University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

426

Pages:

8
Author:

banenetta0l

Question 1 Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations. 𝑦 = 𝑥 − 2 − 1 y= x−2 −1 Answer: To determine: Graph of 𝑦 = 𝑥 − 2 − 1 y= x−2 −1 by applying appropriate transformation. Step 1: Draw the graph of 𝑦 = 𝑥 y= x . Step 2: Shift the graph of 𝑦 = 𝑥 y= x two units to the right to get 𝑦 = 𝑥 − 2 y= x−2 . Step 3: Shift the graph of 𝑦 = 𝑥 − 2 y= x−2 one unit downward along the y-axis to get 𝑦 = 𝑥 − 2 − 1 y= x−2 −1. The graph of 𝑦 = 𝑥 − 2 − 1 y= x−2 −1 is labeled as 𝑌 3 Y3. Question 2 Determine the transformations of the function 𝑓 ( 𝑥 ) f(x) to get the function 𝑔 ( 𝑥 ) = 1 2 𝑓 ( 5 ( 𝑥 − 3 ) ) − 12 g(x)= 2 1 f(5(x−3))−12. (The transformations happen in c, b, a order.) Answer: Given: A function 𝑓 ( 𝑥 ) f(x) is transformed to get 𝑔 ( 𝑥 ) g(x): 𝑔 ( 𝑥 ) = 1 2 𝑓 ( 5 ( 𝑥 − 3 ) ) − 12 g(x)= 2 1 f(5(x−3))−12 Sequence of transformations to transform 𝑓 ( 𝑥 ) f(x) into 𝑔 ( 𝑥 ) g(x): Vertical compression by a scale factor of 1 2 2 1 . Horizontal compression by a scale factor of 1 5 5 1 . Horizontal translation of 3 units to the right. Vertical translation of 12 units downward. Question 3 Area and Perimeter Formulas with Examples: Area: Total space enclosed by the boundary of a plane. Perimeter: Length of the border around any enclosed plane. Answer: Triangle: Equilateral Triangle: Area = 3 4 𝑎 2 , Perimeter = 3 𝑎 Area= 4 3 a 2 ,Perimeter=3a Isosceles Triangle: Area = 𝑏 4 𝑎 2 − 𝑏 2 4 , Perimeter = 2 𝑎 + 𝑏 Area=b 4 4a 2 −b 2 ,Perimeter=2a+b Scalene Triangle: Area = 𝑠 ( 𝑠 − 𝑎 ) ( 𝑠 − 𝑏 ) ( 𝑠 − 𝑐 ) (Heron’s Formula) Area= s(s−a)(s−b)(s−c) (Heron’s Formula) where 𝑠 = 𝑎 + 𝑏 + 𝑐 2 s= 2 a+b+c . Right-Angled Triangle: Area = 1 2 Base × Height , Perimeter = 𝑎 + 𝑏 + 𝑐 . Area= 2 1 Base×Height,Perimeter=a+b+c. Quadrilateral: Square: Area = 𝑎 2 , Perimeter = 4 𝑎 Area=a 2 ,Perimeter=4a Rectangle: Area = 𝑙 × 𝑏 , Perimeter = 2 ( 𝑙 + 𝑏 ) Area=l×b,Perimeter=2(l+b) Parallelogram: Area = 𝑏 × ℎ , Perimeter = 2 ( 𝑎 + 𝑏 ) Area=b×h,Perimeter=2(a+b) Trapezium: Area = 1 2 × ( 𝑎 + 𝑏 ) × ℎ , Perimeter = Sum of all sides Area= 2 1 ×(a+b)×h,Perimeter=Sum of all sides Rhombus: Area = 1 2 × ( Diagonal 1 × Diagonal 2 ) Area= 2 1 ×(Diagonal 1 ×Diagonal 2 ) Circle: Area: 𝜋 𝑟 2 πr 2 Circumference: 2 𝜋 𝑟 2πr Additional Examples and Applications: Percentage Changes: If a quadrilateral’s length and breadth are expanded by 𝑎 % a% and 𝑏 % b%, then the area increase is approximately: Increase = ( 𝑎 + 𝑏 + 𝑎 𝑏 100 ) % Increase=(a+b+ 100 ab )% Product Stability Ratio: To maintain 𝑃 = 𝐴 × 𝐵 P=A×B constant, increasing 𝐴 A requires decreasing 𝐵 B: Δ 𝐵 = Δ 𝐴 1 + Δ 𝐴 ΔB= 1+ΔA ΔA Compound Interest Applications: Question 4 The measure of ∠ 𝐷 𝐵 𝐸 ∠DBE is ( 0.1 𝑥 − 22 ) ∘ (0.1x−22) ∘ and ∠ 𝐶 𝐵 𝐸 = ( 0.3 𝑥 − 54 ) ∘ ∠CBE=(0.3x−54) ∘ . Find the value of 𝑥 x. Answer: By the postulate of angle addition: 𝑚 ∠ 𝐷 𝐵 𝐸 + 𝑚 ∠ 𝐶 𝐵 𝐸 = 𝑚 ∠ 𝐷 𝐵 𝐶 m∠DBE+m∠CBE=m∠DBC Since ∠ 𝐷 𝐵 𝐶 ∠DBC is a right angle ( 9 0 ∘ 90 ∘ ): ( 0.1 𝑥 − 22 ) + ( 0.3 𝑥 − 54 ) = 90 (0.1x−22)+(0.3x−54)=90 Solve for 𝑥 x: 0.4 𝑥 − 76 = 90 0.4x−76=90 0.4 𝑥 = 166 , 𝑥 = 166 0.4 = 415 0.4x=166,x= 0.4 166 =415 Final Answer: 𝑥 = 415 x=415