Questions and Answers #17 Quadratic Function and Equation

Questions and Answers Sheet 17 Quadratic Function and Equation Question #1 Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation for the parabola’s axis of symmetry. Use the parabola to identify the function's domain and range, 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 3 Use the graphing tool to graph the equation, Use the vertex and the y-intercept when drawing the graph. Click to enlarge graph The exis of symmetry is (Type an equation. Simplify your answer) The domain of f is (Type your answer in interval notation.) The range of f is (Type your answer in interval notation.) Answer: Forming whole square to find the vertex: 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 3 ⇒ 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 4 − 4 + 3 ⇒ 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 22 − 4 + 3 ⇒ 𝑓(𝑥) = (𝑥 − 2)2 − 1 ⇒ 𝑦 = 𝑓(𝑥) = (𝑥 − 2)2 − 1 ⇒ 𝑦 + 1 = (𝑥 − 2)2 ⇒ 𝑦 − (−1) = (𝑥 − 2)2 Hence the vertex will be: (2,-1) Finding x intercept: 𝑓(0) = 02 − 4(0) + 3 = 3 The x intercept will be: (0, 3) The axis of symmetry will be: x=2 From graph: the domain will be: (−∝, ∝) The range will be: [−1, ∝) Question #2 Find the coordinates of the x-intercepts of the graph of the equation. (If an answer does not exist, enter DNE.) 𝑦 = −𝑥 2 − 4𝑥 − 11 (x,y)= Find the minimum or maximum value of the quadratic function. 𝑓(𝑥) = −4𝑥 2 + 8𝑥 − 6 Select minimum value maximum value 𝑓(𝑥) = Answer: The given equation is, 𝑦 = −𝑥 2 − 4𝑥 − 11 From the graph it is clear that the x-intercepts (a, 0) of the function does not exists. That is, −𝑥 2 − 4𝑥 − 11 = 0 𝑥= −(−4)±√(−4)2 −4(−1)(−11) 2 = −2 ∓ √7𝑖 Which is imaginary. Therefore the answer is DNE. The maximum or minimum value of the quadratic function is obtained as follows. -4𝑥 2 + 8𝑥-6 = 0 which is of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 As a<0 , the parabola opens downwards and we find the maximum value. 𝑥=− =− 𝑏 2𝑎 −8 2(−4) =1 𝑓(1) = −4(1)2 + 8(1) − 6 =-2 The maximum value is (1,2) Question #3 From the graph of the quadratic function 𝑓(𝑥) = (𝑥 + 2)2 − 9, determine the equation of the symmetry Answer: 𝑓(𝑥) = (𝑥 + 2)2 − 9 = 𝑥 2 + 4𝑥 − 5 𝑦 = 𝑥 2 + 4𝑥 − 5 x y 0 −5 1 0 −2 −9 −5 0 The axis of symmetry is the vertical line that goes through the verter ⇒ 𝑥 = −2 by graph Question #4 Solve the initial value problem for r as a vector function of t Differential equation: (180𝑡 − 16𝑡 2 )𝑗 Initial condition: 𝑟(0) = 100𝑗 𝑑𝑟 𝑑𝑡 = (180𝑡)𝑖 + Answer: Step 1 Given differential equation, 𝑑𝑟 𝑑𝑡 = (180𝑡)𝑖 + (180𝑡 − 16𝑡 2 )𝑗 Initial condition, 𝑟(0) = 100𝑗 Step 2 Differential equation can be written as 𝑑𝑟 = [(180𝑡)𝑖 + (180𝑡 − 16𝑡 2 )𝑗]𝑑𝑡 On integrating both sides, 𝑟(𝑡) = ∫[(180𝑡)𝑖 + (180𝑡 − 16𝑡 2 )𝑗]𝑑𝑡 = 180 × 𝑡2 2 𝑖 + [180 × 𝑡2 2 𝑖 − 16 × here is an integration constant. 