Chapter 2 Homework Pre-Cal (2.5 - 2.7)

2.5 Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value. n=​3; 3 and 5i are​ zeros; f (−1)=104 f(x)=____​ (Type an expression using x as the variable. Simplify your​ answer.)

−x3+3x2−25x+75

2.6 Find the domain of the following rational function. f(x)=4xx+4

Rational functions are quotients of polynomial functions and are expressed as f(x)=p(x)q(x)​, where p and q are polynomial​ functions, and q(x)≠0. The domain of a rational function is the set of all real numbers except the​ x-values that make the denominator zero. In the given​ problem, notice the denominator​ q(x) is x+4. The real number x that makes x+4=0 is −4​, since −4+4=0. The domain is then the set of all real numbers except −4. The domain can be expressed in​ set-builder or interval notation. In​ set-builder notation, the domain of​ f(x) is​ {x| x≠−4​}. To write the domain of​ f(x) in interval​ notation, identify the intervals on either side of the​ exclusion, x≠−4 and use the proper notation. On one side of the​ exclusion, the interval extends from −∞ to −4. On the other​ side, it extends from −4 to ∞. The domain of this function represents the union ​(∪​) of these two sets. ​Thus, in interval​ notation, the domain of​ f(x) is (−∞, −4)∪(−4,∞).

2.5 A model of​ carry-on luggage has a length that is 5 inches greater than its depth. Airline regulations require that the sum of the​ length, width, and depth cannot exceed 49 inches. These​ conditions, with the assumption that this sum is 49 ​inches, can be modeled by the function​ V(x) that gives the​ luggage's volume, in cubic​ inches, in terms of its​ depth, x, in inches. If its volume is 2000 cubic​ inches, determine two possibilities for its depth. Volume = depth • length • width: 49−(depth+length) ​V(x) = x • (x+5) • ​[49−​(x+x+5​)] ​V(x) = x(x+5)(44−2x) Two possibilities for its depth are ___________ inches. ​(Use a comma to separate answers as needed. Round to the nearest tenth of an inch as​ needed.)

20,5.7

2.6 Use the graph of the rational function to complete the following statement. As x → 1−​, f(x) → .

As x → 1−​, f(x) →infinity∞.

2.5 Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value. n=​3; −4 and 7+5i are​ zeros; f (−1)=267 A

According to the Linear Factorization​ Theorem, an​ nth-degree polynomial can be expressed as the product of a nonzero constant and n linear factors. The form of the polynomial is shown​ below, c1, c2, ..., cn are complex numbers​ (possibly real and not necessarily​ distinct). f(x)=anx−c1x−c2⋯x−cn ​Therefore, the​ third-degree polynomial can be written as the product of a nonzero constant and three linear​ factors, f(x)=anx−c1x−c2x−c3. Each zero c of the polynomial can be used to write one linear factor x−c. The linear factor for c=−4 is x+4. The linear factor for c=7+5i is x−7−5i. For a​ third-degree polynomial, there are three zeros. Notice that the coefficients of the polynomial must be real.​ Therefore, for every complex​ zero, the complex conjugate must also be a zero of the polynomial. The third zero is the complex conjugate of 7+5i​, which is c=7−5i. This zero corresponds to the linear factor x−7+5i. ​Therefore, the polynomial can be written as the product of the three linear factors and a nonzero​ constant, as shown below. f(x)=an(x+4)(x−7−5i)(x−7+5i) Simplify this polynomial. f(x) = an(x+4)(x−7−5i)(x−7+5i) = an(x+4)x2−14x+74 Multiply the complex factors. = anx3−10x2+18x+296 Complete the multiplication. To find an​, use the fact that f (−1)=267. Substitute the values into the equation and solve for an. f(x) = anx3−10x2+18x+296 267 = an(−1−10(1)+18(−1)+296) an = 1 ​Therefore, the​ third-degree polynomial​ function, with zeros −4 and 7+5i where f (−1)=267​, is ​f(x)=x3−10x2+18x+296. The graph of ​f(x)=x3−10x2+18x+296 is shown below in a window that has dimensions [−10,10,0] by [−300,300,50]. (−4,0) (−1, 267) x y graph

