2. They break even if they sell 100 tickets.

4.  Ontario’s population (in 100 000s) can be modelled by the relation P = 0.007x 2 + 0.196x + 21.5, where x is the number of years since 1900.  a) Using this model, what was Ontario’s population in 1925?  x = 25  P = 0.007(25) 2 + 0.196(25) + 21.5  = 4.375 + 4.9 + 21.5  P = 30.775  Remember, P is in 100 000s, so P = 30.775 x 100 000  P = 3,077,500 people in Ontario in 1925.

6. x = -150.20 does not make sense, because this means that the population of Ontario would be 15 000 000 in 1750. x = 122.20 makes more sense. This means that we add 122.20 to 1900. So the population of Ontario will be 15 000 000 in 2022

7.  Below is a dog fence attached to a wall for 4 dogs. There is 30m of fence available. What dimensions will give the maximum area? ww www l = 30 – 5w A = lw = w(30 – 5w) = 30w – 5w 2 = -5w 2 + 30w = -5(w 2 - 6w) Try (w-3) 2 = w 2 -6w+9 = -5((w-3) 2 – 9) = -5(w-3) 2 + 45 So vertex is (3, 45) so maximum area is 45m 2.

8.  When solving a problem that involves a quadratic relation:  1. Write a relation to help you solve the problem, e.g. A = lw  2. Use the vertex form to determine the maximum or minimum value  3. Use standard form or factored form to find x-value that corresponds to a given y-value of the relation ◦ You may need to use the quadratic formula to find this