MTH-4111 QUIZ Chap 1 - 8 1. Determine the solution algebraically of the following system: 2. The following graphs illustrate the value of 2 stocks on the.

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MTH-4111 QUIZ Chap Determine the solution algebraically of the following system: 2. The following graphs illustrate the value of 2 stocks on the Toronto Stock Exchange. After starting the year at a value of $3, Stock A is at its lowest value after 4 months at $2 per share. Stock B steadily increases from an initial value of $2 per share to $3.50 after 12 months. At what point(s) does the two stocks have the same value? 2 4 $ 2 month s Stock A Stock B x 2 – 6x + 8 = x + 2 x 2 – 6x – x = 0 x 2 – 7x + 6 = 0 (x – 1)(x – 6) = 0 x – 1 = 0 OR x – 6 = 0 x = 1 OR x = 6 S = { (1,1.5), (6,4)} Equation for Stock A: (0,3); Vertex (4,2) y = a(x – h) 2 + k y = a(x – 4) = a(0 – 4) – 2 = 16a

2 4 $ 2 month s Stock A Stock B Equation for Stock B: (0,2) b = 2; (12,3.5) = x 2 - 8x + 48 = 2x + 32 x 2 - 8x – 2x + 48 – 32 = 0 x x + 16 = 0 (x – 2)(x – 8) = 0 x – 2 = 0 OR x – 8 = 0 x = 2 OR x = 8 These 2 stocks have the same value after 2 months when they both cost $2.25 each and they also have the same value after 8 months when they both cost $3 each.

I IIIIV II 3. From the functions defined below, choose the graph that can represent the result of the given operation.. a) f g f(x) = m 1 x + b when m 1 0 g(x) = m 2 x when m 2 < 0 b) f 2 f 1 f 1 (x) = mx + b 1 when m 0 f 2 (x) = b 2 when b 2 = -b 1 III IIIIV Let m 1 = -1 and b = 1; f(x) = -x + 1 Let m 2 = -1; g(x) = -x f  g = (-x + 1)(-x) = x 2 – x … a > 0 – parabola opens up The zero for f(x) is 1 and the zero for g(x) is 0. These will also be the zeros for f  g. Let m = -1 and b 1 = 1; f 1 (x) = -x + 1 Let b 2 = -1; f 2 (x) = -1 f 2  f 1 = (-1)(-x + 1) = x - 1 … m > 0 – slope is positive The y-intercept for f 2  f 1 is -1.

4. Calculate the perimeter of the following triangle. 2x + 3y = 6 4x - 3y = 9 x - 6y = 11 A B C 4x – 3y = 9 2x + 3y = 6 6x = 15 x = 2.5 2(2.5) + 3y = y = 6 3y = 1 x – 6y = 11 2x + 3y = 6 4x + 6y = 12 5x = 23 x = 4.6 2(4.6) + 3y = y = 6 3y = x – 3y = 9 -8x + 6y = -18 x – 6y = 11 -7x = -7 x = 1 1 – 6y = 11 -6y = 10 Find A: Find C: Find B: Perimeter = = 8.67 units

5. Calculate the area of the following trapezoid. A (-5,2) B (1,-2) C (1,-7) D (-8,-1) h

6.a) Determine the equation of the median from A. b) Determine the perpendicular bisector of side BC. A (2,8) B (7,-2)C (-7,-2) Side BC is horizontal. Its perpendicular bisector would therefore be vertical. The form of an equation for a vertical line is: x = x 1. Its midpoint is (0,-2). The equation of the perpendicular bisector is: x = x 1 x = 0