Chapter 2 Homework Pre-Cal (2.1 - 2.4)

2.2 Consider the function f(x)=3x2−18x−8. a. ​Determine, without​ graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the​ function's domain and its range.

..a. The function has a minimum value. Note: If a>​0, the graph of f opens​ up, the vertex is a minimum​ point, and the function has a minimum value. If a<​0, the graph of f opens​ down, the vertex is a maximum​ point, and the function has a maximum value. Because a>​0, the function has a minimum value. b. The​ minimum/maximum value is −35. It occurs at x=3. c. The domain of f is (−∞,∞). ​(Type your answer in interval​ notation.) The range of f is[−35,∞). ​(Type your answer in interval​ notation.)

2.2 Consider the function f(x)=−3x2+6x−8. a. ​Determine, without​ graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the​ function's domain and its range.

.a. The function has a maximum value. Note: If a>​0, the graph of f opens​ up, the vertex is a minimum​ point, and the function has a minimum value. If a<​0, the graph of f opens​ down, the vertex is a maximum​ point, and the function has a maximum value. Because a>​0, the function has a minimum value. b. The​ minimum/maximum value is −5. It occurs at x=1. c. The domain of f is (−∞,∞). ​(Type your answer in interval​ notation.) The range of f is (−∞,−5]. ​(Type your answer in interval​ notation.)

2.3 Use the leading coefficient test to determine the end behavior of the graph of the given polynomial function. ​f(x)=6x7−4x5+7x4+4 A. Falls left​ & rises right. B. Rises left​ & falls right. C. Falls left​ & falls right. D. Rises left​ & rises right. E. None of the above.

A. Falls left​ & rises right.

2.3 Determine whether the function is a polynomial function. If it​ is, identify the degree. ​g(x)=2x4−πx3+16x2 Choose the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. It is a polynomial. The degree of the polynomial is ____. B. It is not a polynomial.

A. It is a polynomial. The degree of the polynomial is 4

2.2 You have 1000 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the​ river, find the length and width of the plot that will maximize the area. What is the largest area that can be​ enclosed?

Begin by expressing the​ area, A, in terms of x. Since the area of a rectangle is found by the​ formula, Area = length×width​, multiply the length and the width of the plot to find the area. Express the​ area, A, in terms of x. A=length×width =​(1000−​2x)•x =1000x−2x2 Distribute the x. =−2x2+1000x Write with the exponents in descending order. The​ area, A, is a quadratic function of​ x, in the form ax2+bx+c. ​A(x)=−2x2+1000x Since a=−2<​0, the parabola opens​ downward, so the vertex is the maximum point on the graph of A. This means that the maximum value of the area function is the​ y-coordinate of the vertex. The​ x-coordinate of the vertex corresponds to the side labeled x in the figure. Find the vertex. The​ x-coordinate of the vertex of a quadratic function occurs at x=−b2a. x=−b2a=−10002(−2)=250 The width of the plot is x=250 feet. To find the length of the​ plot, look back at the figure and note that the length is labeled 1000−2x. Substitute 250 for x and simplify to find the length. length=1000−2x =1000−​2(250​)=500 feet To find the maximum​ area, either multiply the length from above by the width from​ above, or find the maximum value of the area function​ (the y-coordinate of the​ vertex). Here, evaluate the area​ function, at its vertex. Evaluate ​A(x)=−2x2+1000x at x=250. ​A(250​)=−2(250)2+1000(250)=125,000 ​Therefore, the length and width that maximize the area of the plot are 500 feet and 250 feet. The largest area that can be enclosed is 125,000 square feet.