2.7 Solve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. x3≥13x2

Express the inequality in the form f(x)≤0 or f(x)≥0. Begin by rewriting the inequality so that zero is on the right side. x3−13x2≥0 This inequality is in the form f(x)≥0, where f(x)=x3−13x2. Solve the equation f(x)=0. Find the​ x-intercepts of f(x)=x3−13x2 by solving the equation x3−13x2=0. The​ x-intercepts will be used as boundary points on a number line. x3−13x2 = 0 This equation is of degree 3. x2(x−13) = 0 Factor x2 from both terms. x2 = 0 or x−13 = 0 Set each factor equal to 0. x = 0 x = 13 Solve for x. Locate the boundary points on a number line and separate the line into intervals. The number line with the boundary points is shown as follows. An infinite number line, labeled from negative 2 to 15, has tick marks in increments of 1 where every tick is labeled. There are filled circles at 0 and 13, which are labeled 0 and 13, respectively.-2-10123456789101112131415013 The boundary points divide the number line into three intervals. (−∞, 0) (0, 13) (13, ∞) Choose one representative number within each interval and evaluate f at that number.​ First, pick a representative number in the interval (−∞, 0). Use x=−1. Substitute −1 for x in f(x)=x3−13x2. ​Then, evaluate. f(−1) = (−1)3−13(−1)2 Substitute −1 for x. = −14 Evaluate. ​Therefore, f(x)<0 for all x in (−∞, 0). ​Next, pick a representative number in the interval (0, 13). Use x=1. Substitute 1 for x in f(x)=x3−13x2. ​Then, evaluate. f(1) = (1)3−13(1)2 Substitute 1 for x. = −12 Evaluate. ​Thus, f(x)<0 for all x in (0, 13). ​Finally, pick a representative number in the interval (13, ∞). Use x=14. Substitute 14 for x in f(x)=x3−13x2. ​Then, evaluate. f(14) = (14)3−13(14)2 Substitute 14 for x. = 2744−2548 Evaluate. = 196 ​So, f(x)>0 for all x in (13, ∞). Write the solution​ set, selecting the​ interval(s) that satisfy the given inequality. Based on the work in the previous​ steps, f(x)>0 for all x in (13, ∞). ​However, because the inequality involves ≥​, the solutions of x3−13x2=0 must also be included in the solution set. The solution set of the given​ inequality, x3−13x2≥0, is shown below. {0} ∪ [13, ∞) The graph of the solution set on a real number line is shown as follows.

2.5 The following function is given. f(x)=x3−3x2−49x+147

The Rational Zero Theorem states that if f(x)=anxn+an−1xn−1+⋯+a1x+a0 has integer coefficients and pq ​(where pq is​ reduced) is a rational​ zero, then p is a factor of the constant​ term, a0, and q is a factor of the leading​ coefficient, an. Possible rational zeros=Factors of the constant termFactors of the leading coefficient a. List all rational zeros that are possible according to the Rational Zero Theorem. The constant term is 147. The factors of the constant term are equal to tbe following. ​1, −​1, 3​, −​3, 7​, −​7, 21​, −​21, 49​, −​49, 147​, −147 The lead coefficient is 1. The factors of the lead coefficient are​ 1, and −1. Possible rational zeros = 1,−1,3,−3,21,−21,49,−49,147,−1471,−1 = 1,−1,3,−3,21,−21,49,−49,147,−147 b. Use synthetic division to test several possible rational zeros in order to identify one actual zero. 3 1 −3 −49 147 3 0 −147 1 0 −49 0 The zero remainder shows that 3 is a zero of the polynomial function f(x)=x3−3x2−49x+147. c. Since 3 is a zero of the polynomial​ function, x−3 is a factor of the polynomial. Use this result to write the equation ​f(x)=0 in factored form. ​(x−​3)(x2−​49)=0 Solve the factored equation. x−3=0 or x2−49=0 ​First, solve the linear equation. x−3 = 0 x = 3 Factor to solve for the quadratic equation x2−49=0. x2−49=​(x+​7)(x−​7)=0 Solve both factors for x. x=​7,−7 ​Therefore, the zeros of the function f(x)=x3−3x2−49x+147 are 3,7,−7.

