71 Sect 4. Inequalities Involving Quadratic Function Objective #0: Solving Inequalities using a graph Use the graph to the right to find the following: Ex. 1 a) Find the intervals where f(x) > 0. b) Find the intervals where f(x) 0. c) Find the intervals where f(x) < 0. d) Find the intervals where f(x) 0. a) We are looking for the x-values where the graph is above the x- axis: (, 1) U (4, ) b) Now, we will include the x-ints.: [, 1] U [4, ) c) We are looking for the x-values where the graph is below the x- axis: (, ) U (1, 4) d) Now, we will include the x-ints.: (, ] U [1, 4] Use the graph to the right to find the following: Ex. a) Find the intervals where g(x) > 0. b) Find the intervals where g(x) 0. c) Find the intervals where g(x) < 0. d) Find the intervals where g(x) 0. a) We are looking for the x-values where the graph is above the x- axis: (, ) U (, 1) U (, ) b) Now, we will include the x-ints.: (, 1] U [, ) c) We are looking for the x-values where the graph is below the x- axis: ( 1, ) d) Now, we will include the x-ints.: { } U [ 1, ] 4 1 0-4 - - -1-1 0 1 4 6 - - -4 - f(x) 1 0 - -4 - - -1-1 0 1 4 - - -4 - -6-7 g(x)
7 Objective #1: Solve Inequalities Involving a Quadratic Function. Much in the same way as when solving a quadratic equation, we will need to first get 0 on one side of the inequality so we get something in the form of ax + bx + c > 0 or ax + bx + c < 0. Our next step then will be to graph the quadratic equation y = ax + bx + c. If the inequality involves >, then the intervals where the graph of the function is above the x-axis will be the solution. If the inequality involves <, then the intervals where the graph of the function is below the x-axis will be the solution. If the inequality involves or, then we include the x-coordinates of the x-intercepts as part of our solution. Solve the following inequalities: Ex. x x > 0 Let f(x) = x x We will use the same steps for graphing that we used in section 4.: a) f(x) = x x, a =, b =, & c = (vertex formula) h = b a = ( ) () = 1. k = f(1.) = (1.) (1.) =.1 6. = 6.1 Thus, the vertex is (1., 6.1). The axis of symmetry is x = 1.. Since a > 0, f has a minimum value of 6.1 at x = 1.. b) x-intercepts. Let f(x) = 0: 0 = x x (factor) 0 = (x + 1)(x ) (solve) x + 1 = 0 or x = 0 x = 1 or x = y - intercepts. Let x = 0: f(0) = (0) (0) = So, the x-intercepts are ( 1, 0 ) and (, 0) and the y-intercept is (0, ). c) Since a =, h = 1., & k = 6.1, then the graphing form is f(x) = (x 1.) 6.1
7 d) Let's go through the steps of our general strategy: i) Since a =, the graph ii) iii) is stretched by a factor of. Since a is positive, the graph is not reflected across the x-axis. Since k is 6.1 and h is 1., the graph is shifted down by 6.1 units and to the right by 1. units. Since the inequality was x x > 0, we need the intervals where the graph is above the x-axis which are (, 0.) and (, ). Thus, the solution is (, 0.) U (, ). Ex. 4 x x 0 The graph of f(x) = x x is the same as the one in the previous example, but this time, we are looking at the intervals where the graph is below the x-axis which is ( 0., ). Since the inequality is < or =, we need to include the x-coordinates of the x-intercepts in the solution. Thus, our answer is [ 0., ]. Objective #: Solving Quadratic Inequalities Using a Sign Chart. At the x-intercepts, the graph of a quadratic function must either touch or cross the x-axis. Thus, between any two consecutive x-intercepts, the graph is either above the x-axis (f(x) is positive) or below the x-axis (f(x) is negative). We can then pick a value of x between two consecutive x- intercepts and use it to determine if the function is positive or negative in that interval. We can then select the interval(s) that satisfy the inequality. Procedure for Solving Quadratic Inequalities Using a Sign Chart. 1) Get zero on one side of the inequality and then find the x-intercepts of the quadratic function on the other side. ) Plot the x-values of the x-intercepts on a number line. This will split the number line into intervals. ) Select a test value from each interval and substitute it into the quadratic function to determine the sign (+ or ) of the function in that interval. 4) Examine the inequality found in the first step and determine which intervals from step # satisfy the inequality. If the inequality is or, be sure to include the x-values of the x-intercepts as part of the solution.
