ID: 8768cbea Name: Jed Albarico 1. What is the graph of 𝒙𝟐 − 𝒚𝟐 = 𝟐𝟑? What are its lines of symmetry? What are the domain and range? Given: 𝑥 2 − 𝑦 2 = 23? Required: a. Graph b. Lines of symmetry c. Domain and range Solution: First, identify what conic section is expressed by the equation by checking the standard equations of each. Hint: inspect the x and y values Circle: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 sum of squares Ellipse: (𝑥−ℎ)2 sum of squares Parabola: (𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ) a square and a linear Hyperbola: (𝑥−ℎ)2 difference of squares 𝑎2 𝑎2 + − (𝑦−𝑘)2 𝑏2 (𝑦−𝑘)2 𝑏2 =1 =1 Since the given equation is a difference of squares, ∴ it is a hyperbola. Next, transform the given equation to the standard equation. 𝑥 2 − 𝑦 2 = 23 Given 𝑥 2 −𝑦 2 Divide both sides by 23 23 𝑥2 23 = 23 𝑦2 Simplify, separate terms − 23 = 1 23 (𝑥−0)2 (√23) 2 − (𝑦−0)2 (√23) 2 =1 Rewrite based on the standard equation Having the original equation transformed into the standard equation of a hyperbola, we can now identify the values of the variables which will be needed for further calculations. ℎ = 0; 𝑘 = 0; 𝑎 = √23 = 4.80; 𝑏 = √23 = 4.80 a. Graph First, locate the center of the hyperbola through (ℎ, 𝑘) Center = (0,0) Second, make a rectangle using the center point, and the values of 𝑎 = 4.80 and 𝑏 = 4.80 as guide to plotting in the graph later. Move 𝑎 and 𝑏 units depending on what variable is found at its numerator. (In this problem, it will not matter since a and b are equal). First, from the center, move 𝑎 units (4.80) to the right, then move 𝑏 units (4.80) up and down. Mark these two points {(4.80,4.80),(4.80,-4.80)}. Then, from the center, move 𝑎 units (4.80) to the left, then move 𝑏 units (4.80) up and down again. Mark these two points {(-4.80,4.80),(-4.80,4.80)}.as well. Connect these four points to make a rectangle. Use broken lines or pencil since this rectangle is not really part of the graph. Third, connect the diagonals of this rectangle and extend it as a line. These two lines are the asymptotes of the hyperbola where the graph will try to approach but will not touch. Draw the asymptotes using a pencil or broken line as well. Fourth, identify where will the hyperbola will face through the transverse and conjugate axes. The transverse and conjugate axes are lines parallel to the x- and y- axes and will pass through the center point. Tip: comparing the x and y variables, the positive value is the transverse axis and that is where the hyperbola will face. The transverse axis is parallel to x-axis since x is positive, while the conjugate axis is parallel to y-axis. Passing through the center point (0,0) are the x-axis and y-axis themselves. Fifth. Locate the vertices. Vertices are two points where the transverse axis and the rectangle will meet. Sixth, construct the graph using the vertex, transverse axis, and the asymptotes. Knowing where the hyperbola faces, start at the vertex, then make a curve line approaching the asymptote but not touch it. Draw arrowheads at the end to indicate the graph of hyperbola extends. b. Lines of symmetry In the lines of symmetry, the graphs on both sides are mirror images of each other. Hyperbolas are symmetric in their transverse and conjugate axes. The lines of symmetry are the x- and y- axes. c. Domain and Range Determine the Domain and Range through the graph. The domain of the equation is the set of all possible values of x. Looking at the graph, there are no x-values within the rectangle, and that is within (-4.80,4.80). So the domain is the set of values of x such that x