Questions and Answers #3 Polynomial Graphs

Questions and Answers Sheet 3 Polynomial Graphs Question #1 If the result of a polynomial division is 2𝑥 3 − 4𝑥 2 − 5𝑥 + 23𝑥 + 2 , make at least three definitive statements Answer: Given that, the result of the polynomial division is 2𝑥 3 − 4𝑥 2 − 5𝑥 + 23𝑥 + 2 From the above expression, it is observed that the divisor is (𝑥 + 2), the remainder is 23, and the quotient is 2𝑥 3 − 4𝑥 2 − 5𝑥 Then, the dividend is [(2𝑥 3 − 4𝑥 2 − 5𝑥) × (𝑥 + 2)] + 23. (i) The dividend is has degree higher than the divisor. (ii) The division can be performed by both long division method and remainder method. As the degree of the divisor is 1 so remainder method is easier to use as compared to the long division method. (iii) The dividend is not completely divisible by the divisor as there is remainder present equal to 23. Above mentioned are the three definitive statements regarding the dividend and the divisor. Question #2 𝑓(𝑥) = 𝑥 5 + 2𝑥 4 + 3𝑥 3 − 2𝑥 6 − 9𝑥 2 − 6𝑥 + 4 Is it true or false the degree of this polynomial function is 5? If not, why? What is the degree? Answer: False, because the degree of this polynomial function is the highest degree among the terms, which is 6 (−2𝑥 6 ), in this instnance. Question #3 𝑓(𝑥) = 100(1.25) 𝑥 Is it true or false, the growth factor is 25%? If not, what is and what is 25%? Answer: Given: 𝑓(𝑥) = 100(1.25) 𝑥 The formula: 𝑓(𝑥) = 𝑎(1 + 𝑟) 𝑥 Where the growth factor is 1 + 𝑟, and the growth rate is r. So, in the given, 25% is the growth rate (0.25), and the growth factor is 1 + 𝑟 = 1 + 0.25 = 1.25. Question #4 Simplify the difference quotients between 𝑓(𝑥 + ℎ) − f(x)=sqrt(x^2-7) Answer: f ( x ) = x2 − 7 f ( x + h) = ( x + h) 2 −7 𝑓(𝑥) ℎ and 𝑓(𝑥) − 𝑓(𝑎) 𝑥−𝑎 , if f ( x + h) − f ( x) h ( x + h) = 2 = ( x + h) 2 − 7 − x2 − 7 h − 7 − x2 − 7 h ( x + h) − 7 − 2 ( x + h) − 7 − 2  ( x + h ) 2 − 7  −  x 2 − 7     =  2 h  ( x + h ) − 7 + x2 − 7    ( x + h ) − x2 2 ( x + h ) − 7 + x 2 − 7  2 = = h   h + 2 xh 2 h  c= − 7 + x2 − 7   h + 2x ( x + h) ( x + h) 2 2 − 7 + x2 − 7 f ( x ) = x 2 − 7, f ( a ) = a 2 − 7 f ( x) − f (a) x−a = = = x2 − 7 − a2 − 7 x−a x2 − 7 − a2 − 7 x2 − 7 − a2 − 7  x−a x2 − 7 − a2 − 7 (x 2 ) ( − 7 − a2 − 7 ) ( x − a )  x2 − 7 − a2 − 7   2 2 x −a = ( x − a )  x 2 − 7 − a 2 − 7  ( x − a )( x + a ) = ( x − a )  x 2 − 7 − a 2 − 7  x+a = 2 x − 7 − a2 − 7 f ( x + h) − f ( x) h + 2x = 2 h ( x + h ) − 7 + x2 − 7 f ( x) − f (a) x−a = x+a x2 − 7 − a2 − 7 Question #5 x2 − 7 x2 − 7 1 2   , −1 and 9 + 3i are zeros. 