N 58 Graphical Solutions to Quadratic Functions Subject Content Reference: N6.7h GCSE Maths Number & Algebra

We need to be able to find graphical solutions to quadratic functions.. Example Solve the quadratic equation x 2 + 2x - 5 = 0 using a graph: Step 1: Draw up a table of values for y = x 2 + 2x - 5 x y = x 2 + 2x Step 2: Draw the graph of y = x 2 + 2x x y 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).. here.. and here The approximate solutions of the quadratic equation x 2 + 2x - 5 = 0 are 1.4 and -3.4 When there are two points of intersection, there are two solutions to the equation.. 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.. Step 3:

Exercise 1 x y x y = x 2 + 3x - 5 x y x y = 2x 2 - x - 5 1) Solve the quadratic equation x 2 + 3x - 5 = 0 using a graph:2) Solve the quadratic equation 2x 2 - x - 5 = 0 using a graph:

We need to be able to find graphical solutions to more complex quadratic equations.. Example Solve the quadratic equation x 2 + 2x - 5 = 2x + 1 using graphs: Step 1: Draw up a table of values for y = x 2 + 2x x y = x 2 + 2x Step 3: Draw the graphs of y = x 2 + 2x - 5 and y = 2x x y Now use the graphs to find approximate solutions by looking at the points where the graphs intersect.. here.. and here The approximate solutions of the quadratic equation x 2 + 2x - 5 = 2x + 1 are 2.4 and -2.4 When there are two points of intersection, there are two solutions to the equation.. Be sure to give them both! Step 4: Step 2: Find two values for the linear graph y = 2x x y = 2x Algebraic check: x 2 + 2x - 5 = 2x + 1 becomes x 2 = 6 (taking 2x from and adding 5 to both sides)  x = √6 = ± 2.45

Exercise 2 x y x y = 2x 2 + 3x - 5 y = 2x + 3 x y x y = 2x 2 - 3x + 1 y = 2x - 3 1) Solve the quadratic equation 2x 2 + 3x - 5 = 2x + 3 using graphs:2) Solve the quadratic equation 2x 2 - 3x +1 = 2x - 3 using graphs:

Examples 1) Show that the points where the graphs y = 4x - 3 and y = 5 / 2x intersect are the solutions to the quadratic equation 8x 2 - 6x - 5 = 0: At the points of intersection of these graphs, 4x - 3 = 5 / 2x  8x 2 - 6x = 5 (multiplying both sides by 2x )  8x 2 - 6x - 5 = 0 (subtracting 5 from both sides) answer 2) Find the linear equation to use with the quadratic graph y = x 2 - 2x + 1 to solve the equation y = x 2 + 3x - 1 Let the linear equation be y = mx + c, and this intersects with y = x 2 - 2x + 1 At the points of intersection, x 2 - 2x + 1 = mx + c (subtracting mx + c from both sides)  x 2 - 2x - mx c = 0 Comparing this equation with y = x 2 + 3x - 1 when y = 0 we have x 2 - 2x - mx c = x 2 + 3x - 1  - 2x - mx = 3x so m = -5 and 1 - c = -1 so c = 2 So the linear equation we need to use is y = -5x + 2 answer ( because -2x - - 5x = 3x) ( because = -1)

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 10x 2 - 4x - 3 = 0: 2) Find the linear equation to use with the quadratic graph y = x 2 - 3x + 2 to solve the equation y = x 2 + 2x - 4

Exercise 4 Complete the table of values below, draw the graphs and use them to solve the following equations: x y = x 2 + 3x - 2 y = x 2 - 2x + 3 y = 4 - x 2 y x a) x 2 + 3x - 2 = 0 b) x 2 + 3x - 2 = -3 y x a) x 2 - 2x + 3 = 5 b) x 2 - 2x + 3 = 12 1) 2) y x a) 4 - x 2 = 0 b) x 2 - 2x + 3 = 4 - x 2 3) y x a) 4 - x 2 = x 2 + 3x - 2 b) 4 - x 2 = 2x + 1 4) c) Show, using graphs, there is just one solution to the equation x 2 + 3x - 2 = x 2 - 2x + 3