) = (x− (1 + 2i))(x− (1− 2i))(x− i√

3)(x + i√

3)

2. Find all zeros of the polynomial P defined by

P (x) = x4 − 5x3 + 5x2 − 20x + 4

given that x = 2i is a zero of P .

Solution: Because complex (non-real) zeros of polynomials with real coefficientscome in conjugate pairs, we know that both 2i and 2i are zeros of P (x).We now use synthetic division divide the factors with these zeros out of P (x).

2i 1 −5 5 −20 42i −4− 10i 20 + 2i −4

1 −5 + 2i 1− 10i 2i 0

Now use synthetic division again to divide out the factor (x + 2i).

−2i 1 −5 + 2i 1− 10i 2i−2i −10i −2i

1 −5 1 0

This means P (x) = (x− 2i)(x + 2i)(x2 + 5)(x2 − 5x + 1). To find the remaining twozeros of P , we solve solve

x2 − 5x + 1 = 0.

Page 14

According to the quadratic formula, the solutions are

x =5±

√(5)2 − 4(1)(1)

2=

5±√

21

2

Thus, the zeros of P are 2i, −2i,5−√

21

2and

5 +√

21

2.

3. (a) Find a polynomial P with real coefficients of smallest degree that has zeros i, and 2.

(b) Find a polynomial P with real coefficients of smallest degree that has zeros i, and 2such that P (0) = 4.

Solution: (a) Because complex zeros come in conjugate pairs, the zeros of the poly-nomial of smallest degree would be i, −i, and 2. Therefore the polynomial must havefactors (x − i), (x + i) and (x − 2). So, one possible polynomial of smallest degreewith these factors have is

P (x) = (x− i)(x + i)(x− 2)

= (x2 + 1)(x− 2)

= x3 − 2x2 + x− 2

(b) Building on (a), any polynomial with real coefficients and factors (x− i), (x + i)and (x− 2) will have the form

P (x) = k(x− i)(x + i)(x− 2) = k(x3 − 2x2 + x− 2)

where k is a real number. Thus we need to find k so that P (0) = 4. Then

P (0) = k(03 − 2(02) + (0)− 2) = −2k

Because P (0) = 4, this means −2k = 4 and so k = −2. Therefore,

P (x) = −2(x3 − 2x2 + x− 2)

Page 15

5 Rational Functions

1. Determine the domain of the rational function

F (x) =x2 + 2x− 7

x2 − 7x + 6.

Write your answer in interval notation.

Solution: The domain of F is all numbers x for which the denominator of F is not0. In this case, we factor the denominator as

x2 − 7x + 6 = (x− 1)(x− 6)

and so the domain of F is {x : x 6= 1, x 6= 6} which in interval form we write

domF = (−∞, 1) ∪ (1, 6) ∪ (6,∞)

2. Consder the rational function F defined by

F (x) =x2 − 3x + 2

x2 − 4x + 3

(a) Find the domain of F ; write your answer in interval notation.

(b) Find all x-intercepts of F , if there are any.

(c) Find the y-intercept of F , if there is one.

(d) Write equations of all vertical asymptotes of F , if there are any.

Solution: (a) The domain of F is all numbers x for which the denominator of F isnot 0. In this case, we factor the denominator as

x2 − 4x + 3 = (x− 1)(x− 3)

and so the domain of F is {x : x 6= 1, x 6= 3} which in interval form we have

domF = (−∞, 1) ∪ (1, 3) ∪ (3,∞)

(b) The x-intercepts will occur at the zeros of the numerator of F that are in thedomain of F . In this case, we factor the numerator as

x2 − 3x + 2 = (x− 1)(x− 2)

Thus the possible intercepts occur when x = 1 and x = 2, but because 1 is not in thedomain of F , the graph cannot have an intercept there, and so the only x-interceptis when x = 2, that is (2, 0) is the x-intercept.

Page 16

(c) The y-intercept is always (0, F (0)) as long as 0 is in the domain of F . In this caseit is, and we find

F (0) =02 − 3(0) + 2

02 − 4(0) + 3=

2

3

so the y-intercept is (0, 23).

(d) For asymptotes, we simplify F , that is

F (x) =x2 − 3x + 2

x2 − 4x + 3=

(x− 1)(x− 2)

(x− 1)(x− 3)=

x− 2

x− 3, x 6= 1, x 6= 3

Once F is simplified, the vertical asymptotes will occur where the denominator is 0,in this case where x− 3 = 0. Thus x = 3 is the equation of the vertical asymptote.

3. Let f(x) =x4 − 1

(x− 1)2(x + 1)

(a) Find the domain of f .

(b) Write equations of all vertical asymptotes for the graph of f .

(c) For each vertical asymptote x = c, describe the behavior of the graph of f just to theright and just to the left of the asymptote.

Solution: Given f(x) =x4 − 1

(x− 1)2(x + 1)

(a) The domain of f is {x : x 6= −1, x 6= 1}(b) Simplifying f we have

f(x) =(x− 1)(x + 1)(x2 + 1)

(x− 1)2(x + 1)=

x2 + 1

x− 1, x 6= −1

The vertical asymptote is then x = 1. Note that there is a hole in the graph when xis −1, but no vertical asymptote there.