2.3 Find the zeros for the polynomial function and give the multiplicity for each zero. State whether the graph crosses the​ x-axis or touches the​ x-axis and turns around at each zero. ​f(x)=x3+9x2−4x−36

Determine the​ zero(s), if they exist. The​ zero(s) is/are negative 9 comma negative 2 comma 2−9,−2,2. ​(Type integers or decimals. Use a comma to separate answers as​ needed.) Determine the multiplicities of the​ zero(s), if they exist. Select the correct choice below​ and, if​ necessary, fill in the answer​ box(es) within your choice. A. There are two zeros. The multiplicity of the smallest zero is nothing. The multiplicity of the largest zero is nothing. ​(Simplify your​ answers.) B. There is one zero. The multiplicity of the zero is nothing. ​(Simplify your​ answer.) C. There are three zeros. The multiplicity of the smallest zero is 11. The multiplicity of the largest zero is 11. The multiplicity of the other zero is 11. ​(Simplify your​ answers.) . Determine the behavior of the function at each zero. Select the correct choice below​ and, if​ necessary, fill in the answer boxes within your choice. A. The graph crosses the​ x-axis at all zeros. Your answer is correct. B. The graph crosses the​ x-axis at x=nothing. The graph touches the​ x-axis and turns around at x=nothing. ​(Type integers or decimals. Simplify your answers. Use a comma to separate answers as​ needed.) C. The graph touches the​ x-axis and turns around at all zeros.

2.2 Give the domain and range of the quadratic function whose graph is described. The vertex is (−9,−5) and the parabola opens down.

The domain of f is s(−∞,∞). ​(Type your answer in interval​ notation.) The range of the function is (−∞,−5]. ​(Type your answer in interval​ notation.)

2.2 Give the domain and range of the quadratic function whose graph is described. Maximum=−6 at x=2

The domain of the function i(−∞,∞). ​(Type your answer in interval​ notation.) The range of the function is −∞,−6]. ​(Type your answer in interval​ notation.)

2.2 Write an equation in vertex form of the parabola that has the same shape as the graph of f(x)=10x2​, but with (4,11) as the vertex.

The quadratic function g(x)=a(x−h)2+k, a≠0, is in vertex form. The graph of g is a parabola whose vertex is the point (h,k). The graph of f(x)=10x2 is a parabola with its vertex at (0,0). The new graph must have its vertex at (4,11). For the new​ graph, h=4 and k=11. The​ a-value determines the shape of the parabola. The sign of a determines whether the parabola opens upward or downward. Since the new graph must have the same shape as f(x)=10x2, the​ a-value is known. In the new​ graph, the value of a is 10. ​Now, substitute the values of​ a, h, and k into the vertex form for the equation of a quadratic​ function, g(x)=a(x−h)2+k, to obtain the final answer. g(x)=10(x−4)2+11

2.2 You have 750 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the​ river, find the length and width of the plot that will maximize the area. What is the largest area that can be​ enclosed?

The​ width, labeled x in the​ figure, is 187.5 feet. ​(Type an integer or​ decimal.) The​ length, labeled 750−2x in the​ figure, is 375 feet. ​(Type an integer or​ decimal.) The largest area that can be enclosed is 70312.5 square feet. ​(Type an integer or​ decimal.)

2.4 a. Use synthetic division to show that 2 is a solution of the polynomial equation 16x3+11x2−16x−140=0 b. Use the solution from part​ (a) to solve this problem. The number of​ eggs, f(x), in a female moth is a function of her abdominal​ width, in​ millimeters, modeled by ​f(x)=16x3+11x2−16x+91 What is the abdominal width when there are 231 ​eggs?

