Pre-Calculus 2-6B: Page 161 #’s 27-38odd, 41,49

27) Find an equation in the form y=a(x-x1)(x-x2)

Vertex (0,-2), passing through (3,25) y = 3(x – 0)2 – 2 Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. Vertex (0,-2), passing through (3,25) y = 3(x – 0)2 – 2

31)Vertex (3,1), x-intercept of 0 y = -1/9(x – 3)2 + 1 Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. 31)Vertex (3,1), x-intercept of 0 y = -1/9(x – 3)2 + 1

Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. 33) x-intercepts -3 and 5, highest point has y-coord 4 y = -1/4(x – 1)2 + 4

Find the maximum vertical distance d between the parabola and the line for the green region 35) 6.125

37) D(h)= -0.058h2 + 2.867h – 24.239 24.72 km

Max distance from ground. 424 ft 41) An object is projected vertically upward from the top of a building with an initial velocity of 144 ft/sec. Its distance s(t) in feet above the ground after t sec is given by the equation: h=-16t2 + 144t + 100 Max distance from ground. 424 ft Height of the building. 100 ft

Equation of the parabola y = 1/500x2 + 10 49) One section of a suspension bridge has its weight uniformly distributed between twin towers that are 400 ft apart and rise 90 ft above the horizontal roadway. A cable strung between the tops of the towers has the shape of a parabola, and its center point is 10 feet above the roadway. Suppose coordinate axes are introduced, as shown in the figure. Equation of the parabola y = 1/500x2 + 10 9 equally spaced vertical cables are used to support the bridge. Find the total length of these supports. 282 ft