2.6 se the graph of the rational function to complete the following statement. As x → −1−​, f(x) → .

The notation x → −1− symbolizes​ "x approaches −1 from the left​". Observing the​ graph, as x approaches −1 from the left x → −1−, the function values decrease without bound. That​ is, the function values approach negative infinity. ​Therefore, as x → −1−​, f(x) →−∞. ..... A coordinate system has a horizontal x-axis labeled from negative 10 to 10 in increments of 1 and a vertical y-axis labeled from negative 10 to 10 in increments of 1. A graph has three branches and asymptotes y= negative 1, x = negative 1 and x =1. The first branch is below y equals negative 1 and to the left of x equals negative 1 comma approaching both. The second branch opens upward between the vertical asymptotes comma reaching a minimum at left parenthesis 0 comma 0 right parenthesis . The third branch is below y equals negative 1 and to the right of x equals 1 comma approaching both. Asymptotes are shown as dashed lines. The horizontal asymptote is y=−1. The vertical asymptotes are x=−1 and x=1.

2.5 A model of​ carry-on luggage has a length that is 10 inches greater than its depth. Airline regulations require that the sum of the​ length, width, and depth cannot exceed 40 inches. These​ conditions, with the assumption that this sum is 40 ​inches, can be modeled by the function​ V(x) that gives the​ luggage's volume, in cubic​ inches, in terms of its​ depth, x, in inches. Use the graph of​ V(x) to the right to complete parts​ (a) and​ (b) below. Volume = depth • length • width: 40−(depth+length) ​V(x) = x • (x+10) • ​[40−​(x+x+10​)] ​V(x) = x(x+10)(30−2x) -2020-20003000xy y=V(x) A coordinate system has a horizontal x-axis labeled from negative 20 to 20 in increments of 5 and a vertical y-axis labeled from negative 2000 to 3000 in increments of 1000. From left to right, a curve labeled y equals Upper V left parenthesis x right parenthesis falls in quadrant 2 passing through the point (negative 10, 0) and continues to the minimum in quadrant 3, rises and passes through the origin to a maximum in quadrant 1, and falls passing through (15, 0) continuing through quadrant 4. a. If the volume of the​ carry-on luggage is 2000 cubic​ inches, two possibilities for its depth are 7.8 inches and 10 inches. Identify these values as points on the graph. Choose the correct graph below. Note that the size of the viewing rectangle has been changed. A. -520-20003000xy A coordinate system has a horizontal x-axis labeled from negative 5 to 20 in increments of 5 and a vertical y-axis labeled from negative 2000 to 3000 in increments of 1000. From left to right, a curve rises in quadrant 3 passing through the origin to a maximum in quadrant 1, and falls passing through (15, 0) continuing through quadrant 4. The points (13, 1406) and (7.8, 2000) are plotted on the curve. B. -520-20003000xy A coordinate system has a horizontal x-axis labeled from negative 5 to 20 in increments of 5 and a vertical y-axis labeled from negative 2000 to 3000 in increments of 1000. From left to right, a curve rises in quadrant 3 passing through the origin to a maximum in quadrant 1, and falls passing through (15, 0) continuing through quadrant 4. The points (13.4, 1000) and (3.2, 1000) are plotted on the curve. All coordinate are approximate . C. -520-20003000xy A coordinate system has a horizontal x-axis labeled from negative 5 to 20 in increments of 5 and a vertical y-axis labeled from negative 2000 to 3000 in increments of 1000. From left to right, a curve rises in quadrant 3 passing through the origin to a maximum in quadrant 1, and falls passing through (15, 0) continuing through quadrant 4. The points (10, 2000) and (4, 1203) are plotted on the curve. D. -520-20003000xy A coordinate system has a horizontal x-axis labeled from negative 5 to 20 in increments of 5 and a vertical y-axis labeled from negative 2000 to 3000 in increments of 1000. From left to right, a curve rises in quadrant 3 passing through the origin to a maximum in quadrant 1, and falls passing through (15, 0) continuing through quadrant 4. The points (10, 2000) and (7.8, 2000) are plotted on the curve. b. Use the graph to describe the​ domain, x, where x is the depth of​ carry-on luggage that would be allowed by the regulation. Choose the correct domain below. A. The domain is ​{x | 0<x<15​} or (0,15). B. The domain is ​{x | 7.8<x<30​} or (7.8,30). C. The domain is ​{x | 0<x≤30​} or (0,30]. D. The domain is ​{x | 7.8≤x≤10​} or [7.8,10].