a. One way to show that 2 is a solution is to substitute 2 for x in the equation and obtain 0. An easier way is to use synthetic division and the remainder theorem. The remainder theorem states the following. If the polynomial​ f(x) is divided by x−c​, then the remainder is​ f(c). ​Therefore, let c be 2. If the remainder of the​ division, 16x3+11x2−16x−140 divided by x−2 is​ 0, then 2 is a solution to the given equation. Complete the top row of the synthetic division. 2 16 11 −16 −140 The first step of the process is to write the leading coefficient of the dividend on the bottom row. In other​ words, bring down 16. 2 16 11 −16 −140 ↓ 16 ​Next, multiply c​ (in this​ case, 2​) times the value just written on the bottom row. Type the product in the second column in the second row. 2 16 11 −16 −140 32 16 Add the values in the second column and type the sum in the bottom row. 2 16 11 −16 −140 32 16 43 Now repeat the multiplication and addition procedure for each of the remaining columns. Multiply c times the value just written on the bottom row. Type the product in the third column in the second row. 2 16 11 −16 −140 32 86 16 43 Add the values in the third column and type the sum in the bottom row. 2 16 11 −16 −140 32 86 16 43 70 Multiply c times the value just written on the bottom row. Type the product in the fourth column in the second row. 2 16 11 −16 −140 32 86 140 16 43 70 Add the values in the fourth column and type the sum in the bottom row. 2 16 11 −16 −140 32 86 140 16 43 70 0 Because the remainder is​ 0, the polynomial has a value of 0 when x=2. ​Thus, 2 is a solution of the given equation. b. It is given that​ f(x) represents the number of eggs in the moth and x represents her abdominal width. To find the abdominal width when there are 231 ​eggs, let f(x)=231 and solve for x. 16x3+11x2−16x+91=231 Move the terms to the left side by subtracting 231 from both sides. Then simplify the result. 16x3+11x2−16x+91 =231 16x3+11x2−16x+91−231 =231−231 16x3+11x2−16x−140 =0 From part​ (a), x=2 is a solution to the​ equation, 16x3+11x2−16x−140=0. ​Therefore, x=2 is also a solution to 16x3+11x2−16x+91=231. f(2)=231 It is possible that there are solutions other than x=2. Note that dividing 16x3+11x2−16x−140 by x−2 results in the polynomial 16x2+43x+70. Use the discriminant to determine the type of solutions of 16x2+43x+70=0. The disciminant is b2−4ac=432−4(16)(70)=−2631. Since the discriminant is​ negative, the only real solution of 16x3+11x2−16x+91=231 is x=2. ​Thus, when there are 231 ​eggs, the abdominal width is 2 millimeters.

2.4 a. Use synthetic division to show that 2 is a solution of the polynomial equation below. 12x3+11x2+10x−160=0 b. Use the solution from part​ (a) to solve this problem. The number of​ eggs, f(x), in a female moth is a function of her abdominal​ width, in​ millimeters, modeled by the equation below. ​f(x)=12x3+11x2+10x−49 What is the abdominal width when there are 111 ​eggs?

a. The number 2 is a solution to the equation because the remainder of the​ division, 12x3+11x2+10x−160 divided by x−2​, is 0. b. The abdominal width is 2 millimeters.

2.2 An athlete whose event is the shot put releases a shot. When the shot whose path is shown by the graph to the right is released at an angle of 55​°, its​ height, f(x), in​ feet, can be modeled by f(x)=−0.02x2+1.4x+5.7, where x is the​ shot's horizontal​ distance, in​ feet, from its point of release. Use this model to solve parts​ (a) through​ (c) and verify your answers using the graph. Shot Put's Horizontal DistanceShot Put's Height A coordinate system has a horizontal x-axis labeled Shot Put's Horizontal Distance from 0 to 100 in increments of 10 and a vertical y-axis labeled Shot Put's Height from 0 to 50 in increments of 5. A parabola that opens downwards begins at (0, 6), has vertex (35, 30), and stops at the point (74, 0). All coordinates are approximate.

