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MTH-112 Quiz 7 Name: #: Please write your name in the provided space. Simplify your answers. Show your work. 1. Find the domain of f (x) = your answer as an interval.) √ x2 + 9. (Write (f) The x−intercepts (if exist): (g) The relative minimum, if exists (write your answer as an ordered pair): (h) The relative maximum, if exists (write your answer as an ordered pair): Domain = 2. A quadratic function is graphed below. Draw the axis of symmetry using a dotted line. Find the following using the graph. Write “None” if necessary. y (i) In which interval, if any, is the function increasing? 2 1 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 (j) In which interval, if any, is the function decreasing? 6 -1 -2 -3 -4 (k) In which interval, if any, is the function a constant? -5 (a) The vertex (write your answer as an ordered pair): 3. Use the quadratic function f (x) = 3(x+4)2 −27 to find the following. (b) The equation of the axis of symmetry: (a) Does the graph open upward or downward? (upward / downward) (c) The domain (write your answer as an interval): (b) The vertex (write your answer as an ordered pair): (d) The range (write your answer as an interval): (c) The equation of the axis of symmetry: (e) The y−intercept (if exists): 13 MTH-112 Quiz 7 - Solutions Words in italics are for explanation purposes only (not necessary to write in the tests or quizzes). √ x2 + 9. (Write (d) The range (write your answer as an interval): (−∞, 1] The domain is the set of real numbers which x can be replaced with, and get a meaningful y value. In a square root function, the inside of the radical cannot be negative. Since the inside of the given function is always positive, the domain is all real numbers. (When you square x, it is non-negative. So x2 is always non-negative. When you add 9 to that, it is positive. So x2 + 9 is always positive.) The range of a parabola is always either from the y−coordinate of the vertex to positive infinity (if the parabola opens upward), or negative infinity to the y−coordinate of the vertex (if it opens downward). 1. Find the domain of f (x) = your answer as an interval.) Domain: = (−∞, ∞) 2. A quadratic function is graphed below. Draw the axis of symmetry using a dotted line. Find the following using the graph. Write “None” if necessary. 2 x -4 -3 -2 -1 (f) The x−intercepts (if exist): (3, 0) (1, 0) and The x−intercepts are the points where the graph crosses the x−axis. Read the x−intercepts from the graph. A relative minimum is a point on the graph where the graph turns from decreasing to increasing. 1 -5 The y−intercept is the point where the graph crosses the y−axis. Read the y−intercept from the graph. (g) The relative minimum, if exists (write your answer as an ordered pair): y -6 (e) The y−intercept (if exists): (0, −3) 1 2 3 4 5 6 -1 Relative minimum: None. -2 (h) The relative maximum, if exists (write your answer as an ordered pair): -3 -4 A relative maximum is a point on the graph where the graph turns from increasing to decreasing; the same as the vertex. -5 (a) The vertex (write your answer as an ordered pair): (2, 1) (2, 1) Read the vertex from the graph. (i) In which interval, if any, is the function increasing? (−∞, 2) (b) The equation of the axis of symmetry: x=2 The equation of the axis of symmetry is always x = the x−coordinate of the vertex. From left to right, if the graph is going up, it is increasing. (j) In which interval, if any, is the function decreasing? (2, ∞) From left to right, if the graph is going down, it is decreasing. (c) The domain (write your answer as an interval): (−∞, ∞) The domain of a parabola is always all real numbers (the graph is defined from the negative infinity to positive infinity). 