is greater than or equal to 4.80 and less than or equal to -4.80, or: 𝔻 = {𝑥|𝑥 ≥ 4.80 ∪ 𝑥 ≤ −4.80} The range of the equation is the set of all possible values of y. Looking at the graph, all y-values will have values. So the range is the set of values of y such that y is a real number, or ℝ = {𝑦|𝑦 ∈ ℝ} 2. Divide the polynomials 𝒙𝟐 + 𝟐𝒙𝟐 𝒚 − 𝟐𝒙𝒚 + 𝟐𝒙𝒚𝟐 − 𝟑𝒚𝟐 and 𝒙 + 𝒚 and explain each step. Check your solution. Given: 𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 divided by 𝑥 + 𝑦 Required: Quotient Solution: Dividing polynomials is like dividing numbers. Usually, the long method is employed to get the quotient of the two polynomials. 𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 Write the expression. 𝒙 + 𝑦√𝒙𝟐 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 Look at the first expression in both dividend and divisor. Divide them first in order to get the multiplier. 𝑥2 =𝒙 𝑥 Multiply this multiplier to the divisor to get the product. (𝑥 + 𝑦) ∗ 𝒙 = 𝒙𝟐 + 𝒙𝒚 𝒙 𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝒙𝟐 + 𝒙𝒚 −3𝑥𝑦 Going back to the long method, align the product under the same term of the dividend, then do subtraction. Write the multiplier at the quotient. 𝑥 𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝑥2 + 𝑥𝑦 2 2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝒙 𝒙 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝑥2 + 𝑥𝑦 𝟐 𝟐𝒙 𝒚 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 Bring down or copy other terms of the dividend. Look at the first expression in both the difference and divisor. Divide them first in order to get the multiplier. 2𝑥 2 𝑦 = 𝟐𝒙𝒚 𝑥 Multiply this multiplier to the divisor to get the product. (𝑥 + 𝑦) ∗ 𝟐𝒙𝒚 = 𝟐𝒙𝟐 𝒚 + 𝟐𝒙𝒚𝟐 𝒙 + 𝟐𝒙𝒚 𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝒙𝟐 + 𝒙𝒚 2 2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝟐𝒙𝟐 𝒚 + 𝟐𝒙𝒚𝟐 𝟎 𝟎 𝑥 + 2𝑥𝑦 𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝑥2 + 𝑥𝑦 2 2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 2𝑥 2 𝑦 + 2𝑥𝑦 2 −3𝑥𝑦 − 3𝑦 2 Going back to the long method, align the product under the same term of the dividend, then do subtraction. Add the multiplier to the quotient. Bring down or copy other terms of the dividend. 𝑥 + 2𝑥𝑦 𝒙 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝑥2 + 𝑥𝑦 2 2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 2𝑥 2 𝑦 + 2𝑥𝑦 2 −𝟑𝒙𝒚 − 3𝑦 2 Look at the first expression in both the difference and divisor. Divide them first in order to get the multiplier. −3𝑥𝑦 = −3𝑦 𝑥 Multiply this multiplier to the divisor to get the product. (𝑥 + 𝑦) ∗ −𝟑𝒚 = −𝟑𝒙𝒚 − 𝟑𝒚𝟐 𝑥 + 2𝑥𝑦 − 𝟑𝒚 𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 𝑥2 + 𝑥𝑦 2 2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 2𝑥 2 𝑦 + 2𝑥𝑦 2 −3𝑥𝑦 − 3𝑦 2 −𝟑𝒙𝒚 − 𝟑𝒚𝟐 𝟎 𝟎 2 2 2 2 𝑥 + 2𝑥 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 − 3𝑦 𝑥+𝑦 = 𝒙 + 𝟐𝒙𝒚 − 𝟑𝒚 Going back to the long method, align the product under the same term of the dividend, then do subtraction. Add the multiplier to the quotient. If no other term remains and the difference is zero, the quotient is the final answer. 