Questions and Answers #13 Quadratic Function and Equation

Questions and Answers Sheet 13 Quadratic Function and Equation Question #1 Consider the following equations: 𝑓(𝑥) = 5𝑥 2 + 4 and 𝑔(𝑥) = 2𝑥 x 0 1 2 3 4 5 6 7 8 9 10 f ( x) G ( x) a) As x increases, will the value of f(x) always be greater than the value of g(x)? b) Will an exponential function eventually always succeed a quadratic function? Answer: Step 1 We have 𝑓(𝑥) = 5𝑥 2 + 4, 𝑔(𝑥) = 2𝑥 x f ( x) g ( x) 0 4 1 1 9 2 2 2 24 2 =4 3 49 23 = 8 4 5  42 + 4 = 84 24 = 16 5 5  52 + 4 = 129 25 = 32 6 5  62 + 4 = 184 26 = 64 7 5  7 2 + 4 = 249 27 = 128 8 5  82 + 4 = 324 28 = 256 9 5  92 + 4 = 409 29 = 512 10 5 102 + 4 = 504 210 = 1024 Step 2 a) No, x=9 , and x=10 the value of f(x) is less than value of g(x), other interval [9, 10] 𝑓(𝑥) = 5𝑥 2 + 4, 𝑔(𝑥) = 2𝑥 when x=9 𝑓(9) = 5 × 92 + 4 = 409 and 𝑔(9) = 22 = 512 𝑓(9) < 𝑔(9) Step 3 Other the interval [0, 10] , the exponentional function enequally succed the quadratic function from x=9 onwards The value or 𝑔(𝑥) = 2𝑥 begin to exceed the values of 𝑓(𝑥) = 5𝑥 2 + 4 with in interval [9, 10] Question #2 A quadratic function goes through the points (−3, 3), (−1, 3) , and (−2, 4) . What is the axis of symmetry? Its vertex? Its equation? Answer: We know that general form of quadratic function 1) 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 vertex of the quadratic graph ℎ= −𝑏 2𝑎 , 𝑘 = 𝑓(ℎ) (h, k) vertex ∵ f(x) goes through (−3, 3), (−1, 3) and (−2, 4) then 2) 3=9a-3b+c 3) 3=a-b+c 4) 4=4a-2b+c Equation (2) - equation (3) 0=8a-2b ⇒ 8𝑎 = 2𝑏 ⇒ 𝑏 = 4𝑎 Put value of in equation (3) and (4) 3=a-4a+c=-3a+c 4=4a-8a+c=-4a+c 5) ∴ 3 = −3𝑎 + 𝑐 6) 4=-4a+c Equation (5) - equation (6) 3-4=-3a+4a ⇒ 𝑎 = −1, 𝑏 = −4 Value of a put in equation (5) 3 =3+𝑐 ⇒𝑐 =0 Hence 𝑓(𝑥) = −𝑥 2 − 4𝑥 = −(𝑥 2 + 4𝑥 + (2)2 − (2)2 ) ⇒ 𝑓(𝑥) = −(𝑥 + 2)2 + 4 Vertex is: (−2, 4) axis of symmetry: x=-2 Question #3 Use a system of linear equations to find the quadratic function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 that satisfies the given conditions. Solve the system using matrices. 𝑓(1) = −1, 𝑓(2) = −4, 𝑓(3) = −11 𝑓(𝑥) =? Answer: Step 1 Given: 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑓(1) = −1 which means value of function is -1 at x=1 −1 = 𝑎(1)2 + 𝑏(1) + 𝑐 1) -1=a+b+c f(2)=-4 −4 = 𝑎(2)2 + 𝑏(2) + 𝑐 2) -4=4a+2b+c f(3)=-11 −11 = 𝑎(3)2 + 𝑏(3) + 𝑐 3) -11=9a+3a+c Step 2 Writing coefficient of a, b and c in form of a matrix for all 3 equation, as 1 1 1  a   −1   4 2 1 b  =  −4       9 3 1  c   −11 It is of the form Ax=B 1) ⇒ 𝑥 = 𝐴−1 𝐵 Where a  x =  b   c  1 1 1 A =  4 2 1 9 3 1  −1  B =  −4   −11 1 1 1 adjA (cofactor of A) A = = | A |=  4 2 1 = 1(−1) − 1(−5) + 1(−6) = −1 + 5 − 6 = −2 | A| | A| 9 3 1  −1 2 −1 adjA =  5 −8 3   −6 6 −2  −1 Step 3  1   −1 2 −1  2 1  5  A−1 = 5 −8 3  =  −   2 (−2)  −6 6 −2    3  Put in (1) 1  2   3 4 − 2 −3 1   −1  1  a   2 b  = − 5    2  