14 (k) In which interval, if any, is the function a constant? None From left to right, if the graph stays at the same level, it is constant. MTH-112 Quiz 7 - Solutions Name: 3. Use the quadratic function f (x) = 3(x+4)2 −27 to find the following. #: (a) Does the graph open upward or downward? ( upward / downward) Read from the function; the x−coordinate of the vertex is the number inside the parenthesis with opposite sign; the y−coordinate of the vertex is the number subtracted or added at the end. Since a = 3 is a positive number, the graph opens upward. (c) The equation of the axis of symmetry: x = −4 (b) The vertex (write your answer as an ordered pair): (−4, −27) The equation of the axis of symmetry is always x = the x−coordinate of the vertex. MTH-112 Quiz 8 Please write your name in the provided space. Simplify your answers. Show your work. 1. Use the quadratic function f (x) = 3(x+4)2 −27 to find the following. (b) The vertex (write your answer as an ordered pair): (a) The range (write your answer as an interval): (b) The y−intercept (if exists): (c) The equation of the axis of symmetry: (d) The range (write your answer as an interval): (c) The x−intercepts (if exist): (e) The y−intercept (if exists): (f) The x−intercepts (if exist): 2. Use the quadratic function f (x) = x2 − 8x + 12 to find the following. (a) Does the graph open upward or downward? 15 3. Is f (x) = x6 + 16 x + (yes / no) 1 6 a polynomial function? If the function is a polynomial, find the degree of the polynomial. MTH-112 Quiz 8 - Solutions Words in italics are for explanation purposes only (not necessary to write in the tests or quizzes). 1. Use the quadratic function f (x) = 3(x+4)2 −27 to find the following. (b) The vertex (write your answer as an ordered pair): (4, −4) (a) The range (write your answer as an interval): [−27, ∞) The range of a parabola is always either from the y−coordinate of the vertex to positive infinity (if the parabola opens upward), or negative infinity to the y−coordinate of the vertex (if it opens downward). (b) The y−intercept (if exists): (0, 21) The y−intercept is the point where the graph crosses the y−axis. Since the x−coordinate of the y−intercept is zero, find f (0), which is the same as the y− value when x is 0. f (0) = 3(0 + 4)2 − 27 = 21 (c) The x−intercepts (if exist): (−7, 0) (−1, 0), The x−intercepts are the points where the graph crosses the x−axis. Since the y−coordinate of the x−intercepts is zero, write 0 for y, and solve for x. 2 0 = 3(x + 4) − 27 27 = 3(x + 4)2 9 = (x + 4)2 ±3 = x + 4 − −8 =4 2(1) b . 2a To find the y−coordinate of the vertex, find f (4), which is the same as the y− value when x is 4. f (4) = (4)2 − 8(4) + 12 = −4 (c) The equation of the axis of symmetry: x=4 The equation of the axis of symmetry is always x = the x−coordinate of the vertex. (d) The range (write your answer as an interval): [−4, ∞) The range of a parabola is always either from the y−coordinate of the vertex to positive infinity (if the parabola opens upward), or negative infinity to the y−coordinate of the vertex (if it opens downward). (e) The y−intercept (if exists): (0, 12) The y−intercept is the point where the graph crosses the y−axis. Since the x−coordinate of the y−intercept is zero, find f (0), which is the same as the y− value when x is 0. f (0) = 02 − 8(0) + 12 = 12 −4 ± 3 = x −1 = x The x−coordinate of the vertex is − When the function is given in the form f (x) = ax2 + bx + c, the y−coordinate of the y−intercept is c. −7 = x 2. Use the quadratic function f (x) = x2 − 8x + 12 to find the following. (f) The x−intercepts (if exist): (2, 0), (6, 0) The x−intercepts are the points where the graph crosses the x−axis. Since the y−coordinate of the x−intercepts is zero, write 0 for y, and solve for x. The function is given in the form, f (x) = ax2 + bx + c, where a = 1, b = −8, and c = 12. (a) Does the graph open upward or downward? Upward Since a = 1 is a positive number, the graph opens upward. 