74 Solve the following inequalities: Ex. x + x 10 Step #1: Get zero on one side. x + x 10 (subtract 10 from both sides) x + x 10 0 Let f(x) = x + x 10 Find the x-intercepts of x + x 10: x + x 10 = 0 (factor) (x )(x + ) = 0 (solve) x = 0 or x + = 0 x = or x = Step #: Plot the x-values of the x-intercepts on a number line: Intervals: (, ) (, ) (, ) Step #: Pick a test value in each interval and substitute into the quadratic function f(x) = x + x 10. Intervals: (, ) (, ) (, ) f( ) f(0) f() = ( ) + ( ) 10 = (0) + (0) 10 = () + () 10 = 14 = 10 = 4 f is positive on f is negative on f is positive on (, ). (, ). (, ). Step #4: We are looking for when x + x 10 0 which is where f(x) is positive or zero. The intervals that satisfy this inequality are (, ) & (, ). Since the inequality is, we need to include the x-values of the x-intercepts so our solution is (, ] U [, ). Ex. 6 x + 10x < 0 Step #1: Let f(x) = x + 10x Find the x-intercepts of x + 10x : x + 10x = 0 (factor)
7 (x 10x + ) = 0 (x ) = 0 x = 0 or x = Step #: Plot the x-value of the x-intercept on a number line: Intervals: (, ) (, ) Step #: Pick a test value in each interval and substitute into the quadratic function f(x) = x + 10x. Intervals: (, ) (, ) f(0) f(6) = (0) + 10(0) = (6) + 10(6) = = 1 f is negative on f is negative on (, ). (, ). Step #4: We are looking for when x + 10x < 0 which is where f(x) is negative. The intervals that satisfy this inequality are (, ) & (, ). Since the inequality is <, we do not include the x-values of the x-intercepts so our solution is (, ) U (, ). Ex. 7a x + 10x 0 Ex. 7b x + 10x 0 Ex. 7c x + 10x > 0 a) We are looking for when x + 10x 0 which is where f(x) is negative or zero. The intervals that satisfy this inequality are (, ) & (, ). Since the inequality is, we do include the x-values of the x-intercepts which gives us all real numbers. Thus, our answer is (, ). b) We are looking for when x + 10x 0 which is where f(x) is positive or zero. There are no intervals that satisfy this inequality. Since the inequality is, we do include the x-values of the x-intercept which gives us just x =. Thus, our answer is {}. c) We are looking for when x + 10x > 0 which is where f(x) is positive. There are no intervals that satisfy this inequality. Since the inequality is >, we do not include the x-values of the x-intercept which means there is no solution. So, our answer is no solution or.
Ex. 8 x(x + ) + 8 < x Step #1: We need to simply and get 0 on one side of the inequality: x(x + ) + 8 < x (distribute) x + x + 8 < x (add x to both sides) x + 4x + 8 < 0 Let f(x) = x + 4x + 8 Find the x-intercepts of x + 4x + 8. 0 = x + 4x + 8, a = 1, b = 4, & c = 8 (quadratic formula) x = (4) (4) 4(1)(8) (1) So, there are no x-intercept. = 4± 16 = 4± 16 is not real Step #: Since there are no x-values of the x-intercepts to plot, then we only have one interval, (, ). We can pick 0 as our test value. Step # 76 f(0) = (0) + 4(0) + 8 = 8 so f(x) is positive on (, ). Step #4: Since the inequality is x + 4x + 8 < 0, we are looking at the intervals where f(x) is negative which there are none. Thus, our answer is no solution or. Find the Domain of the following: Ex. 9 x + 4x+ 8 This function is defined when x + 4x + 8 0, so we are looking for the intervals where f(x) is positive or zero. From the previous example, we saw that f(x) was always positive so the domain is (, ). Ex. 10 1 x x This function is defined when x x > 0, so we are looking for the intervals where the graph is above the x-axis. From example #, we saw this was the intervals (, 0.) and (, ). Thus, the domain is (, 0.) U (, ).