Chapter 5: Exercises with Solutions

Section 5.2Vertex Form59. First complete the square to transform the function into vertex form a(x−h) 2 +k:f(x) = (1/2)x 2 − 4x + 5= 1 (x 2 − 8x + 10 )2= 1 (x 2 − 8x + 16 − 16 + 10 )2= 1 ((x 2 − 8x + 16 ) − 16 + 10 )2= 1 ()(x − 4) 2 − 62= 1 2 (x − 4)2 − 3Compare the quadratic function f(x) = 1 2 (x − 4)2 − 3 with f(x) = a(x − h) 2 + k andnote that h = 4 and k = −3. Hence, the vertex is located at (h, k) = (4, −3). Theaxis of symmetry is a vertical line through the vertex with equation x = 4. Make atable to find two points on either side of the axis of symmetry. Plot them and mirrorthem across the axis of symmetry. Use all of this information to complete the graph off(x) = 1 2 (x − 4)2 − 3.y10f(x)= 1 2 (x−4)2 −3x y = 1 2 (x − 4)2 − 33 −2.52 −1(4,−3)x10x=4To find the domain of f, mentally project every point of the graph onto the x-axis, asshown on the left below. This covers the entire x-axis, so the domain= (−∞, ∞). Tofind the range, mentally project every point of the graph onto the y-axis, as shown onthe right below. The shaded interval on the y-axis is range= [−3, ∞).Version: Fall 2007