16 0 = x2 − 8x + 12 0 = (x − 2)(x − 6) x−2=0 x−6=0 x=2 x=6 MTH-112 Quiz 8 - Solutions 3. Is f (x) = x6 + 16 x + ( yes / no) 1 6 Name: a polynomial function? Since all the exponents of the variable are nonnegative integers, the function is a polynomial. #: If the function is a polynomial, find the degree of the polynomial. The degree is the highest power of the variable. Degree: 6 MTH-112 Quiz 9 Please write your name in the provided space. Simplify your answers. Show your work. 1. Use the quadratic function f (x) = x2 − 2x − 24 to find the following. 2. Find the zeros of f (x) = 11x(x−22)33 (x+44)55 , and the multiplicity of each zero. (a) The equation of the axis of symmetry: Zero (b) The range (write your answer as an interval): Multiplicity 3. Find the zeros of f (x) = 5x2 + x3 , and the multiplicity of each zero. Zero Multiplicity (c) The x−intercepts (if exist): 4. Is the graph of the above function symmetric about y axis, origin, or neither? (y axis / origin / neither) 17 MTH-112 Quiz 9 - Solutions Words in italics are for explanation purposes only (not necessary to write in the tests or quizzes). 1. Use the quadratic function f (x) = x2 − 2x − 24 to find the following. The factor x − 22 gives the zero 22, and the multiplicity of that zero is the exponent of the factor, which is 33. The function is of the form f (x) = ax2 + bx + c, where a = 1, b = −2, and c = −24. The factor x + 44 gives the zero −44, and the multiplicity of that zero is the exponent of the factor, which is 55. (a) The equation of the axis of symmetry: x=1 The equation of the axis of symmetry is always x = the x−coordinate of the vertex. −b to find the x−coordinate of the verUse 2a tex. x= −(−2) −b = =1 2a 2(1) Zero Multiplicity 0 1 22 33 −44 55 (b) The range (write your answer as an interval): [−25, ∞) 3. Find the zeros of f (x) = 5x2 + x3 , and the multiplicity of each zero. Since a = 1 is a positive number, the graph opens upward. Therefore, the range is from the y−coordinate of the vertex to infinity. Replace x of the function with 1 (the x−coordinate of the vertex) to find the y−coordinate of the vertex. The zeros are the x values for which f (x) is zero. Write 0 for f (x), and solve for x. 5x2 + x3 = 0 x2 (5 + x) = 0 5x2 = 0 f (1) = (1)2 − 2(1) − 24 = −25 (c) The x−intercepts (if exist): (−4, 0) (6, 0) and x2 − 2x − 24 = 0 (x − 6)(x + 4) = 0 x=6 x = −5 x=0 The x−intercepts are the points where the graph crosses the x−axis. Since the y−coordinate of the x−intercepts is zero, write 0 for f (x), and solve for x. x−6=0 5+x=0 x+4=0 The factor 5x2 gives the zero 0, and the multiplicity of that zero is the exponent of the factor, which is 2. The factor 5 + x gives the zero −5, and the multiplicity of that zero is the exponent of the factor, which is 1. Zero Multiplicity 0 2 −5 1 x = −4 2. Find the zeros of the polynomial function f (x) = 11x(x − 22)33 (x + 44)55 , and the multiplicity of each zero. 4. Is the graph of the above function symmetric about y axis, origin, or neither? (y axis / origin / neither ) The factor 11x gives the zero 0, and the multiplicity of that zero is the exponent of the factor, which is 1. The function is neither even nor odd. Therefore, it is neither symmetric about y− axis nor about the origin. 18 MTH-112 Quiz 10 Name: #: Please write your name in the provided space. Simplify your answers. Show your work. 1. Use the quadratic function f (x) = −x2 + 4x + 4 to find the following. 4. Use the function f (x) = 2x3 + x2 − 12x + 9 to find the following. (a) The equation of the axis of symmetry: (a) Find all possible rational zeros of the function. (b) The range (write your answer as an interval): (c) The x−intercepts (if exist): 3 Factor of the leading coefficient. Factor of the constant term. (b) Find one actual zero of the function using synthetic substitution. (c) Write a factor of the function using the zero found in part (b). 