a) D. -520-20003000xy A coordinate system has a horizontal x-axis labeled from negative 5 to 20 in increments of 5 and a vertical y-axis labeled from negative 2000 to 3000 in increments of 1000. From left to right, a curve rises in quadrant 3 passing through the origin to a maximum in quadrant 1, and falls passing through (15, 0) continuing through quadrant 4. The points (10, 2000) and (7.8, 2000) are plotted on the curve. b) A. The domain is ​{x | 0<x<15​} or (0,15).

2.5 Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. n=​4; 2i and 5i are​ zeros; f(−1)=130 ​f(x)=___________ ​(Type an expression using x as the variable. Simplify your​ answer.)

x Superscript 4 Baseline plus 29 x squared plus 100 x4+29x2+100

2.5 Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value. n=​3; −4 and 6+2i are​ zeros; f (−1)=159 f(x)=______________ ​(Type an expression using x as the variable. Simplify your​ answer.)

x cubed minus 8 x squared minus 8 x plus 160 x3−8x2−8x+160

2.5 Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value. n=​3; 1 and 3i are​ zeros; f (−2)=117

According to the Linear Factorization​ Theorem, an​ nth-degree polynomial can be expressed as the product of a nonzero constant and n linear factors. The form of the polynomial is shown​ below, c1, c2, ..., cn are complex numbers​ (possibly real and not necessarily​ distinct). f(x)=anx−c1x−c2⋯x−cn ​Therefore, the​ third-degree polynomial can be written as the product of a nonzero constant and three linear​ factors, f(x)=anx−c1x−c2x−c3. Each zero c of the polynomial can be used to write one linear factor x−c. The linear factor for c=1 is x−1. The linear factor for c=3i is x−3i. For a​ third-degree polynomial, there are three zeros. Notice that the coefficients of the polynomial must be real.​ Therefore, for every complex​ zero, the complex conjugate must also be a zero of the polynomial. The third zero is the complex conjugate of 3i​, which is c=−3i. This zero corresponds to the linear factor x+3i. ​Therefore, the polynomial can be written as the product of the three linear factors and a nonzero​ constant, as shown below. f(x)=an(x−1)(x−3i)(x+3i) Simplify this polynomial. f(x) = an(x−1)(x−3i)(x+3i) = an(x−1)x2+9 Multiply the complex factors. = anx3−x2+9x−9 Complete the multiplication. To find an​, use the fact that f (−2)=117. Substitute the values into the equation and solve for an. f(x) = anx3−x2+9x−9 117 = an(−2)3−(−2)2+9(−2)−9 an = −3 Then substitute an=−3 and multiply. f(x) = anx3−x2+9x−9 = −3x3+3x2−27x+27 The graph of ​f(x)=−3x3+3x2−27x+27 is shown here. -33-150150xy(1,0)(−2,117) x y graph

2.6 Use the graph of the rational function to complete the following statement. As x → ∞​, f(x) → . ..... A coordinate system has a horizontal x-axis labeled from negative 10 to 10 in increments of 1 and a vertical y-axis labeled from negative 10 to 10 in increments of 1. A graph has three branches and asymptotes y= 3, x = negative 3 and x =4. The first branch is to the left of x equals negative 3 comma falls from right to left passing through y equals 3 comma then rises approaching y equals 3 from below. The second branch opens downward between the vertical asymptotes comma reaching a maximum at left parenthesis 0 comma 0 right parenthesis . The third branch is above y equals 3 and to the right of x equals 4 comma approaching both. Asymptotes are shown as dashed lines. The horizontal asymptote is y=3. The vertical asymptotes are x=−3 and

As x → ∞​, f(x) →3.