7  Find a polynomial function with real coefficients. Answer: ( ( )) ( ( 2   x −  ( x − ( −1) ) x − 9 − 3i x − 9 + 3i 7  2 2 2 2  =  x 2 + x − x −   ( x − 9 ) − 3i   7 7   5 2  =  x 2 + x −  ( x 2 − 18x + 81 + 3) 7 7  )) =  x − 72  ( x + 1) ( x − (9 − 3i )) ( x − (9 + 3i )) ( ) ( )( ) 1 2 x + 5 x − 2 x 2 − 18 x + 84 7 1 4 = x − 13 x 3 − 8 x 2 + 456 x − 168 7 1 = x 4 − 13 x 3 − 8 x 2 + 456 x − 168 is the polynomial function with real coefficients for the given zeros. 7 = ( ) ( ) Question #6 a) By inspection, find a particular solution of 𝑦 ′′ + 2𝑦 = 10 b) By inspection, find a particular solution of 𝑦 ′′ + 2𝑦 = −4𝑥 c) Find a particular solution of 𝑦 ′′ + 2𝑦 = −4𝑥 + 10 d) Find a particular solution of 𝑦 ′′ + 2𝑦 = 8𝑥 + 5 Answer: Step 1 a) By inspection we see that the particular solution of the DE 𝑦 ′′ + 2𝑦 = 10 is 𝑦𝑝1 = 5 ′′ ∵ 𝑦𝑝1 + 2𝑦𝑝1 = 0 + 2(5) = 10 b) By inspection we see that the particular solution of the DE 𝑦 ′′ + 2𝑦 = −4𝑥𝑖𝑠𝑦𝑝2 = −2𝑥 ′′ ∵ 𝑦𝑝2 + 2𝑦𝑝2 = 0 + 2(−2𝑥) = −4𝑥 c) By Superposition principle we see that the particular solution of the DE 𝑦 ′′ + 2𝑦 = −4𝑥 + 10 is the superposition of the solutions in parts (a) and (b) ∴ 𝑦𝑝 = 𝑦𝑝1 + 𝑦𝑝2 ⇒ 𝑦𝑝 = 5 − 2𝑥 ∵ 𝑦𝑝′′ + 2𝑦𝑝 = 0 + 2(5 − 2𝑥) = −4𝑥 + 10 d) Be Superposition principle we see that the particular solution of the DE 𝑦 ′′ + 2𝑦 = 8𝑥 + 5 = 1 −2(−4𝑥) + (10) is the superposition of the solutions in parts (a) and (b) 2 1[2} 1 ∴ 𝑦𝑝 = − 2𝑦𝑝2 ⇒ 𝑦𝑝 = (5) − 2(−2𝑥) 𝑦 𝑝1 2 5 ⇒ 𝑦𝑝 = + 4𝑥 2 5 ∵ 𝑦𝑝′′ + 2𝑦𝑝 = 0 + 2 ( + 4𝑥) = 8𝑥 + 5 Answer: 2 a) 𝑦𝑝1 = 5 b) 𝑦𝑝2 = −2𝑥 c) 𝑦𝑝 = 5 − 2𝑥 5 d) 𝑦𝑝 = + 4𝑥 2 Question #7 Find the points on the ellipse 4𝑥 2 + 𝑦 2 = 4 that are farthest away from the poin (1,0). (𝑥, 𝑦) = (? ) (smaller y-value) (𝑥, 𝑦) = (? ) (largen y-value) Answer: Step 1 Given: ellipse 4𝑥 2 + 𝑦 2 = 4 (given) To Find:- (𝑥; 𝑦)- smaller value (𝑥; 𝑦) - langer value farthest away from point (1,0) 1. Concept of maxima minima 2. Distance behocen 2 point formula. (𝑥1 , 𝑦1 )(𝑥2 , 𝑦2 ) 𝐷 = √(𝑥2 − 𝑥1 )2 + (𝑦2 , −𝑦1 )2 Step 2 Solution 4𝑥 2 + 𝑦 2 = 4 𝑦 2 + 4 − 4𝑥 2 (1) Let (𝑥, 𝑦) be any point on the wave 4𝑥 2 + 𝑦 2 = 4 Distance of the point (𝑥, 𝑦) from (1,0) 𝐷 = √(𝑥 − 1)2 + 𝑦 2 Squwing both cides 𝐷2 = 𝑥 2 + 1 − 2𝑥 + 𝑦 2 𝐷2 = 𝑥 2 + 𝑦 2 − 2𝑥 + 1 ∴ 𝑦 2 = 4 − 4𝑥 2 given Step 3 𝐷2 = 𝑥 2 + 1 − 2𝑥 + 4 − 4𝑥 2 𝐷2 = −3𝑥 2 − 2𝑥 + 5 Suppose 𝐷2 = 𝑝 𝑝 = −3𝑥 2 − 2𝑥 + 5 𝑑𝑝 𝑑𝑥 = −6𝑥 − 2 Find csitical point by equating 𝑑𝑝 𝑑𝑥 =0 −6𝑥 = 2 1 𝑥=− 3 ∴ 𝑑2𝑝 = −6 < 0 𝑑𝑥 2 −1 ∴𝑥= is a point of maxima. 