From the equation f (x) (x 3)2 1 we can see that i) The vertex is (3, 1). ii) The axis of symmetry is x 3. iii) a 1 so the parabola opens upward. iv) Since a 1, the graph is the same width as y x2. Find some other points on the parabola. Use the axis of symmetry. To fi nd the x-intercepts, let f (x) 0 and solve for x. Use either form of the equation. We will use f (x) (x 3)2 1. 1 1 1 5 5 5 2 2 2 0 (x 3)2 1 Let f (x) 0. 1 (x 3)2 Subtract 1. 1 x f(x) 2 2 1 5 11 x 3 Square root property 3 i x 11 i; subtract 3. y 1 Because the solutions to f (x) 0 are not real numbers, there are no x-intercepts. To fi nd the y-intercept, let x 0 and solve for f (0). f (x) (x 3)2 1 f (0) (0 3)2 1 f (0) 9 1 10 The y-intercept is (0, 10). The domain is (q, q), and the range is 1, q). b) Step 1: The coeffi cient of x2 is 1 2 x . Multiply both sides of the equation (including the g(x)) by 2 so that the coeffi cient of x2 will be 1. g(x) 1 2 x2 4x 6 2g(x) 2a 1 2 x2 4x 6b Multiply by 2. 2g(x) x2 8x 12 Distribute. Step 2: Separate the constant from the variable terms using parentheses. 2g(x) (x2 8x) 12 Step 3: Complete the square for the quantity in parentheses. 1 2 (8) 4 (4)2 16 Add 16 inside the parentheses, and subtract 16 from the 12. 2g(x) (x2 8x 16) 12 16 2g(x) (x2 8x 16) 4 Step 4: Factor the expression inside the parentheses. 2g(x) (x 4)2 4 Write out each step as you are reading the examples, and be sure you understand each step! 660 CHAPTER 10 Quadratic Equations and Functions www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
To see the actual publication please follow the link above