2.6 Use the graph of the rational function to complete the following statement. As x → −3−​, f(x) →_______ . ..... -10-8-6-4-2246810-10-8-6-4-2246810xy A coordinate system has a horizontal x-axis labeled from negative 10 to 10 in increments of 1 and a vertical y-axis labeled from negative 10 to 10 in increments of 1. A graph has three branches and asymptotes y= 1, x = negative 3 and x =2. The first branch is above y equals 1 and to the left of x equals negative 3 comma approaching both. The second branch opens downward between the vertical asymptotes comma reaching a maximum at left parenthesis 0 comma 0 right parenthesis . The third branch is to the right of x equals 2 comma falls from left to right passing through y equals 1 comma then rises approaching y equals 1 from below. Asymptotes are shown as dashed lines. The horizontal asymptote is y=1. The vertical asymptotes are x=−3 and

As x → −3−​, f(x) →infinity∞.

2.5 Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. n=​4; 3i and 4i are​ zeros; f(−1)=170 B

Because the polynomial function has imaginary zeros and real​ coefficients, the conjugates of the given zeros must also be zeros. Find the conjugates of the given zeros. Recall that the conjugate of a+bi is a−bi. The conjugate of 3i is −3i. The conjugate of 4i is −4i. ​Therefore, four zeros of the polynomial function are 3i​, −3i​, 4i​, and −4i. The fundamental theorem of algebra states that a polynomial function of degree n has at most n distinct zeros.​ Thus, these four zeros are all of the zeros of the fourth degree polynomial f(x). Use these zeros and the linear factorization theorem shown below to find​ f(x). If f(x)=anxn+an−1xn−1+...+a1x+a0, where n≥1 and an≠0, then f(x)=anx−c1x−c2...x−cn, where c1, c2, ​..., cn are complex numbers​ (possibly real and not necessarily​ distinct). In other​ words, an​ nth-degree polynomial can be expressed as the product of a nonzero constant and n linear​ factors, where each linear factor has a leading coefficient of 1. To find​ f(x), substitute the four imaginary zeros into the factored form of a​ fourth-degree polynomial. ​f(x) = anx−c1x−c2x−c3x−c4 = an(x−3i)(x+3i)(x−4i)(x+4i) Expand this expression. ​f(x) = an(x−3i)(x+3i)(x−4i)(x+4i) = anx2+9x2+16 Multiply the complex conjugates. = anx4+25x2+144 Complete the multiplication. Now find the value of an. Since f(−1)=170​, substitute −1 for x in the function f(x)=anx4+25x2+144 and set the result equal to 170. Then solve for an. f(−1) = 170 an(−1)4+25(−1)2+144 = 170 Substitute −1 for x. an•170 = 170 Simplify. an = 1 Solve for an. ​Finally, substitute this value into the formula for​ f(x). f(x)=1x4+25x2+144 or equivalently ​f(x)=x4+25x2+144

2.5 A model of​ carry-on luggage has a length that is 6 inches greater than its depth. Airline regulations require that the sum of the​ length, width, and depth cannot exceed 20 inches. These​ conditions, with the assumption that this sum is 20 ​inches, can be modeled by the function​ V(x) that gives the​ luggage's volume, in cubic​ inches, in terms of its​ depth, x, in inches. If its volume is 240 cubic​ inches, determine two possibilities for its depth. Volume = depth • length • width: 20−(depth+length) ​V(x) = x • (x+6) • ​[20−​(x+x+6​)] ​V(x) = x(x+6)(14−2x)