3 ∴ 𝑝 has a maximum value at 𝑥 = −1 3 Function i c equation of ellipse given. 4𝑥 2 + 𝑦 2 = 4 −1 2 𝑦 2 = 4 − 4𝑥 2 ⇒ 4 − 4 ( ) 3 1 𝑦2 ⇒ 4 − 4 ( ) 9 𝑦2 ⇒ 32 9 𝑦= ±4√2 3 ∴ poins are : − (−43 , points are : − ( −1 4√2 3 , 3 −4√2 3 ) - smaiier value ) - larger value Question #8 The curve from the graph of a degree 4 polynomial. It goes through the point (5, −160) . Find the polynomial. 𝑓(𝑥) = Answer: Given (5, −160) from the graph,the polynomial has −3,0,0,4, 𝑓(𝑥) = 𝑎(𝑥 + 3)(𝑥 − 0)(𝑥 − 0)(𝑥 − 4) It paves through (5,160), 𝑓(5) = −160 ⇒ −160 = 𝑎(5 + 3)(5)(5)(5 − 4)[∵ 𝑥 = 5, 𝑓(5) = −160]. ⇒ −20 = 𝑎(8)(25)(1) ⇒ −20 = 25𝑎 ⇒𝑎= −4 5 ∴ the polynomial is, −4 𝑓(𝑥) = − ((𝑥 + 3)(𝑥)(𝑥)(𝑥 − 4)) 5 −4 = ((𝑥 3 + 3𝑥 2 )(𝑥 − 4)) 5 −4 4 (𝑥 − 4𝑥 3 + 3𝑥 3 − 12𝑥 2 ) = 5 −4 4 𝑓(𝑥) = 𝑥 − 𝑥 3 − 12𝑥 2 5 −4𝑥 4 4𝑥 3 48𝑥 2 𝑓(𝑥) = + + 5 5 5 Question #9 In Figure 3, is graph (A) or graph (B) the graph of an antiderivative of 𝑦 = 𝑓(𝑥) ? Answer: Step1 Option Ais the correct graph. Use the fundamental theorem for calculus. If 𝑓(𝑥) = ∫ 𝑔(𝑥) Then, 𝑑 𝑑𝑥 𝑓(𝑥) = 𝑔(𝑥) Therefore, f(x) is the graph for the derivative for either A or B. Step2 Check if the f(x) is possible derivative for A using the graphical approach. The function of the graph Ais of polynomial of degree greeter than 2 as the graph of the derivative ie. f(x) is curve and not straight line. The function for graph 𝐴 = 6𝑥 sin 𝑥 + 6 cos 𝑥 − 8 Derivative for graph 𝐴 = 6 sin 𝑥 + 6𝑥 cos 𝑥 − 6 sin 𝑥 = 6𝑥 cos 𝑥 Step3 Plot the graphs for A and the derivative of A. Step4 The red graph matches with the graph for 𝑦 = 𝑓(𝑥) . Thus, 𝑦 = 𝑓(𝑥) is the derivative for graph A. Hence, graph A is the antiderivative for 𝑦 = 𝑓(𝑥) and the option Ais the correct choice. Question #10 Evaluate the integral to find the area between the curves: 𝑓(𝑥) = 𝑥 𝑔(𝑥) = 𝑥 2 − 2𝑥 Answer: If p(x) and q(x) are continuous functions over an interval [𝑚, 𝑛] and R is the region between the graphs of p(x) and q(x) and bounded dy the linex x-m and x-n respectively.,then the area of R is determined as 𝑛 Area = ∫𝑚|𝑝(𝑥) − 𝑞(𝑥)|𝑑𝑥 . Both the functions are polynomial functions and hence continuous. Graph of the given functions: Intersection points: (0,0), (3,3) 3 Area of between the curves = ∫0 |𝑓(𝑥) − 𝑔(𝑥)|𝑑𝑥 3 = ∫0 |𝑥 − (𝑥 2 − 2𝑥)|𝑑𝑥 3 = ∫ 𝑥 − 𝑥 2 𝑑𝑥 0 𝑥2 =| 2 2 − 𝑥3 3 | 3 0 3 3 3 0 2 03  =  − − ( − ) 2 3  2 3 9 = | − 9| 2 9 = = 4.5 sq.units 2