c    3  1  2   −1   3  4 −  −4  2  −11 −3 1    −1 11   1 − 2 + 4 − 2  a     −2   b  =  5 − 16 + 33  =  3    2 2     c     −2   −3 + 12 − 11   ⇒ 𝑎 = −2, 𝑏 = 3, 𝑐 = −2 is the required solution 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑓(𝑥) = −2𝑥 2 + 3𝑥 − 2 Question #4 How to solve step by step: consider the quadratic function 𝑓(𝑥) = 𝑥 2 − 3𝑥 − 18 What is the smallest x-intercept? Answer: Step 1 𝑓(𝑥) = 𝑥 2 − 3𝑥 − 18 x-intercept is a point on the graph where y=0 𝑥 2 − 3𝑥 − 18 = 0 For a quadratic equation of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 the solutions are 𝑥1, 2 = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 For a=1, b=-3, c=-18 𝑥1, 2 = 𝑥1, 2 = 𝑥1, 2 = 𝑥1, 2 = −(−3)±√(−3)2 −4×1(−18) 2×1 3±√9+72 2 3±√81 2 3+9 3−9 2 , 2 [∵ √81 = 9] 𝑥1, 2 = 6, − 3 Smallest x-intercept ⇒ (−3, 0) Largest x-intercept ⇒ (6, 0) Step 2 y-intercept is the point on the graph where x=0 𝑦 = 02 − 3 × 0 − 18 y=0-0-18 y=-18 y-intercept ⇒ (0, − 18) Question #5 Find the quadratic equation whose roots are 𝛼+5 𝛽 and 𝛽+5 𝛼 where 𝛼 and 𝛽 are roots of the quadratic 2 equation 2𝑥 + 7𝑥 − 3 = 0 Answer: Step 1 Given quadratic equation is 2𝑥 2 + 7𝑥 − 3 = 0 𝛼 and 𝛽 are the roots of the given quadratic equation. We know that sum of roots of a quadratic equation = − So, 𝛼 + 𝛽 = − 7 2 Product of roots = −3 ⇒ 𝛼𝛽 = 𝑏 𝑎 𝑐 𝑎 2 Step 2 Now, we have to find quadratic equation whose roots are 𝛼+5 𝛽 and 𝛽+5 𝛼 The quadratic equation whose roots are given then equation of quadratic is given as: 𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0 Therefore, the required quadratic equation is: 𝛼+5 𝛽+5 𝛼+5 𝛽+5 𝑥2 − ( + )𝑥 + ( × )=0 𝛽 𝛼 𝛽 𝛼 𝛼(𝛼 + 5) + 𝛽(𝛽 + 5) 𝛼(𝛽 + 5) + 5(𝛽 + 5) 𝑥2 − ( )𝑥 + ( )=0 𝛼𝛽 𝛼𝛽 𝛼 2 + 5𝛼 + 𝛽 2 + 5𝛽 𝛼𝛽 + 5𝛼 + 5𝛽 + 25 𝑥2 − ( )𝑥 +( )=0 𝛼𝛽 𝛼𝛽 (𝛼 + 𝛽)2 − 𝛼𝛽 + 5(𝛼 + 𝛽) 𝛼𝛽 + 5(𝛼 + 𝛽) + 25 𝑥2 − ( )𝑥 + ( )=0 𝛼𝛽 𝛼𝛽 Step 3 Now, putting the value of (𝛼 + 𝛽) and 𝛼𝛽 in above equation, we get 7 2 𝑥2 − ( −3 −7 −3 (− ) −2× +5× 2 2 2 3 − )𝑥 + ( 49 3 35 2 𝑥2 − ( 4 − 35 +3− 2 3 2 − )𝑥 + ( 9 7 − 2 +5×−2+25 ) 3 − =0 2 −2− 2 +25 − 3 2 )=0 6 𝑥 2 − ( 43) 𝑥 + ( 3) = 0 − 2 2 − 2 3 𝑥 − 𝑥−4=0 2 2𝑥 2 − 3𝑥 − 8 = 0 Hence, required quadratic equation is 2𝑥 2 − 3𝑥 − 8 = 0 Question #6 A teenager who is 5 feet tall throws an object into the air. The quadratic function 𝑓(𝑥) = −16𝑥 2 + 64𝑥 + 5 is where f(x) is the height of the object in feet and x is the time in seconds. Answer: Given that 𝑓(𝑥) = −16𝑥 2 + 64𝑥 + 5 now to complete the square of f(x) 𝑓(𝑥) = −16 (𝑥 2 − 4𝑥 − 5 16 ) 5 = −16 (𝑥 2 − 4𝑥 − 16 5 = 16 ((𝑥 − 2)2 − 16 = −16 ((𝑥 − 2)2 − + (−2)2 − (−2)2 ) − (−2)2 ) 5 − (−2)2 ) 16 = −16(𝑥 − 2)2 + 69 The function in vertex form is 𝑓(𝑥) = −16(𝑥 − 2)2 + 69 The object hits the ground after bout f(x)=0 = −16(𝑥 − 2)2 + 69 = 0 −16(𝑥 − 2)2 = −69 69 (𝑥 − 2)2 = 16 2 For (𝑔(𝑥)) = 𝑓(𝑎) the solution are 𝑔(𝑥) = √𝑓(𝑎), −√𝑓(𝑎) 𝑥= √69 4 + 2, 𝑥 = − √69 +2 4 Question #7 Use a system of linear equations to find the quadratic function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 that satisfies the given conditions. Solve the system using matrices. 