Since​ V(x) represents the volume of the luggage with depth x​ inches, set​ V(x) equal to 240. ​V(x)=x(x+6)(14−2x)=240 Now solve for x. Begin by expanding the expression x(x+6)(14−2x). x(x+6)(14−2x) = 240 −2x3+2x2+84x = 240 Then subtract 240 from both sides. −2x3+2x2+84x = 240 −2x3+2x2+84x−240 = 0 Notice that the terms of the resulting polynomial have a greatest common factor​ (GCF) that is not 1. Divide both sides of the equation by this​ GCF, −2. −2x3+2x2+84x−240 = 0 x3−x2−42x+120 = 0 Now use the rational zero theorem to find this polynomial​ equation's rational roots. The rational zero theorem states that if f(x)=anxn+an−1xn−1+⋯+a1x+a0 has integer coefficients and pq ​(where pq is reduced to lowest​ terms) is a rational zero of​ f, then p is a factor of the constant term a0 and q is a factor of the leading coefficient an. To find all possible rational roots of x3−x2−42x+120=​0, begin by finding all factors of the constant​ term, 120. The factors of 120 are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60, and ±120. Now find all factors of the leading​ coefficient, 1. The factors of 1 are ±1. By the rational zero​ theorem, all possible rational roots of x3−x2−42x+120=0 are shown below. Factors of the constant term, 120Factors of the leading coefficient, 1 ​Therefore, all possible rational roots of the polynomial equation are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60, and ±120. Graph the function ​f(x)=x3−x2−42x+120. The graph is shown to the right. 35-0.51xy x y graph Use this graph to find the actual rational roots of the polynomial equation. A real root of the polynomial equation will correspond to an​ x-intercept of the graph of the polynomial. Determine which of the possible rational roots corresponds to one of the​ x-intercepts. The actual rational zero is 4. 35-0.51xy(4,0) x y graph To find the other​ zeros, factor the polynomial x3−x2−42x+120 using synthetic division. Since 4 is a​ zero, x−4 is a factor of the polynomial Start by setting up the synthetic division. 4 1 −1 −42 120 Now bring down the number in the first column. 4 1 −1 −42 120 ↓ 1 Multiply by 4 to find the number in the middle row. Then add the numbers in the second column. 4 1 −1 −42 120 4 1 3 Repeat this process to find the number in the bottom row of the next column. 4 1 −1 −42 120 4 12 1 3 −30 Find the number in the bottom row of the last column. 4 1 −1 −42 120 4 12 −120 1 3 −30 0 The bottom row gives the quotient of x3−x2−42x+120÷(x−4). The degree of the first term of the quotient is one less than the degree of the first term of the dividend. Use the bottom row to factor the polynomial x3−x2−42x+120 on the left side of the equation from above. x3−x2−42x+120 = 0 (x−4)x2+3x−30 = 0 Set each factor equal to​ zero, and solve the resulting equations. x−4 = 0 or x2+3x−30 = 0 x = 4 The trinomial x2+3x−30 cannot be factored. Use the quadratic formula. x = −3±32−4(1)(−30)2(1) Substitute into the quadratic formula. = −3±1292 Multiply and subtract under the radical. ≈ −7.2​,4.2 Simplify. ​Thus, the solution set to the original equation is {−7.2, 4.2, 4}. Since x represents depth and depth cannot be​ negative, discard the value x=−7.2. ​Therefore, the two possibilities for the depth of the luggage are 4 inches and 4.2 inches.

2.7 Solve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. x3≥12x2 Solve the inequality. What is the solution​ set? Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

Solve the inequality. What is the solution​ set? Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. The solution set is StartSet 0 EndSet union left bracket 12 comma infinity right parenthesis{0}∪[12,∞). ​(Simplify your answer. Type your answer in interval notation. Type an exact​ answer, using radicals as needed. Use integers or fractions for any numbers in the​ expression.)

2.6 Use the graph of the rational function to complete the following statement. As x → −2−​, f(x) → .

The notation x → −2− symbolizes​ "x approaches −2 from the​ left." Observing the​ graph, as x approaches −2 from the left ​(x → −2−​), the function values increase without bound. That​ is, the function values approach infinity. ​Therefore, as x → −2−​, f(x) →∞.

