Questions and Answers #4 Polynomial Graphs

Questions and Answers Sheet 4 Polynomial Graphs Question #1 Write an equation for a rational function with: Vertical asymptotes at 𝑥 = −2 and 𝑥 = −4 x intercepts at 𝑥 = 1 and 𝑥 = 3 Horizontal asymptote at y = 8 Answer: Step 1 ⇒ A function of the form 𝑓(𝑥) 𝑛(𝑥) where f(x) and n(x) are polynomials is called a rational function. ⇒ The graphs of rational functions can be recognized by the fact that they often break into two or more parts. These parts go out of the coordinate system along an imaginary straight line called an asymptote. Step 2 As we have vertical asymptote at 𝑥 = −2 and 𝑥 = 4 . We have in denominator, 𝑛(𝑥) = (𝑥 + 2)(𝑥 + 4) 𝑛(𝑥) =2 + 6𝑥 + 8 f(x) which is in the numerator must be of the same degree as the denominator. As we have x intercepts at 𝑥 = 1 and 𝑥 = 3 So, 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 3) Rational function, 𝑅(𝑥) = 𝑘𝑓(𝑥) 𝑛(𝑥) = 𝑘(𝑥−1)(𝑥−3) 𝑥 2 +(6𝑥+8) As we have the horizontal asymptote at 𝑦 = 8 So at 𝑥 = 0, 𝑅(0) = 8 8= 𝑘(0−1)(0−3) (02 +6𝑥0+8) ⇒8= 3𝑘 8 ⇒𝑘= 64 3 Hence, rational function, 64 )( x − 1)( x − 3) 64( x 2 − 4 x + 3) 3 R( x) =  R( x) = ( x 2 + 6 x + 8) 3( x 2 + 6 x + 8) ( Question #2 Graph 𝑦 = 𝑒 𝑥 and 𝑦 = 1 + 𝑥 + 𝑥 2 /2 + 𝑥 3 /6 + 𝑥 4 /24 in the same viewing rectangle. Answer: Two functions are given that, first is 𝑦 = 𝑒 𝑥 and anotheris 𝑦 = 1 + 𝑥 + 𝑥2 2 + 𝑥3 6 + 𝑥4 24 . The graph of exponential functional 𝑦 = 𝑒 𝑥 is: The graph of polynomial functional 𝑦 = 1 + 𝑥 + 𝑥2 2 + 𝑥3 6 + 𝑥4 24 .. The graphs of both functions in the same viewing rectangle are: Question #3 Use the intermediate value theorem to determine whether the function 𝑓(𝑥) = 𝑥 3 + 2𝑥 − 4 has a root or not. between 𝑥 = 1 and 𝑥 = 2 . Ifyes, then find the root to five decimal places. Answer: Step 1 Given, 𝑓(𝑥) = 𝑥 3 + 2𝑥 − 4 over interval, [1,2] Definition of Intermediate value theorem: If f(x) be the continuous function on the closed interval [𝑎, 𝑏] then, f(a) and f(b) has opposite sides in the interval, then there must be some value 𝑥 = 𝑐 lies on the interval [𝑎, 𝑏] for which 𝑓(𝑐) = 0 Step 2 Plug the endpoints of intervals in the given function: Plug 𝑥 = 1 we get, 𝑓(1) = 13 + 2(1) − 4 = 1+2−4 = 3−4 = −1 Plug 𝑥 = 2 we get, 𝑓(2) = 23 + 2(2) − 4 = 8+4−4 =8 From the above definition of intermediate value theorem we get, 𝑓(1) < 0, 𝑓(2) > 𝑂 then f(x) is continuous on the closed interval [1,2], there must be some value 𝑥 = 𝑐 lies on the interval [1,2] for which 𝑓(𝑐) = 0 Step 3 Put 𝑥 = 𝑐 on the given function and equate to zero we get, 𝑓(𝑐) = 𝑐 3 + 2𝑐 − 4 = 0 The below graph is the graph of the function, From the graph, we get at 𝑓(𝑥) = 0 we get a value of 𝑥 = 1.17951 Therefore roots of the given function are, 𝑥 = 1.17951 The given equation is the equation of polynomial with degree 3 so it has 3 roots. Now we get one of the roots of the polynomial from the graph we clearly see that the other two graphs are imaginary. Therefore the roots are 𝑥 = 1.17951 Question #4 For 𝑓(𝑥) = 3𝑥 4 − 12𝑥 3 + 5 find the following. (A) 𝑓 ′ (𝑥) (B) The slope of the graph of f at 𝑥 = 2 (C) The equation of the tangent line at 𝑥 = 2 (D) The value(s) of x where the tangent line is horizontal (A) 𝑓′(𝑥) = ____ (B) At x=2 , the slope of the graph of f is ___ (C) At x = 2 , the equation of the tangent line is y = ___ (D) The tangent line is horizontal at 𝑥 = _____ (Use a comma to separate answers as needed.) Answer: Step 1 Given: 𝑓(𝑥) = 3𝑥 4 − 12𝑥 3 + 5 Step 2 A) To find 𝑓’(𝑥) Differentiate the function with respect to x 𝑑 𝑑 (3𝑥 4 − 12𝑥 3 + 5) 𝑓(𝑥) = 𝑑𝑥 𝑑𝑥 𝑓 ′ (𝑥) = 12𝑧 3 − 36𝑥 2 Question #5 How do you graph the function 𝑓(𝑥) = 𝑥? Answer: See below Explanation: f ( x) = x : x  Let's think for a moment about what this means. '' f is function of x that is equal to the value x for all real numbers x '' The only way this is possible is if f(x) is a straight line through the origin with a slope of 1 . In slope/intercept form: 𝑦 = 1𝑥 + 0 We can visualise f(x) from the graph below. graph 𝑥[−10,10, −5,5] Question #6 How to find the intervals of increasing and decreasing using the first derivative given 𝑦 = −2𝑥 2 + 4𝑥 + 3 ? Answer: Use the power rule: 𝑑 𝑑𝑥 = −2(2)𝑥 2−1 + 4(1)𝑥 1−1 + 0 And remember that 𝑥 0 = 1 and a derivative of a constant is zero 𝑓 ′ (𝑥) = −4𝑥 + 4 Now, factor −4(𝑥 − 1) = 0 𝑥−1=0 𝑥=1 Now you pick numbers in between the interval and test them in the derivative. If the number is positive, the function is increasing and if it's negative the function is decreasing. For example, pick 0 a number from the left 𝑓 ′ (0) = 4 This means that from (−1, ∞) the function is decreasing And from (∞, 1) the function is increasing. Question #7 Consider the polynomial function 𝑔(𝑥) = −𝑥 5 − 9𝑥 3 − 8𝑥. Factor 𝑔(𝑥) completely and determine all of its complex zeros. Which of the zeros of g are not x-intercepts of the graph of 𝑦 = 𝑔(𝑥)? Why do these zeros not correspond to a point on the x-axis? In general, what can be said about the similarities and differences between the zeros and the x-intercepts of the graphs of polynomial functions? Answer: Step 1 Given 𝑔(𝑥) = −𝑥 5 − 9𝑥 3 − 8𝑥 To find the zeros we put 𝑔(𝑥) = 0 −𝑥 5 − 9𝑥 3 − 8𝑥 = 0 −𝑥(𝑥 4 + 9𝑥 2 + 8) = 0 −𝑥(𝑥 4 + 8𝑥 2 + 𝑥 2 + 8) = 0 −𝑥(𝑥 2 (𝑥 2 + 8) + 1(𝑥 2 + 8)) = 0 −𝑥(𝑥 2 + 1)(𝑥 2 + 8) = 0 The zeros are 𝑥 = 0, 𝑥 2 + 1 = 0 and 𝑥 2 + 8 = 0 𝑥 2 + 1 = 0 ⇒ 𝑥 2 = −1 ⇒ 𝑥 = ±𝑖 and 𝑥 2 + 8 = 0 ⇒ 𝑥2 = −8 ⇒ 𝑥 + ±2√2𝑖 Hence the zeros are 𝑥 = 0, 𝑥 = ±𝑖, and 𝑥 = ±2√2𝑖 Step 2 𝑥 = ±𝑖 and 𝑥 = ±2√2𝑖 are the two roots which are not the x-intercepts of the graph 𝑦 = 𝑔(𝑥). This is because the above mentioned 4 roots are imaginary or complex roots in nature which cannot be defined in the real plane hence these cannot be included in the x-intercepts. Step 3 In general whenever any polynomial function have real roots or zeros then the zeros are also the xintercept but when the roots of a polynomial function are imaginary or complex then those complex points cannot be represented in a real number line or graph or plane due to which these cannot be the xintercept. Question #8 1. Is 𝑥 2 − 2√5𝑥 + 𝑥 a polynomial? If not, state a reason. 2. Is −2020𝑥 a polynomial? If not, state a reason. 2 3. Is 𝑥 + 3𝑥 + 1 a polynomial? If not, state a reason. 3 4. Is 1/𝑥 2 + 𝑟/𝑥 3 + 𝑟/𝑥 4 a polynomial? If not, state a reason. 5. Is 𝜋 a polynomial? If not, state a reason. 6. Is 𝑥3√2 + 3𝑥 2 a polynomial? If not, state a reason. 7. Is 𝑥 3 + 2𝑥 + 1 a polynomial? If not, state a reason. 8. Is −2𝑥 −3 + 𝑥 3 a polynomial? If not, state a reason. 9. Is 1 − 4𝑥 2 a polynomial? If not, state a reason. Answer: Step 1 Hello. Since your question has multiple sub-parts, we will solve first three sub-parts for you. If you want remaining sub-parts to be solved, then please resubmit the whole question and specify those sub-parts you want us to solve. Polynomials: Let, 𝑎0 , 𝑎1 , 𝑎2 … … … … … … . 