𝑟(𝑡) = 90𝑡 2 𝑖 + [90𝑡 2 − 16 3 𝑡 ]𝑗 3 +𝐶 Step 3 Apply the initial condition, 𝑟(0) = 100𝑗 𝑟(0) = 90𝑡 2 𝑖 + [90𝑡 2 − 16 3 𝑡 ]𝑗 3 16 100𝑗 = 90(0)𝑖 + [90(0) − 3 +𝐶 (0)] 𝑗 + 𝐶 𝐶 = 100𝑗 Therefore, 16 2 𝑡 ] 𝑗 + 100𝑗 3 16 𝑟(𝑡) = 90𝑡 2 𝑖 + [90𝑡 2 − 𝑡 2 + 100] 𝑗 𝑟(𝑡) = 90𝑡 2 𝑖 + [90𝑡 2 − 3 Step 4 Hence, solution of initial value problem is 𝑟(𝑡) = 90𝑡 2 𝑖 + [90𝑡 2 − 16 2 𝑡 3 + 100] 𝑗 Question #5 Consider the function below. 𝑓(𝑥) = 𝑥 2 𝑒 −𝑥 (a) Find the exact value of the minimum of f for 𝑥 ≥ 0 . Find the exact value of the maximum of f for 𝑥 ≥ 0 . (b) Find the exact value of x at which f increases most rapidly. Answer: Step 1 Given the function 𝑓(𝑥) = 𝑥 2 𝑒 −𝑥 Step 2 First find the critical points, the first derivative of the function to zero 𝑓(𝑥) = 𝑥 2 𝑒 −𝑥 𝑑 𝑑 𝑓 ′ (𝑥) = 𝑒 −𝑥 𝑥 2 + 𝑥 2 𝑒 −𝑥 𝑑𝑥 𝑑𝑥 𝑓 ′ (𝑥) = 2𝑥𝑒 −𝑥 − 𝑥 2 𝑒 −𝑥 𝑓 ′ (𝑥) = 𝑥𝑒 −𝑥 (2 − 𝑥) 𝑓 ′ (𝑥) = 0 therefore, 𝑡3 3 𝑖] 𝑗 + 𝐶 𝑥𝑒 −𝑥 (2 − 𝑥) = 0 x=0, x=2 Step 3 To find the exact maximum or minimum, plug the value of critical point into the function f(0)=0 𝑓(2) = (2)2 𝑒 −2 = 0.5413 therefore, the exact value of the minimum f(x)=0 4 the exact value of the maximum 𝑓(𝑥) = 2 𝑒 Question #6 A family of cubic functions have zeros of -2 and 4 ± √7. 4) Write the equation ofthe family of cubic Functions in factored form. The equation may contain a ‘nonfactorable quadratic but should contain no radicals. Show your work, b) Write the equation for the member which goes through the point (2, -72). Show your work, c) Describe the end behavior for the member in b). Answer: Step 1 Family of cubic functions that have the roots a, b and c is 𝑦 = 𝐷(𝑥 − 𝑎)(𝑥 − 𝑏)(𝑥 − 𝑐), here D is a constant. Given roots are: 𝑎 = −2, 𝑏 = 4 + √7 and 𝑐 = 4 − √7 . (a) 𝑦 = 𝐷(𝑥 − (−2)) (𝑥 − (4 + √7)) (𝑥 − (4 − √7)) = 𝐷(𝑥 + 2) ((𝑥 − 4) − √7) ((𝑥 − 4) + √7) = 𝐷(𝑥 + 2)((𝑥 − 4)2 − 7)(𝐴 − 𝐵)(𝐴 + 𝐵) = 𝐴2 − 𝐵2 = 𝐷(𝑥 + 2)(𝑥 2 + 16 − 8𝑥 − 7)(𝐴 − 𝐵)2 = 𝐴2 + 𝐵2 − 2𝐴𝐵 = 𝐷(𝑥 + 2)(𝑥 2 − 8𝑥 + 9) Step 2 (b) Substitute x = 2 and y=-72 into 𝑦 = 𝐷(𝑥 + 2)(𝑥 2 − 8𝑥 + 9) and solve for D. −712 = 𝐷(2 + 2)(22 − 8(2) + 9) 6=D For cubic equation which goes through point (2, -72) substitute D=6 into 𝑦 = 𝐷(𝑥 + 2)(𝑥 2 − 8 + 9) we get, 𝑦 = 6(𝑥 + 2)(𝑥 2 − 8𝑥 + 9) The graph 𝑦 = 6(𝑥 + 2)(𝑥 2 − 8𝑥 + 9) shows the cubic equation passing through the point (2, -72) which have the three roots −2,4 + √7 ≈ 6.646 and 4 − √7 ≈ 1.354. Question #7 Graph the quadratic function 𝑓(𝑥) = (𝑥 + 4)2 . Give the (a) vertex, (b) axis, (c) domain, and (d) range. Then determine (e) the largest open interval of the domain over which the function is increasing and (f) the largest open interval over which the function is decreasing. (a) The vertex is (-4,0). (b) The axis is Answer: Domain set of all input values where function defined when f'>0 then function increasing when f'<0 then function decreasing 𝑦 = (𝑥 + 4)2 b) Axis is x+4=0 x=-4 The axis is x=-4 c) Domain 𝑦 = (𝑥 + 4)2 Defmen everywhere