(c) If x is a little larger than 1, then

f(x) =x2 + 1

x− 1=

positive num

very small positive number= a very large positive number

and so the graph of f increases without bound just to the right of x = 1.

If x is a little smaller than 1, then

f(x) =x2 + 1

x− 1=

positive num

very small negative number= a very large negative number

and so the graph of f decreases without bound just to the left of x = 1.

4. Let f(x) =x2 − 8x + 16

x2 − 7x + 12.

Page 17

(a) State the domain of f , and simplify f if possible.

(b) Find equations for the vertical asymptotes for the graph of f .

(c) For each vertical asymptote found in part (b), determine the behavior of f just tothe right and just to the left of the vertical asymptote. Confirm your answer by creatinga table of values.

(d) Find all values of c for which there is a “hole” in the graph of f above x = c.

Solution: (a) The domain of f is {x : x 6= 3, x 6= 4}, and we simplify f as

f(x) =(x− 4)(x− 4)

(x− 3)(x− 4)=

x− 4

x− 3x 6= 3, x 6= 4

(b) The zeros in the denominator of the simplified rational function provide thelocations of the vertical asymptotes. Therefore, the graph of f has vertical asymptotex = 3.

(c) For numbers x slightly bigger than 3 we have

f(x) =(x− 4)

(x− 3)≈ −1

a small positive number= a very large negative number.

Therefore, f decreases without bound just to the right of the asymptote.

For numbers x slightly smaller than 3 we have

f(x) =(x− 4)

(x− 3)≈ −1

a small negative number= a very large positive number.

Therefore, f increases without bound just to the left of the asymptote.

A table of values is as follows

x 2.9 2.99 2.999 3 3.001 3.01 3.1(x− 4)

(x− 3)11.00 101.00 1001.00 — −999.00 −99.00 −9.00

(d) There would be a hole in the graph above x = 4 since 4 is not in the domain off , but there is no vertical asymptote at x = 4.

5. The rational function f(x) =3x2 − 7x + 4

(x− 1)is graphed below to the right, and the rational

function g(x) =x2 + 5x + 6

3(x + 4)is graphed below to the left.

Page 18

−7 −5 −3 −1 1 3 5 7

−7

−5

−3

−1

1

3

5

7

x

y

−6 −4 −2 2 4 6

−16

−12

−8

−4

4

8

12

16

x

y

f(x) =3x2 − 7x + 4

x− 1g(x) =

x2 + 5x + 6

3(x + 4)

Explain why the function f has a hole in the graph where its denominator is 0, whereasthe function g has a vertical asymptote where its denominator is 0.

Solution: We simplify f as follows

f(x) =3x2 − 7x + 4

x− 1=

(3x− 4)(x− 1)

x− 1= 3x− 4, x 6= 1

Thus the limit limx→1

f(x) = limx→1

3x− 4 = −1; the limit is a real number, there cannot

be a vertical asymptote, however 1 is not in the domain of f , so there is a hole in itsgraph above 1 on the x-axis.

Observe that

g(x) =x2 + 5x + 6

3(x + 4)=

(x + 3)(x + 2)

3(x + 4)

therefore the denominator x + 4 cannot be cancelled with a factor in the numerator,thus when x gets close to −4 the numerator gets close to 2 while the denominatorgets close to 0 resulting in values of g that approach −∞ or∞ depending on whetherx < −4 or x > −4. Thus g has a vertical asymptote where the denominator is 0.

6. Consider the rational function f(x) =3x2 − 10x + 8

x− 2.

(a) Does f have a vertical asymptote at x = 2?

(b) Sketch the graph of f .

Solution: (a) We need to factor the numerator of f and see if (x− 2) is a factor ofthe numerator that can cancel with the denominator

f(x) =3x2 − 10x + 8

x− 2=

(3x− 4)(x− 2)

x− 2= 3x− 4, x 6= 2

Page 19

because the graph of f is the line y = 3x − 4, except when x = 2, it does not haveany vertical asymptotes, its graph is the line y = 3x − 4 with a hole in the graphabove x = 2.

(b) A graph is as follows

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

f

x

y

7. Find all horizontal asymptotes, or explain why there is no horizontal asymptote, for therational functions F , G and H defined below.

(a) F (x) =18x8 + 11x5 − 8

8− 28x5 + 5x11

(b) G(x) =18x8 + 11x5 − 8

5x8 − 28x5 + 8

(c) H(x) =−8 + 11x5 + 18x8

8− 28x5 + 5x6

Solution: (a) The horizontal asymptote is y = 0 because the degree of the numeratorof F is 8 and this is less than the degree of the denominator of F which is 11.

(b) The degrees of the numerator and denominator of G are both 8, and so thehorizontal asymptote is determined by the ratio of the leading coefficients of G, thatis the horizontal asymptote has equation

y =18

5

(c) There is no horizontal asymptote because the degree of the numerator of H is 8and this is greater than the degree of the denominator of H which is 6.

