Chapter 5: Exercises with Solutions

Section 5.4The Quadratic Formula20yf(x)=(x−2) 2 −1220yf(x)=(x−2) 2 −12x10x10(2,−12)19. First, complete the square:f(x) = x 2 + 6x − 3= x 2 + 6x + 9 − 9 − 3= ( x 2 + 6x + 9 ) − 9 − 3= (x + 3) 2 − 12Read off the vertex as (h, k) = (−3, −12).through the vertex with equation x = −3.The axis of symmetry is a vertical lineTo find the x-intercepts algebraically, set x 2 +6x−3 = 0 and use the quadratic formulawith a = 1, b = 6 and c = −3:x = −b ± √ b 2 − 4ac2a= −6 ± √ (6) 2 − 4(1)(−3)2(1)= −6 ± √ 36 + 122= −6 ± √ 482So the x-intercepts are ((−6 − √ 48)/2, 0) and ((−6 + √ 48)/2, 0). These are approximatedon your calculator in (a) and (b).(a)(b)Version: Fall 2007