2 2. Given f (x) = x − 4x + 5x − 3, find f (2) using synthetic substitution. (d) Factor the function completely using the factor found in part (c), and the quotient found in part (b). f (2) = . . . . . .. 3. Determine whether 3 is a zero of f (x) = x3 − 4x2 + 5x − 3 using synthetic substitution. (yes / no) (e) Find all actual zeros of the function using the factors found in part (d). 19 MTH-112 Quiz 10 - Solutions Words in italics are for explanation purposes only (not necessary to write in the tests or quizzes). 1. Use the quadratic function f (x) = −x2 + 4x + 4 to find the following. √ b2 − 4ac 2a p −4 ± 42 − 4(−1)(4) = 2(−1) √ −4 ± 32 = −2 √ −4 ± 4 2 = −2 √ x=2−2 2 x= 2 The function is of the form f (x) = ax + bx + c, where a = −1, b = 4, and c = 4. (a) The equation of the axis of symmetry: The equation of the axis of symmetry is always x = the x−coordinate of the vertex. −b to find the x−coordinate of the verUse 2a tex. −b ± √ x=2+2 2 √ √ (2 − 2 2, 0) and (2 + 2 2, 0) −4 −b = =2 x= 2a 2(−1) 2. Given f (x) = x3 − 4x2 + 5x − 3, find f (2) using synthetic substitution. x=2 (b) The range (write your answer as an interval): Since a = −1 is a negative number, the graph opens downward. Therefore, the range is from negative infinity to the y−coordinate of the vertex. Replace x of the function with 2 (the x−coordinate of the vertex) to find the y−coordinate of the vertex. 1 2 1 −4 5 −3 2 −4 2 −2 1 −1 f (2) is the remainder, which is −1. f (2) = −1. 3. Determine whether 3 is a zero of f (x) = x3 − 4x2 + 5x − 3 using synthetic substitution. (yes / no ) f (2) = −22 + 4(2) + 4 = −4 + 8 + 4 =8 1 (−∞, 8]. 3 1 (c) The x−intercepts (if exist): The x−intercepts are the points where the graph crosses the x−axis. Since the y−coordinate of the x−intercepts is zero, write 0 for y, and solve for x. 0 = −x2 + 4x + 4 −4 5 −3 3 −3 6 −1 2 3 Since the remainder is not zero, 3 is not a zero of the function. 4. Use the function f (x) = 2x3 + x2 − 12x + 9 to find the following. (a) Find all possible rational zeros of the function. Since this cannot be factored, use quadratic formula. 20 MTH-112 Quiz 10 - Solutions Name: When x = c is a zero, x − c is a factor. Factor of the leading coefficient. Factor of the constant term. 2 9 1 1 ±1 2 ± 1 2 3 9 ±3 ± 3 2 x−1 (d) Factor the function completely using the factor found in part (c), and the quotient found in part (b). ±9 ± #: 9 2 The quotient in part (b) is 2x2 + 3x − 9. f (x) = 2x3 + x2 − 12x + 9 = (x − 1)(2x2 + 3x − 9) (b) Find one actual zero of the function using synthetic substitution. 2 1 2 1 − 12 9 2 3 −9 3 −9 0 = (x − 1)(2x − 3)(x + 3) (e) Find all actual zeros of the function using the factors found in part (d). To find zeros, set each factor equal to zero, and solve for x. x=1 (c) Write a factor of the function using the zero found in part (b). x = 1, x = 3 , x = −3 2 MTH-112 Quiz 11 Please write your name in the provided space. Simplify your answers. Show your work. 1. Let 1 and −3i be two zeros of a third degree polynomial function f (x), and f (2) = 26. (a) What is the other zero of the function? (b) Write the factored form of the function using ‘a’ as the leading coefficient. Then multiply the factors (FOIL). f (x) = (d) Find the polynomial function (do not leave your answer in factored form, multiply all factors). 2. If (−3, 0) and (5, 0) are the x− intercepts of the graph of a quadratic function, what is the equation of the axis of symmetry? 3. What are the zeros of the quadratic function in problem 2? (c) Find the value of a, using the given condition f (2) = 26. 4. Find the domain of f (x) = Domain = 21 2x2 − 72 . − x − 12 x2 MTH-112 Quiz 11 - Solutions Words in italics are for explanation purposes only (not necessary to write in the tests or quizzes). 1. Let 1 and −3i be two zeros of a third degree polynomial function f (x), and f (2) = 26. 