GCSE Maths Number & Algebra N 58 Graphical Solutions to Quadratic Functions Subject Content Reference: N6.7h We need to be able to find graphical solutions to quadratic functions . . Example Solve the quadratic equation x2 + 2x - 5 = 0 using a graph: Step 1: Draw up a table of values for y = x2 + 2x - 5 y= x2 x -3 -2 -1 0 1 2 3 + 2x - 5 -2 -5 -6 -5 -2 3 10 Step 2: Draw the graph of y = x2 + 2x - 5 . . y Step 3: Now use the graph to find approximate solutions by looking at the points where the graph intersects the line y = 0 (i.e. the x-axis) . . 10 here 5 The approximate solutions of the quadratic equation x2 + 2x - 5 = 0 are 1.4 and -3.4 -3 -2 -1 -5 0 1 2 3 x When there are two points of intersection, there are two solutions to the equation . . . . and here Be sure to give them both! Remember - a quadratic graph is symmetrical. Here, if we just connect the plotted points, the graph only intersects the line y = 0 once - but it can easily be extended because of its symmetry . . learn Exercise 1 1) Solve the quadratic equation x2 x -3 -2 2) Solve the quadratic equation 2x2 + 3x - 5 = 0 using a graph: -1 0 y = x2 + 3x - 5 1 2 3 x -3 -2 - x - 5 = 0 using a graph: -1 0 y = 2x2 - x - 5 y y x x 1 2 3 We need to be able to find graphical solutions to more complex quadratic equations . . Example Solve the quadratic equation x2 + 2x - 5 = 2x + 1 using graphs: Step 1: Draw up a table of values for y = x2 + 2x - 5 . . x -3 -2 -1 0 1 2 3 + 2x - 5 -2 -5 -6 -5 -2 3 10 x -3 -2 -1 0 1 2 3 y = 2x + 1 -5 y= Step 2: Find two values for the linear graph y = 2x + 1 . . y x2 1 Step 3: Draw the graphs of y = x2 + 2x - 5 and y = 2x + 1 . . 10 Step 4: Now use the graphs to find approximate solutions by looking at the points where the graphs intersect . . here 5 The approximate solutions of the quadratic equation x2 + 2x - 5 = 2x + 1 are -3 -2 -1 0 1 2 3 x 2.4 and -2.4 When there are two points of intersection, there are two solutions to the equation . . -5 . . and here Be sure to give them both! Algebraic check: x2 + 2x - 5 = 2x + 1 becomes x2 = 6 (taking 2x from and adding 5 to both sides) x = √6 = ± 2.45 learn Exercise 2 1) Solve the quadratic equation 2x2 x -3 -2 + 3x - 5 = 2x + 3 using graphs: -1 0 1 2 3 2) Solve the quadratic equation 2x2 x y = 2x2 + 3x - 5 y = 2x2 - 3x + 1 y = 2x + 3 y = 2x - 3 y -3 -2 - 3x +1 = 2x - 3 using graphs: -1 0 y x x 1 2 3 Examples 1) Show that the points where the graphs y = 4x - 3 and y = 5/2x intersect are the solutions to the quadratic equation 8x2 - 6x - 5 = 0: At the points of intersection of these graphs, 4x - 3 = 5/2x 8x2 - 6x = 5 (multiplying both sides by 2x) 8x2 - 6x - 5 = 0 (subtracting 5 from both sides) 2) Find the linear equation to use with the quadratic graph answer y = x2 - 2x + 1 to solve the equation y = x2 + 3x - 1 Let the linear equation be y = mx + c, and this intersects with y = x2 - 2x + 1 At the points of intersection, x2 - 2x + 1 = mx + c x2 - 2x - mx + 1 - c = 0 (subtracting mx + c from both sides) Comparing this equation with y = x2 + 3x - 1 when y = 0 we have x2 - 2x - mx + 1 - c = x2 + 3x - 1 - 2x - mx = 3x so m = -5 and 1 - c = -1 so c = 2 (because -2x - - 5x = 3x) (because 1 - 2 = -1) So the linear equation we need to use is y = -5x + 2 answer Exercise 3 1) Show that the points where the graphs y = 5x - 2 and y = 3/2x intersect are the solutions to the quadratic equation 10x2 - 4x - 3 = 0: 2) Find the linear equation to use with the quadratic graph y = x2 - 3x + 2 to solve the equation y = x2 + 2x - 4 Exercise 4 Complete the table of values below, draw the graphs and use them to solve the following equations: x y= x2 -3 -2 -1 0 1 2 3 1) y a) x2 + 3x - 2 = 0 b) x2 + 3x - 2 = -3 a) 4 - x2 = x2 + 3x - 2 + 3x - 2 y = x2 - 2x + 3 y = 4 - x2 2) x y x a) x2 - 2x + 3 = 5 b) x2 4) - 2x + 3 = 12 y x b) 4 3) y x a) 4 - x2 = 0 b) x2 - 2x + 3 = 4 - x2 - x2 = 2x + 1 c) Show, using graphs, there is just one solution to the equation x2 + 3x - 2 = x2 - 2x + 3

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