Chapter 5: Exercises with Solutions

Section 5.2Vertex Formy10y10(3,5)x10x1053. First complete the square to transform the function into vertex form a(x−h) 2 +k:f(x) = −x 2 − 10x − 21= −(x 2 + 10x + 21)= −[x 2 + 10x + 25 − 25 + 21]= −[(x 2 + 10x + 25) − 25 + 21]= −[(x + 5) 2 − 25 + 21]= −[(x + 5) 2 − 4]= −(x + 5) 2 + 4Compare the quadratic function f(x) = −(x + 5) 2 + 4 with f(x) = a(x − h) 2 + k andnote that h = −5 and k = 4. Hence, the vertex is located at (h, k) = (−5, 4). Theaxis of symmetry is a vertical line through the vertex with equation x = −5. 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) = −(x + 5) 2 + 4.x=−5y10(−5,4)x y = −(x + 5) 2 + 4−4 3−3 0x10f(x)=−(x+5) 2 +4Version: Fall 2007