So. Domain 𝑥 ∈ (−∞, ∞) d) Ranges 𝑦 = (𝑥 + 4)2 ≥ 0 Range [0, ∞) e) 𝑑𝑦 𝑑𝑥 = 2(𝑥 + 4) ≥ 0 x=-4 Function decreasing when 𝑑𝑦 𝑑𝑥 >0 𝑥 ∈ (−4, ∞) f) Function decreasing when 𝑥 ∈ (−∞, −4) 𝑑𝑦 𝑑𝑥 <0 Question #8 Graph each quadratic function using the steps from the picture. Based on the graph determine the domain and range of the quadratic function. |𝐺(𝑥) = −9𝑥 2 + 18𝑥 + 7 Answer: Step 1 The given quadratic function 𝑦 = 9𝑥 2 + 18𝑥 + 7 Since the coefficient of 𝑥 2 is negative, therefore parabola opens downward Arrange the x terms by completing the squares Step 2 7 y = −9( x 2 − 2 x − ) 9 7 y = −9(( x − 1) 2 − 1 − ) 9 16 y = −9(( x − 1) 2 − ) 9 𝑦 = −9(𝑥 − 1)2 + 16 𝑦 − 16 = −9(𝑥 − 1)2 Step 3 Vertex (1, 16) axis of symmetry x=1 y-intercept, x=0 y=7 Step 4 Discriminant 𝑏 2 − 4𝑎𝑐 = 364 − 4 × (−9) × 7 = 616 Therefore parabola has 2 intercepts on x-axis 9𝑥 2 + 18𝑥 + 7 = 0 𝑥= −1 3, 7 3 Question #9 Graph each quadratic function using the steps from the picture. Based on the graph determine t and range of the quadratic function. 𝐹(𝑥) = 4𝑥 2 − 4𝑥 − 3 Answer: Step 1 To draw graph of the quadratic equation 𝐹(𝑥) = 4𝑥 2 − 4𝑥 − 3 Step 2 Compare given function with 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 a=4 b=-4 c=-3 1) Since a>0 graph is open upwards Step 3 2) vertex = ( − − − 𝑏 2𝑎 𝑏 2𝑎 =− b b , F (− )) 2a 2a (−4) 2×4 = 0.5 𝐹(0.5) = 4(0.5)2 − 4(0.5) − 3 F(0.5)=-4 vertex=(0.5,-4) axis of symmetry 𝑥=− 𝑏 2𝑎 x=0.5 Step 4 3) y-intercept F(0)=0-0-3 F(0)=-3 y-intercept is (0, -3) Step 5 4) 𝑏 2 = 4𝑎𝑐 = (−4)2 − 4(4)(−3) 𝑏 2 − 4𝑎𝑐 = 16 + 48 𝑏 2 − 4𝑎𝑐 = 64 𝑏 2 − 4𝑎𝑐 > 0 Thus, F(x) has two x-intercept F(x)=0 4𝑥 2 − 4𝑥 − 3 − 0 4𝑥 2 − 6𝑥 + 2𝑥 − 3 = 0 2𝑥(2𝑥 − 3) + 1(2𝑥 − 3) = 0 (2𝑥 + 1)(2𝑥 − 3) = 0 1 3 𝑥=− , 2 2 x=-0.5,1.5 x-intercept are (-0.5,0) and (1.5,0) Domain of the given function is (−∞, ∞) Range of the given function is [−4, ∞) Question #10 Sketch the graph of the quadratic function 𝑓(𝑥) = −2𝑥 2 + 6𝑥 − 3. On the graph, show the values of x and y intercepts, vertex, and axis of the quadratic function. Give the domain and range of the function also. Answer: Step 1 The given function is 𝑓(𝑥) = −2𝑥 2 + 6𝑥 − 3 Step 2 Here, a=-2,b=6,c=-3 . Obtain the vertex as follows: −b −b , f ( )) 2a 2a −6 −b =( , f ( )) 2(−2) 2a 3 3 = ( , f ( )) 2 2 3 3 3 = ( , −2( ) 2 + 6( ) − 3) 2 2 2 3 3 2 3 = ( , −2( ) + 6( ) − 3) 2 2 2 3 3 =( , ) 2 2 3 3 Thus, the vertex is ( , ). 2 2 ( Note that, the axis of symmetry is a vertical line 𝑥 = −𝑏 2𝑎 . Step 3 Obtain the y-intercept as follows: 𝑦 = −2(0)2 + 6(0) − 3 =-3 Thus, the y-intercept is (0,-3). Obtain the x-intercept as follows: 0 = −2𝑥 2 + 6𝑥 − 3 𝑥1,2 = = = −6±√62 −4(−2)(−3) 2(−2) −6±√12 −4 3−√3 3+√3 2 , 2 = 0.63397, 2.36602 Thus, the x-intercepts are (0.63397,0) and (2.36602. 0). From the above figure, its observed that the domain of the function is (−∞, ∞) and the range of the function is (−∞, 1.5].