8. Find all slant and horizontal asymptotes for the rational function

F (x) =8x3 − 6x2 − 3x + 7

2x2 + 1

Page 20

Solution: There are no horizontal asymptotes because the degree of the numeratoris larger than the degree of the denominator. There is a slant asymptote because thedegree of the numerator is the degree of the denominator plus 1. The slant asymptoteis found by long division as follows.

4x − 32x2 + 1 8x3 − 6x2 −3x + 7

−(8x3 + 4x)− 6x2 − 7x−(−6x2 − 3)

− 7x + 10

Therefore,

F (x) = 4x− 3 +10− 7x

2x2 + 1

and the slant asymptote is the line y = 4x− 3.

9. Use the rational function F defined below to answer the following questions.

F (x) =(x + 2)(x− 2)

x− 3

(a) Find the domain of F . Express answer in interval notation.

(b) Find the x-intercept(s) of F , if there are any.

(c) Find the y-intercept(s) of F , if there are any.

(d) Find the horizontal asymptote(s) of F , if there are any.

(e) Find the slant asymptote(s) of F , if there are any.

(f) Find all vertical asymptotes of F , if there are any. For each vertical asymptote,determine the behavior of F just to the right, and just to the left of the asymptote.

(g) Use the information from (a) through (f), along with plotting some additional pointsas necessary to sketch the graph of F along with all of its asymptotes.

Solution: (a) The domain of f is (−∞, 3) ∪ (3,∞).

(b) The x intercepts are all x-values in the domain where the numerator of f is zero.So (x + 2)(x− 2) = 0, that means the intercepts are (−2, 0) and (2, 0).

(c) There is a y-intercept since 0 is in the domain of F , and it is (0, F (0)) where

F (0) =(0 + 2)(0− 2)

0− 3=

4

3

That is, the y-intercept is (0, 43).

(d) There is no horizontal asymptote because the degree of the numerator of F islarger than degree of the denominator of F (i.e. 2 > 1).

Page 21

(e) There is a slant asymptote, and it is found by performing the division on F (eitherlong, or synthetic in this case), and you can check

F (x) =(x + 2)(x− 2)

x− 3= x + 3 +

5

x− 3

so the slant asymptote is y = x + 3.

(f) The vertical asymptote is x = 3 (since the denominator of F is 0 when x = 3 butthe numerator is not).

To check the behavior the graph of F just to the left of 3, notice that when x isslightly smaller than 3 we get

F (x) ≈ 5

small negative numberwhich is a large negative number

and so the graph of F decreases without bound to the left of x = 3.

To check the behavior the graph of F just to the right of 3, notice that when x isslightly larger than 3 we get

F (x) ≈ 5

small positive numberwhich is a large positive number

and so the graph of F increases without bound to the right of x = 3.

(g) A graph of f is as follows, the asymptotes are the dashed lines.

−8 −6 −4 −2 2 4 6 8

−20

−16

−12

−8

−4

4

8

12

16

20

x

y

10. Let f(x) =x2 − 10x + 25

x2 − 6x + 5.

(a) Simplify f and find its domain.

(b) Find equations for the vertical asymptote(s) for the graph of f .

Page 22

(c) Find the x- and y-intercepts of the graph of f .

(c) For each vertical asymptote found in part (b), determine the behavior of f just tothe right and just to the left of the vertical asymptote.

(d) Find all values of c for which there is a “hole” in the graph of f above x = c.

(e) Find all horizontal asymptote of f .

(f) Use the information above and plot additional points a necessary to graph f .

Solution: (a) The domain of f is {x : x 6= 1, x 6= 5} (see the simplified form of fbelow):

f(x) =(x− 5)(x− 5)

(x− 1)(x− 5)=

x− 5

x− 1x 6= 1, x 6= 5

(b) The zeros in the denominator of the simplified rational function provide thelocations of the vertical asymptotes. Therefore, the graph of f has vertical asymptotex = 1, there will be a hole in the graph at x = 5.

(c) The x-intercept is (5, 0) and the y-intercept is (0, 5).

(c) When x is close to, but larger than 1 we have

f(x) =(x− 5)

(x− 1)≈ −4

small positive number= large negative number.

When x is close to, but smaller than 1 we have

f(x) =(x− 5)

(x− 1)=

−4

small negative number= large positive number.

Thus the graph decreases without bound as x approaches 1 from the right and in-creases without bound as x approaches 1 from the left.

(d) There would be a hole in the graph above x = 5 (since 5 is not in the domain off , but there is no vertical asymptote at x = 5).

(e) The degrees of the numerator and denominator are the same, so the horizontalasymptote is found by setting y equal to the ratio of the leading coefficients, that isy = 1

(f) A graph of f along with its asymptotes is as follows

Page 23

−10−8 −6 −4 −2 2 4 6 8 10

−10

−8

−6

−4

−2

2

4

6

8

10

x

y

Page 24