Note: These are not meant to be complete answers as you would complete them on the exam itself; rather, they are to be used as a guide towards correct solutions.

1. • x is about -2 or x is about 3.
   • f(-1) is about -1.
   • You need to make the top of the graph only go to about 1.5, make the bottom of the graph only go to -2, move the graph one unit to the right, then shift the whole graph up 3 units.
   • Domain: -4 <= x <= 5
     Range: -4 < = y < = 3
   • f is increasing from x = -4 until x = -2.
   • The same as when it is concave down.
   • f(g(2)) is about 1.8, g(f(3)) is about 3, and (g+f)(2), which is defined to be f(2) + f(2), is about 1.8.

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2. • The missing point could be (2,6).
   • The missing point could be (2,4*sqrt(2)).
   • If you draw a cosine graph whose lowest point is (1,4) and whose highest point is (3,8), the missing point could be (2,6) (Draw it!)
   • To do this one, switch the x and y coordinates, then figure out what the missing point might be if it were exponential, then switch the x and y coordinates back again. Answer: (4,4^(1/5)).

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3. • A trig graph. Possible equation: y = 5.5cos(pi*x/2).
   • An exponential graph. Possible equation: y = 2.5^x.
   • A cubic graph. Possible equation: y = 1.6(x-2)^2(x+1).

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4. The cubic has three roots; by hunting a lot we see that they're around x=1, x=-8, x=116 (very roughly!) We see that 2^x crosses this graph twice near the two roots: at x = -8.563 (to three decimal places) and at x = 1 (exactly, I think.). After that 2^x starts climbing very quickly, and the cubic becomes negative until x is about 116. By this time 2^x is about 10^35. Since exponential functions dominate cubics eventually, we can safely conclude that the cubic function will never catch up, and we've found the only two solutions.

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5. • M = 500(1+.082)^t
   • Solving algebraically, t = log(3)/log(1.082). Solving graphically, we graph M = 500(1+.082)^t and M = 1500, and see when they cross. They cross at the point (13.9398,1500) (to 4 decimal places), so the answer found graphically would be 13.9398 years.
   • 500(1+.082/12)^12 - 500 is about $42.58.

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6. I tried to sketch the graph so that when the was narrow, the height was going up quickly, when the the vase was wide, the height was going up slowly, and when the vase was of roughly constant height, the height was going up at a constant rate. One such graph might be:
   • The height is increasing most quickly when the vase is at its most narrow; by my graph that happens at about t = seconds.
   • The height is increasing when the vase is at its widest; by my graph, that happens when t = 0 seconds.
   • This would happen when the graph is curving most sharply up, which seems to be the whole middle section!

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7. The equation of the parabola is y = -(x-2)(x+2). Then the coordinates of the two points are (1.75, 0.9375) and (-sqrt(3.25),0.75). Then an equation of the line is y - 0.9375 = (0.1875/(1.75+\sqrt(3.25))(x-1.75).