Steps 4 and 5: Find additional points, plot the points, and sketch the graph. Vertex (2, 1) x y 5 y x2 4x 3 5 5 5 x y 2 1 0 3 1 0 3 0 4 3 Vertex S y-int. S x-int. S x-int. S YOU TRY 3 Graph y x2 6x 8. Sometimes we need to use the quadratic formula to fi nd the x-intercepts of the graph. EXAMPLE 4 In-Class Example 4 Graph y x2 6x 7. Answer: Graph y x2 2x 2. Solution Step 1: Find the vertex. x b 2a 2 2(1) 2 2 1 a 1 b 2 Substitute x 1 into the equation to fi nd the y-coordinate of the vertex. y (1)2 2 (1) 2 1 2 2 3 The vertex of the parabola is (1, 3). Step 2: Find the y-intercept by substituting 0 for x and solving for y: y (0)2 2 (0) 2 2 The y-intercept is (0, 2). Step 3: Find the x-intercepts by substituting 0 for y and solving for x: 0 x2 2x 2 We will solve this equation using the quadratic formula. x (2) 2(2)2 4(1)(2) 2(1) Let a 1, b 2, c 2. x 2 112 2 2 213 2 2(1 13) 2 1 13 The x-intercepts are (1 13, 0) and (1 13, 0). y x2 6x 7 x 6 y 4 6 (3 √2, 0) 4 (3 √2, 0) Vertex (3, 2) www.mhhe.com/messersmith SECTION 10.4 Graphs of Quadratic Equations 635
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