2. If (−3, 0) and (5, 0) are the x− intercepts of the graph of a quadratic function, what is the equation of the axis of symmetry? (a) What is the other zero of the function? Whenever a complex number is a zero, the conjugate of that is also a zero. Since the graph is symmetric about the axis of symmetry, the x− intercepts are also symmetric about the axis of symmetry. Other zero: 3i (b) Write the factored form of the function using ‘a’ as the leading coefficient. Then multiply the factors (FOIL). -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Since 1 is a zero, x − 1 is a factor. Since −3i is a zero, x + 3i is a factor. x=1 Since 3i is a zero, x − 3i is a factor. The axis of symmetry: x = 1 Therefore, the factored form is f (x) = a(x − 1)(x + 3i)(x − 3i) 3. What are the zeros of the quadratic function in problem 2? First, multiply the factors, starting with (x + 3i)(x − 3i). The zeros of the function are the x− values for which the function values (the y− coordinates) are equal to zero. That is, the x− coordinates of the x− intercepts. f (x) = a(x − 1)(x + 3i)(x − 3i) = a(x − 1)(x2 − 3ix + 3ix − 9i2 ) = a(x − 1)(x2 − 9(−1)) = a(x − 1)(x2 + 9) Zeros: −3, 5 = a(x3 − x2 + 9x − 9) (c) Find the value of a, using the given condition f (2) = 26. 4. Find the domain of f (x) = Replace x with 2, and set it equal to 26. f (2) = a (2)3 − (2)2 + 9(2) − 9 2x2 − 72 . − x − 12 x2 The domain is all real numbers except the zeros of the bottom polynomial. (The zeros of the bottom polynomial make the bottom zero, and hence the function is undefined.) Find the zeros of the bottom polynomial first. 26 = a(8 − 4 + 18 − 9) 26 = a(13) a=2 (d) Find the polynomial function (do not leave your answer in factored form, multiply all factors). Replace a with −1 and rewrite the function found in part (b). x2 − x − 12 = 0 (x − 4)(x + 3) = 0 x−4=0 x=4 x+3=0 x = −3 Therefore, the domain is all real numbers except 4 and −3. f (x) = a(x3 − x2 + 9x − 9) = 2(x3 − x2 + 9x − 9) = 2x3 − 2x2 + 18x − 18 Domain = (−∞, −3) ∪ (−3, 4) ∪ (4, ∞) 22 MTH-112 Quiz 12 Name: #: Please write your name in the provided space. Simplify your answers. Show your work. 1. Use the rational function f (x) = to find the following. 2x2 − 72 − x − 12 (c) The y− intercept (write your answer as an ordered pair): x2 (a) The equation(s) of the vertical asymptote(s): (d) The x− intercept(s) (write your answer(s) as an ordered pair or ordered pairs): (b) The equation of the horizontal asymptote: (e) Graph the function. Draw all asymptotes using dotted lines. y 11 10 9 8 7 6 5 4 3 2 1 x -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 23 2 3 4 5 6 7 8 9 10 11 MTH-112 Quiz 12 - Solutions Words in italics are for explanation purposes only (not necessary to write in the tests or quizzes). Use the rational function f (x) = the following. 2x2 − 72 to find x2 − x − 12 the y−intercept is zero, find f (0), which is the same as the y− value when x is 0. f (0) = 1. The equation(s) of the vertical asymptote(s): Set the bottom equal to zero, and solve for x. 2(0)2 − 72 −72 = =6 (0)2 − 0 − 12 −12 y− intercept: (0, 6) x2 − x − 12 = 0 4. The x− intercept (write your answer as an ordered pair): (x − 4)(x + 3) = 0 x−4=0 x+3=0 x = −3 x=4 The x−intercepts are the points where the graph crosses the x−axis. Since the y−coordinate of the x−intercepts is zero, write 0 for y, and solve for x. Vertical asymptotes: x = 4 and x = −3 2. The equation of the horizontal asymptote: The top degree is 2, and the bottom degree is 2. The top degree is equal to the bottom degree. Therefore the horizontal asymptote is y= 2x2 − 72 x2 − x − 12 0 = 2x2 − 72 0= top leading coefficient 2 = bottom leading coefficient 1 72 = 2x2 36 = x2 Horizontal asymptote: y = 2 ±6 = x 3. The y− intercept (write your answer as an ordered pair): x−intercepts: (6, 0) and (−6, 0) The y−intercept is the point where the graph crosses the y−axis. Since the x−coordinate of 5. Graph the function. Draw all asymptotes using dotted lines. y 10 8 6 4 y=2 2 x -10 -8 -6 -4 -2 2 4 -2 -4 -6 -8 -10 x = −3 x=4 24 6 8 10