𝑓(1) = −1, 𝑓(2) = −2, 𝑓(3) = −7 𝑓(𝑥) =? Answer: Step 1 By using the given information, Solving the quadratic equation for x=1 𝑓(1) = 𝑎(1)2 + 𝑏 × 1 + 𝑐 -1=a+b+c Solving the quadratic equation for x=2 , 𝑓(2) = 𝑎(2)2 + 𝑏 × 2 + 𝑐 -2=4a+2b+c Solving the quadratic equation for x=3 , 𝑓(3) = 𝑎(3)2 + 𝑏 × 3 + 𝑐 -7=9a+3b+c Step 2 Solving the equations to get the value of a, b and c, i) a=-1-b-c -2=4(-1-b-b)+2b+c -2=-4-b-4c+2b+c 2=-2b-3c ii) 𝑐 = − ( 2𝑏+2 3 ) Substituting the value of in equation (i) then, 𝑎 = −1 − 𝑏 − (− 𝑎= 𝑎= 2𝑏+2 3 ) −3−3𝑏+2𝑏+2 3 −1−𝑏 3 Step 3 Substituting the value of a and c in expression of x=3 then, 9a+3b+c=-7 9( −1−𝑏 3 ) + 3𝑏 − ( 2𝑏+2 3 ) = −7 -9-9b+9b-2b-2=-21 -2b=-10 b=5 Then the value of a and c is, a=-2, c=-4 Hence the quadratic equation is, 𝑓(𝑥) = −2𝑥 2 + 5𝑥 − 4 Question #8 a) State a rational equation with 2 vertical asymptotes and an oblique asymptote. b) State a rational equation with a quadratic numerator, a quadratic denominators, a vertical asymptote x= -2 and a horizontal asymptote at y= 3 Answer: Step 1 a) Since the vertical asymptote is 2, therefore the value of denominator must be zero when x=2. Since the equation must have an oblique asymptote, therefore the degree of numerator must be greater than degree of denominator. Therefore the rational equation is, 𝑦 = 𝑥 2 +4 𝑥−2 Step 2 b) Since the vertical asymptote is x=-2 , therefore the value of denominator must be zero when x=-2 . Since the horizontal asymptote is y=3 , therefore the ratio of leading coefficients of numerator and denominator must be 3 Therefore the rational equation is, 𝑦= 3𝑥 2 +18𝑥+27 𝑥 2 +4𝑥+4 Question #9 Use the vertex (h, k) and a point on the graph (x, y) to find the general form of the equation of the quadratic function. (ℎ, 𝑘) = (−6, − 1), (𝑥, 𝑦) = (−9, 8) Answer: Step 1 Formula The quadratic function one have to use here is 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 Step 2 Substitute the values It is given that, h=-6, k=-1, x=-9, y=8 2 8 = 𝑎(−9 − (−6)) + (−1) 8 = 𝑎(−9 + 6)2 − 18 = 𝑎(9) − 1 9=9a a=1 Step 3 Back substituting the value of a Substituting the values of h, k and a 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘 2 𝑓(𝑥) = (𝑥 − (−6)) − 1 𝑓(𝑥) = (𝑥 + 6)2 − 1 𝑓(𝑥) = 𝑥 2 + 12𝑥 + 36 − 1 𝑓(𝑥) = 𝑥 2 + 12𝑥 + 35 Question #10 If -5 is a root of the quadratic equation 2𝑥 2 + 𝑝𝑥 − 15 = 0 and the quadratic equation 𝑝(𝑥 2 + 𝑥) + 𝑘 = 0 has equal roots. Find the value of k Answer: 2𝑥 2 + 𝑝𝑥 − 15 = 0 b-5 is root of the quadratic equation then it will satisfied it. 2(−5)2 + 𝑝(−5) − 15 = 0 50-15=5p 35=5p p=7 Now p=7 𝑝(𝑥 2 + 𝑥) + 𝑘 = 0 has equal roots 7(𝑥 2 + 𝑥) + 𝑘 = 0 7𝑥 2 + 7𝑥 + 𝑘 = 0 b roots are equal then 𝑏 2 − 4𝑎𝑐 = 0 (7)2 − 4 × 7 × 𝑘 = 0 49=28k 49 =𝑘 28 𝑘= 7 4