3. Suppose a number from 1 to 100 is selected at random. What is the probability that a multiple of 4 or 5 is chosen? Explain each step. Given: Population: 1 to 100 Sample: multiple of 4 OR 5 Required: Probability of picking multiple of 4 OR 5 Solution: 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 The formula of the probability is: 𝑃 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 Based on the problem, the total number of all values in the population is 100 (total numbers from 1-100). For the number of possible values, consider this Venn Diagram: Multiples of 4 and 5 Multiples of 4 only Multiples of 5 only Numbers 1-100 Picking multiples of 4 OR 5 means getting the multiples of 4 and multiples of 5, which could be obtained by dividing those from 100. To get total multiples of 4 from 1 to 100: 100 = 25 4 To get total multiples of 5 from 1 to 100: 100 = 20 5 Total multiples of 4 + Total multiples of 5: 25 + 20 = 𝟒𝟓 However, there are also multiples of both 4 and 5 which will be recounted if not being careful. To account for the double count, we will find the multiple of the least common multiple of 4 and 5. Least common multiple: 4 ∗ 5 = 20 To get total multiples of 20 from 1 to 100: 100 =𝟓 20 There are 5 numbers which have overlapped since they are multiples of 4 and 5. We need to account for this by subtracting it to the total multiples of 4 and multiples of 5. So, to get the multiples of 4 OR 5: 25 + 20 − 5 = 𝟒𝟎 Substituting to the probability formula: 𝑃= 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 40 = = 0.40 ≈ 40% 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 100 There is 40% chance of picking a multiple of 4 or 5 from 1 to 100. 4. Prove that 𝟏𝟐 + 𝟐𝟐 + 𝟑𝟐 + ⋯ + 𝒏𝟐 = Given: 12 + 22 + 32 + ⋯ + 𝑛2 = Required: Proof 𝒏(𝒏+𝟏)(𝟐𝒏+𝟏) 𝟔 . Explain each step. 𝑛(𝑛+1)(2𝑛+1) 6 Solution: Eq 1 Eq 2 Eq 3 Eq 4 Eq 5 Eq 6 Eq 7 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) Cubic difference identity 𝑛3 − (𝑛 − 1)3 = [𝑛 − (𝑛 − 1)][𝑛2 + 𝑛 ∗ (𝑛 − 1) + (𝑛 − 1)2 ] Substituting 𝑎 → 𝑛, 𝑏 → 𝑛 − 1 where 𝑛 is any natural number to Eqn 1 𝑛3 − (𝑛 − 1)3 = [𝑛 − 𝑛 + 1][𝑛2 + 𝑛2 − 𝑛 + 𝑛2 − 2𝑛 + 1] Applying distributive properties 𝑛3 − (𝑛 − 1)3 = [1][3𝑛2 − 3𝑛 + 1] Combining like terms 𝑛3 − (𝑛 − 1)3 = 3𝑛2 − 3𝑛 + 1 Simplifying right side of the equation (𝑛 − 1)3 − [(𝑛 − 1) − 1]3 = 3(𝑛 − 1)2 − 3(𝑛 − 1) + 1 Substituting 𝑛 → 𝑛 − 1 to Eqn 2 (𝑛 − 1)3 − [𝑛 − 2]3 = 3(𝑛 − 1)2 − 3(𝑛 − 1) + 1 Combining like terms (𝑛 − 2)3 − [(𝑛 − 2) − 1]3 = 3(𝑛 − 2)2 − 3(𝑛 − 2) + 1 Substituting 𝑛 → 𝑛 − 2 to Eqn 2 (𝑛 − 2)3 − [𝑛 − 3]3 = 3(𝑛 − 2)2 − 3(𝑛 − 2) + 1 Combining like terms 23 − (2 − 1)3 = 3(2)2 − 3(2) + 1 Substituting 𝑛 → 2 to Eq 2 23 − 13 = 3(2)2 − 3(2) + 1 Combining like terms 13 − (1 − 1)3 = 3(1)2 − 3(1) + 1 Substituting 𝑛 → 1 to Eq 2 13 − 03 = 3(1)2 − 3(1) + 1 Combining like terms 𝑛3 − (𝑛 − 1)3 = 3𝑛2 − 3𝑛 + 1 + (𝑛 − 1)3 − [𝑛 − 2]3 = 3(𝑛 − 1)2 − 3(𝑛 − 1) + 1 + (𝑛 − 2)3 − [𝑛 − 3]3 = 3(𝑛 − 2)2 − 3(𝑛 − 2) + 1 + … Adding Eq 2 to 6 for all possible values of 𝑛 + 3 3 2 − 1 = 3(2)2 − 3(2) + 1 + 13 − 03 = 3(1)2 − 3(1) + 1 Eq 7.1 Eq 7.2 [𝑛3 − (𝑛 − 1)3 ]+[ (𝑛 − 1)3 − [𝑛 − 2]3 ]+[ (𝑛 − 2)3 − [𝑛 − 3]3 ]+…+[ 23 − 13 ]+[ 13 − 03 ] Getting the sum of the left side of the equation, combining like terms 𝑛3 − 0 ⇒ 𝑛3 Simplifying left side of the equation 3𝑛2 + 3(𝑛 − 1)2 + 3(𝑛 − 2)2 + ⋯ + 3(2)2 + 3(1)2 Getting the sum of the first terms of the right side of the equation 3[𝑛2 + (𝑛 − 1)2 + (𝑛 − 2)2 + ⋯ + (2)2 + (1)2 ] Factor out common multiple 𝑛 3 ∑ 𝑛2 𝑛=1 Rewriting to summation notation −3𝑛 − 3(𝑛 − 1) − 3(𝑛 − 2) − ⋯ − 3(2) − 3(1) Getting the sum of the second terms of the right side of the equation −3[𝑛 + (𝑛 + 1) + (𝑛 + 2) + ⋯ + 2 + 1] Factor out common multiple 𝑛 Eq 7.3 Rewriting to summation notation −3 ∑ 𝑛 𝑛=1 1 + 1 + 1 + ⋯+ 1 + 1 ⟹ 𝑛 Getting the sum of the third terms of the right side of the equation Rewriting 𝑛 𝑛 3 2 𝑛 = 3∑𝑛 −3∑𝑛+𝑛 𝑛=1 𝑛 𝑛=1 𝑛3 = 3 ∑ 𝑛2 − 3 𝑛=1 𝑛 3 ∑ 𝑛2 = 𝑛3 + 𝑛=1 𝑛 3 ∑ 𝑛2 = 𝑛=1 𝑛 3 ∑ 𝑛2 = 𝑛=1 𝑛(𝑛 + 1) +𝑛 2 3𝑛(𝑛 + 1) −𝑛 2 Simplifying Eq 7 with Eq 7.1, 7.2, and 7.3 Summation of natural numbers identity Transpose 2𝑛3 + 3𝑛(𝑛 + 1) − 2𝑛 2 Simplifying right side of the equation to a single fraction 2𝑛3 + 3𝑛2 + 3𝑛 − 2𝑛 2 Distributive property 𝑛 2𝑛3 + 3𝑛2 + 𝑛 3∑𝑛 = 2 2 𝑛=1 𝑛 3 ∑ 𝑛2 = 𝑛=1 𝑛 3 ∑ 𝑛2 = 𝑛=1 𝑛 ∑ 𝑛2 = 𝑛=1 Combine like terms 𝑛(2𝑛2 + 3𝑛 + 1) 2 Factor out common multiple 𝑛(𝑛 + 1)(2𝑛 + 1) 2 Factor binomials 𝑛(𝑛 + 1)(2𝑛 + 1) 6 Dividing both sides by 3 ∎ 5. Are the matrices A and B inverses? Explain each step. 