a. What is the maximum height of the shot and how far from its point of release does this​ occur? The graph of a function written in the form f(x)=ax2+bx+c has a vertex at the point −b2a,f −b2a. The​ y-value of the vertex gives the maximum height. ​First, identify the values of​ a, b, and c by comparing the function f(x)=−0.02x2+1.4x+5.7 to the form f(x)=ax2+bx+c. a=−0.02​, b=1.4​, c=5.7 Calculate the distance from the point of release that the maximum height occurs. x = −b2a = −1.42(−0.02) Substitute a=−0.02 and b=1.4. = 35 Simplify. ​Therefore, the shot reaches its maximum height at x=35, or at a distance of 35 feet from its point of release. To find the maximum​ height, evaluate the function at x=35. f(x) = −0.02x2+1.4x+5.7 f(35) = −0.02(35)2+1.4(35)+5.7 Substitute x=35. = 30.2 Simplify. The maximum height is 30.2 ​feet, which occurs 35 feet from the point of release. b. What is the​ shot's maximum horizontal​ distance, to the nearest tenth of a​ foot, or the distance of the​ throw? To find the​ shot's maximum horizontal​ distance, use the quadratic formula to find the​ x-intercepts of the function f(x)=−0.02x2+1.4x+5.7. Substitute the values a=−0.02, b=1.4, and c=5.7 into the quadratic formula. x = −b±b2−4ac2a = −1.4±1.42−4(−0.02)(5.7)2(−0.02) Now solve for x. First simplify the formula with addition in the numerator. x = −1.4+1.42−4(−0.02)(5.7)2(−0.02) ≈ −3.9 Simplify and round to the nearest tenth. Make sure to use values that correspond to horizontal distances from the point of​ release, or x≥0. ​Therefore, reject the solution x=−3.9. Now simplify the formula with subtraction in the numerator. x = −1.4−1.42−4(−0.02)(5.7)2(−0.02) ≈ 73.9 Simplify and round to the nearest tenth. ​Therefore, the maximum horizontal distance of the shot is approximately 73.9 feet. c. From what height was the shot​ released? The shot was released at the​ origin, so find the height of the graph at the origin by evaluating the function at x=0. f(0) = −0.02(0)2+1.4(0)+5.7 = 5.7 ​Therefore, the shot was released at a height of 5.7 feet.

2.2 Write an equation in vertex form of the parabola that has the same shape as the graph of f(x)=7x2​, but with (4,2) as the vertex.

g(x)= 7 left parenthesis x minus 4 right parenthesis squared plus 2 7(x−4)2+2

2.3 Find the zeros for the polynomial function and give the multiplicity for each zero. State whether the graph crosses the​ x-axis or touches the​ x-axis and turns around at each zero. ​f(x)=6​(x+5​)(x−3)2 Determine the​ zero(s). The​ zero(s) is/are ______ ​(Type integers or decimals. Use a comma to separate answers as​ needed.) Determine the multiplicities of the​ zero(s). Select the correct choice below​ and, if​ necessary, fill in the answer​ box(es) within your choice. A. There are two zeros. The multiplicity of the smallest zero is __. The multiplicity of the largest zero is __. ​(Simplify your​ answers.) B. There is one zero. The multiplicity of the zero is nothing. ​(Simplify your​ answer.) C. There are three zeros. The multiplicity of the smallest zero is nothing. The multiplicity of the largest zero is nothing. The multiplicity of the other zero is nothing. ​(Simplify your​ answers.) Determine the behavior of the function at each zero. Select the correct choice below​ and, if​ necessary, fill in the answer boxes within your choice. A. The graph touches the​ x-axis and turns around at all zeros. B. The graph crosses the​ x-axis at all zeros. C. The graph crosses the​ x-axis at x=__ and touches the​ x-axis and turns around at x=__. ​(Simplify your answers. Type integers or decimals. Use a comma to separate answers as​ needed.)

negative 5 comma 3 −5,3. A. There are two zeros. The multiplicity of the smallest zero is 1. The multiplicity of the largest zero is 2. C. The graph crosses the​ x-axis at x=−5 and touches the​ x-axis and turns around at x=3.

2.4 Use synthetic division to divide ​f(x)=x3+2x2−5x−6 by x−2. Use the result to find all zeros of f. ​(x3+2x2−5x−6​)÷​(x−2​)=__________ ​(Do not factor. If there is a​ remainder, type your answer in the form quotient+remainderdivisor.​) The zeros of f are ________. ​(Use a comma to separate answers as​ needed.)

x squared plus 4 x plus 3 x2+4x+3 −1,2,−3 negative 1 comma 2 comma negative 3

2.2 Write an equation in vertex form of the parabola that has the same shape as the graph of f(x)=4x2 or g(x)=−4x2​, but with the given minimum. Minimum=0 at x=12 ​

​h(x)= 4 left parenthesis x minus 12 right parenthesis squared 4(x−12)^2