2.5 A model of​ carry-on luggage has a length that is 12 inches greater than its depth. Airline regulations require that the sum of the​ length, width, and depth cannot exceed 28 inches. These​ conditions, with the assumption that this sum is 28 ​inches, can be modeled by the function​ V(x) that gives the​ luggage's volume, in cubic​ inches, in terms of its​ depth, x, in inches. Use the graph of​ V(x) to the right to complete parts​ (a) and​ (b) below. Volume = depth • length • width: 28−(depth+length) ​V(x) = x • (x+12) • ​[28−​(x+x+12​)] ​V(x) = x(x+12)(16−2x) -1616-20002000xy y=V(x) A coordinate system has a horizontal x-axis labeled from negative 16 to 16 in increments of 4 and a vertical y-axis labeled from negative 2000 to 2000 in increments of 1000. From left to right, a curve labeled y equals Upper V left parenthesis x right parenthesis falls in quadrant 2 passing through the point (negative 12, 0) and continues to the minimum in quadrant 3, rises and passes through the origin to a maximum in quadrant 1, and falls passing through (8, 0) continuing through quadrant 4.

a. If the volume of the​ carry-on luggage is 510 cubic​ inches, two possibilities for its depth are 3.9 inches and 5 inches. Identify these values as points on the graph. Use the given information to determine the values of x and​ V(x) for the two possibilities. Start with the case when the volume of the luggage is 510 cubic inches and the depth is 3.9 inches. x1=3.9 ​, ​V(x1​)=510 Use these values to write an ordered pair of the form​ (x,V(x)) for this possibility. ​(3.9​,510​) Now determine the values of x and​ V(x) for the case when the volume of the luggage is 510 cubic inches and the depth is 5 inches. x2=5 ​, ​V(x2​)=510 Use these values to write an ordered pair for this possibility. ​(5​,510​) Now plot the ordered pairs (3.9,510) and (5,510) on the given graph. The points are plotted in the graph to the right. Note that the size of the viewing rectangle has been changed. -416-10002000xy A coordinate system has a horizontal x-axis labeled from negative 4 to 16 in increments of 4 and a vertical y-axis labeled from negative 1000 to 2000 in increments of 1000. From left to right, a curve rises in quadrant 3 passing through the origin to a maximum in quadrant 1, and falls passing through (8, 0) continuing through quadrant 4. The points (5, 510) and (3.9, 510) are plotted in red on the curve. b. Use the graph to describe the​ domain, x, where x is the depth of​ carry-on luggage that would be allowed by the regulation. The​ length, width,​ depth, and volume of the luggage must be positive. First determine where the volume is positive using the graph of​ V(x). Since​ V(x) represents the volume of the​ luggage, the volume is positive when the graph is above the​ x-axis. Use the graph to determine the real zeros of​ V(x). Each real zero appears as an​ x-intercept of the graph of​ V(x). The real zeros are −12​, ​0, and 8. Use these values and the graph to determine where​ V(x) is positive. It is positive in the domain ​{x | x<−12 or 0<x<8​} or (−∞,12)∪(0,8). -1616-20002000xy A coordinate system has a horizontal x-axis labeled from negative 16 to 16 in increments of 4 and a vertical y-axis labeled from negative 2000 to 2000 in increments of 1000. From left to right, a red curve falls in quadrant 2 passing through the point (negative 12, 0) and continues in blue to the minimum in quadrant 3, rises and passes through the origin and continues in red to a maximum in quadrant 1, and falls passing through (8, 0) continuing in blue through quadrant 4. The points (negative 12, 0), (0, 0), and (8, 0) are plotted. Since x represents​ depth, it must also be positive.​ Therefore, do not include the interval {x | x<−12 } in the domain of​ V(x). This leaves {x | 0<x<8}. The length of the luggage is x+12 and its width is 16−2x Both of these values are positive in the domain {x | 0<x<8}. ​Therefore, realistic domain of​ V(x) is {x | 0<x<8}.