𝑎𝑛 and "n" be a non-negatine integer A polynomial in x is an expression of the form, 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ … … … … … … … … + 𝑎1 𝑥 + 𝑎0 Step 2 A polynomial is function having nonnegative integral power, Consider the polynomials, 𝑥 3 + 2√5𝑥 + 𝑥 It is not polynomials because polynomial have only integral power, 1 𝑥 3 + 2√5𝑥 + 𝑥 = 𝑥 3 + 2(5𝑥)2 + 𝑥 1 It is not a polynomials because term 2(5𝑥)2 having fractional power. Step 3 Consider the polynomials, − 2020𝑥 It is linear polynomials. Hence −2020𝑥 is polynomials Step 4 Consider the polynomials, 2 𝑥 3 + 3𝑥 + 1 It is not a polynomial because the power of “x” (2/3) having fractional power. A polynomial contains only integer power of “x” Question #9 6.When a polynomial 𝑥4 + 𝑥3 − 10𝑥2 − 1 is divided by another polynomial, the quotient is 𝑥3 − 3𝑥2 + 2𝑥 − 8 and the remainder is 31. Find the other polynomial. Please type the answer, not in writing thank you Answer: Given Polynomial (dividend) 𝑥 4 + 𝑥 3 − 10𝑥 2 − 1 Remainder is 31 Polynomial (quotient) is 𝑥 3 − 3𝑥 2 + 2𝑥 − 8 To find the other polynomial(divisor) Using division Formula 𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 = 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 × 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 + 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 Rewriting the formula ⇒ 𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 − 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 = 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 × 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 ⇒ (𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 − 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟)/𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 = 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 Hence 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 = (𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 − 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟)/𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 Plugin the known values 𝐷𝑖𝑣𝑖𝑠𝑜𝑟 = (𝑥 4 + 𝑥 3 − 10𝑥 2 − 1 − 31)/(𝑥 3 − 3𝑥 2 + 2𝑥 − 8) ⇒ 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 = (𝑥 4 + 𝑥 3 − 10𝑥 2 − 1 − 32)/(𝑥 3 − 3𝑥 2 + 2𝑥 − 8) Factorising numerator ⇒ 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 = (𝑥3 − 3𝑥2 + 2𝑥 − 8)(𝑥 + 4)/(𝑥3 − 3𝑥2 + 2𝑥 − 8) Canceling 𝑥 3 − 3𝑥 2 + 2𝑥 − 8 from numerator and denominator ⇒ 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 = (𝑥 + 4) Hence other polynomial (𝑑𝑖𝑣𝑖𝑠𝑜𝑟) = 𝑥 + 4 Conclusion Hence the other polynomial which gives quotient 𝑥 3 − 3𝑥 2 + 2𝑥 − 8 and remainder 31 on dividing polynomial 𝑥 4 − 𝑥 3 − 10𝑥 2 − 1 is 𝑥 + 4. Hence the other polynomial (divisor) is 𝑥 + 4 Question #10 Use the table method to find the quotient. Then, substitute the value 𝑦 = 2 into each polynomial, and show that the polynomials represent integers with a true division equation. Show your work for both tasks! ( 12𝑦 4 +4𝑦 3 −31𝑦 2 −9𝑦+9 4𝑦 2 −9 ) Answer: Step 1 The given expression is (4 y 2 − 9)12 y 4 + 4 y 3 − 31 y 2 − 9 y + 9( − 3y2 + y −1 ) quotient 12 y+4 − 27 2 4 y3 − 4 y 2 − 9 y + 9 3 − 4 y 2 + 9 + − 4 y−2 + 9 − 4 y+ − 9 y 0 quotient = 3𝑥22 + 2 − 1 Now put 𝑦 = 2 quotient 3 × 22 + 21 = 3×4+1 = 13 expression ⇒ 12×24 +4𝑥23 −31𝑥22 −9𝑥2+9 4×22 −9 12 × 16 + 4 × 8 − 31 × 4 − 18 + 9 4×4−9 = (= 12 × 16 + 4 × 8 − 31 × 4 − 18 + 9 ) 4×4−9 (= 224 − 133 ) 7 (= 91 91 7 ) 13 7 21 21 verified 0