## Answers

1) To find the perfect square trinomial, add the square of half the x-coefficient . {{{a^2-8a+(8/2)^2}}} Simplify. {{{highlight(a^2-8a+16 = (a-4)^2)}}} 2) Solve by completing the square: {{{t^2-3t-7 = 0}}} Add 7 to both sides. {{{t^2-3t = 7}}} Now complete the square in t by adding the square of half the t-coefficient to both sides. This is {{{(3/2)^2 = 9/4}}}.

{{{t^2-3t+9/4 = 7+(9/4)}}} Factor the left side and simplify the right side. {{{(t-(3/2))^2 = (28/4)+(9/4)}}} {{{(t-(3/2))^2 = 37/4}}} Take the square root of both sides. {{{t-(3/2) = (sqrt(37))/2}}} or {{{t-(3/2) = -(sqrt(37))/2}}} Add {{{3/2}}} to both sides of each solution. {{{highlight_green(t = (3+sqrt(37))/2)}}} or {{{highlight_green(t = (3-sqrt(37))/2)}}} 3) Use the formula for the height of an object propelled upward with an initial velocity of {{{v[0]}}} ft/sec from an initial height of {{{h[0]}}} feet. {{{G = 32 ft/(sec)^2}}}: {{{h(t) = -(1/2)Gt^2+v[0]t+h[0]}}} Substitute {{{v[0] = 64}}} and {{{h[0] = 80}}} {{{highlight(h(t) = -16t^2+64t+80)}}} First find the time t at which the object reaches its maximum height.

Remember that this equation, when graphed, will show a parabola that opens downward so that the vertex of the graph will be the maximum point of the curve or the time of maximum height, t. The vertex of this parabola is given by {{{t = -b/2a}}} where a = -16 and b = 64, so... {{{t = -(64)/2(-16)}}} {{{t = 64/32}}} {{{highlight(t =2)}}}seconds. The object will attain its maximum height in 2 seconds. To find the maximum height of the object, subtitute t = 2 into the beginning equation {{{highlight(h(t))}}}and solve for h.

{{{h(2) = -16(2)^2+64(2)+80}}} {{{h(2) = -64+128+80}}} {{{highlight(h(2) = 144)}}}feet is the maximum height attained by the object...