𝟏 𝑨 = [−𝟐 −𝟏 𝟎 𝟏 𝟏 𝟐 𝟏] 𝟐 −𝟏 𝑩 = [−𝟑 𝟏 Given: 1 𝐴 = [−2 −1 Required: Check if inverses −𝟐 𝟐 −𝟒 𝟓 ] 𝟏 −𝟏 0 2 −1 −2 1 1], 𝐵 = [−3 −4 1 2 1 1 2 5] −1 Solution: To check that matrices A and B are inverses, their product must be equal to the identity matrix. 1 0 𝐴 × 𝐵 = 𝐼 = [0 1 0 0 0 0] 1 Write side by side the two matrices. A 3x3 matrix multiplied to another 3x3 matrix will yield a product which is a 3x3 matrix also. 𝑎 1 0 2 −1 −2 2 𝐴 × 𝐵 = [−2 1 1] [−3 −4 5 ] = [𝑑 𝑔 −1 1 2 1 1 −1 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 Next, get the dot product of the first row of A and the first column in B by multiplying each corresponding (i.e., first element multiplied to the first element) and so on, then get the sum. 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (1)(−1) + (0)(−3) + (2)(1) = −1 + 0 − 2 = 1 Fill the answer in the element of the product matrix where the factors meet. 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (First row and first column 𝑎) 1 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 Then, get the dot product of the second row of A and the first column in B, then fill the answer where these two meet. 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (−2)(−1) + (1)(−3) + (1)(1) =2−3+1=0 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (Second row and first column 𝑑) 1 [0 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 Repeat this for the third row of A and the first column of B. 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (−1)(−1) + (1)(−3) + (2)(1) =1−3+2=0 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (Third row and first column 𝑔) 1 𝑏 [0 𝑒 0 ℎ Moving on the second column of B and back at the first row of A, 𝑐 𝑓] 𝑖 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (1)(−2) + (0)(−4) + (2)(1) = −2 + 0 + 2 = 0 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (First row and second column 𝑏) 1 0 [0 𝑒 0 ℎ 𝑐 𝑓] 𝑖 The second row of A again, 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (−2)(−2) + (1)(−4) + (1)(1) =4−4+1=1 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (Second row and second column 𝑒) 1 0 [0 1 0 ℎ 𝑐 𝑓] 𝑖 And the third row of A. 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (−1)(−2) + (1)(−4) + (2)(1) =2−4+2=0 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (Third row and second column 𝑏) 1 0 [0 1 0 0 𝑐 𝑓] 𝑖 Moving to the third column of B and back at the first row of A, 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (1)(2) + (0)(5) + (2)(−1) =2+0−2=0 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (First row and third column 𝑐) 1 0 [0 1 0 0 0 𝑓] 𝑖 The second row of A again, 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (−2)(2) + (1)(5) + (1)(1) = −4 + 5 + 1 = 0 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (Second row and third column 𝑓) 1 0 [0 1 0 0 0 0] 𝑖 And finally, the third row of A. 1 0 2 −1 −2 2 [−2 1 1] [−3 −4 5 ] −1 1 2 1 1 −1 = (−1)(2) + (1)(5) + (2)(−1) = −2 + 5 − 2 = 1 𝑎 [𝑑 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓] 𝑖 (Third row and second column 𝑖) 1 0 [0 1 0 0 0 0] 1 Since the product of A and B is the identity matrix, then A and B are inverses of each other. 1 0 2 −1 −2 2 1 𝐴 × 𝐵 = [−2 1 1] [−3 −4 5 ] = [0 −1 1 2 1 1 −1 0 0 0 1 0] = 𝐼 0 1