2.5 The following function is given. f(x)=x3−3x2−4x+12 a. List all rational zeros that are possible according to the Rational Zero Theorem. ____________ ​(Use a comma to separate answers as​ needed.) b. Use synthetic division to test several possible rational zeros in order to identify one actual zero. One rational zero of the given function is ___. ​(Simplify your​ answer.) c. Use the zero from part ​(b​) to find all the zeros of the polynomial function. The zeros of the function f(x)=x3−3x2−4x+12 are _____. ​(Simplify your answer. Type an integer or a fraction. Use a comma to separate answers as​ needed.)

a. 1,−1,2,−2,3,−3,4,−4,6,−6,12,−12 b.3 c. 3,2,−2

2.6 a. Find the slant asymptote of the graph of the rational function. b. Follow the​ seven-step strategy and use the slant asymptote to graph the rational function. f(x)=x2−100x

a. Recall that the graph of a rational function has a slant asymptote if the numerator and denominator have no common factors and the degree of the numerator is one more than the degree of the denominator. There are no common factors in the numerator and the denominator of the given rational function. For the given rational function the degree of the numerator is one more than the degree of the denominator. Since the degree of the​ numerator, 2, is exactly one more than the degree of the​ denominator, 1, and x is not a factor of x2−100​, the graph of f has a slant asymptote. To find the equation of the slant​ asymptote, divide x into x2−100. x ​+ 100x x x2−100 The equation of the slant asymptote is y=x. b. To graph the rational​ function, follow the​ seven-step strategy. The first step of the strategy to graph the rational function is to determine the symmetry of the graph of f. To determine the symmetry of the graph of​ f, substitute −x for x in the given rational function and simplify. f(−x) = (−x)2−100(−x) Substitute −x for x. = −x2−100x The graph of f is symmetric with respect to the origin because f(−x)=−f(x). Now find the​ y-intercept by evaluating​ f(0). f(0) = (0)2−1000 Substitute 0 for x. = −1000 There is a 0 in the​ denominator, so​ f(0) is undefined. ​Thus, there is no​ y-intercept. Next find the​ x-intercepts by solving the equation p(x)=0​, where​ p(x) is the numerator of​ f(x). Solve the equation p(x)=0​, where​ p(x) is the numerator of​ f(x). x2−100 = 0 Set the numerator equal to 0. x2 = 100 Add 100 to both sides. Use the square root property. x2 = 100 x = ​10,−10 ​Therefore, the​ x-intercepts are 10 and −10. The graph passes through (10,0) and (−10,0). Find the any vertical​ asymptote(s) by first determining any common factors. Note that the numerator and the denominator of f(x)=x2−100x have no common factors. Since there are no common factors in the numerator and the​ denominator, solve the equation ​q(x)=​0, where​ q(x) is the denominator of​ f(x). ​Therefore, the equation of the vertical asymptote is x=0. Find the horizontal asymptote using the rule for determining the horizontal asymptote of a rational function. Note that the degree of the​ numerator, 2, which is greater than the degree of the​ denominator, 1, for the given rational function. ​Thus, there is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator. Now plot points between and beyond​ x-intercepts at 10 and −10 and vertical asymptote x=0. Evaluate the function at −14​, −8​, 8​, 14​, and 16. x −14 −8 8 14 16 f(x)=x2−100x −487 92 −92 487 394 ​Finally, use the information obtained in the previous steps to graph the function between and beyond the vertical asymptotes. The graph is shown to the right. -30-151530-30-151530xy A coordinate system has a horizontal x-axis labeled from negative 30 to 30 in increments of 3 where every fifteenth tick is labled and a vertical y-axis labeled from negative 30 to 30 in increments of 3 where every fifteenth tick is labled. The xy plane contains two branches and two dashed lines. The first dashed line is vertical and intersects the x-axis at the origin. The second dashed line is diagonal and passes through the origin and rises from the left of the plane in the third quadrant to the right of the plane in the first quadrant. The left branch is above the diagonal dashed line and approaches positive infinity from below, as x approaches 0 from the left. The right branch is below the diagonal dashed line and approaches negative infinity from above, as x approaches 0 from the right.