quadratic function questions

These questions were on the June 2008 final exam. 3) The rule of function f is f (x) = a (x + 7)² + k , with a > 0 and k < 0. Of these 3 statements: 1. Function f has two zeros. 2. Function f has a minimum. 3. Function f is decreasing over the interval ] – ∞ , 7]. which ones are true? Describe or draw the picture: 3) solution: A) 1 and 2 only are true. There are 2 zeros and a minimum. The 3rd statement would be true if the the interval was ] – ∞ , – 7]. 11) A line and a parabola intersect at points A and C. Their equations are:5x – 4y + 48 = 0 and y = 0.25x² – 7x + 41 Line segment AC is the hypotenuse of right triangle ABC, with legs //to the axes. What are the coordinates of point B?A B C . . x y Plan of Action:2008 FINAL EXAM QUESTION ON QUADRATIC FUNCTION11) Solution: We need the x value at A and the y value at C for the coordinates of B. First, find the points of intersection of the 2 curves: (I use ¼ instead of 0.25) From 5x – 4y + 48 = 0, we get y = and set it = ¼ x² – 7x + 41 54 x + 12 Multiply through by 4 and collect terms to get: x² – 33x + 116 = 0 This factors as: (x – 4)(x – 29) = 0.The x-value at A = 4 To find the y value at C, set x = 29 in either equation: (line is easier) y = 54 (29) + 12 = 48.25 The coordinates of B are (4, 48.25) 14) The rule of function g is g(x) = p x² + r x – 36, where p and r are not equal to zero. Function g is positive over the intervals ] − ∞, − 6] 4 [10, +∞[ . What is the value of p? Plan of Action: Soln: We have a parabola opening upwards with zeros at – 6 and 10, so we substitute – 6 and 10 for x and 0 for g(x) and get 2 equations in p and r : 0 = p ( – 6)² + r ( – 6) – 36 , and 0 = p ( 10)² + r (10) – 36 We turn these into: 36 = 36p – 6r which becomes 6 = 6p – r and 36 = 100p + 10r which becomes 3.6 = 10p + r Now we add them together to eliminate “r”: We get 9.6 = 16 p so p = 9.6 ÷ 16 = 0.6 2008 FINAL EXAM QUESTION ON QUADRATIC FUNCTION17) The area of the rectangle is the polynomial 5x² + 38x – 63. The diagonal measures 52 cm. Find the perimeter. 52 cm A = 5x² + 38x – 63 Plan of Action: 18) f (x) = – 2/3 x + 6. V is the vertex of the parabola and the x-intercept of f. We know that f (63) = g(16). Find the rule of correspondence for the parabola. V P f(x) = –2 x + 6 3 f(63) = g(16) g(x) f(x) Plan of Action: what does f (63) = g(16) mean ????? 2008 FINAL EXAM QUESTION ON QUADRATIC FUNCTION17) Solution: 5x² + 38x – 63 = (5x – 7)(x + 9), so we set base b = 5x – 7 and height h = x + 9. Now, b² + h² = 52² which means (5x – 7)² + (x + 9)² = 2704 We expand to get: 25x² – 70x + 49 + x² + 18x + 81 – 2704 = 0 We collect terms: 26x² – 52x – 2574 = 0 (all constants are multiples of 26) We divide by 26: x² – 2x – 99 = 0 and then we factor (x – 11)(x + 9) = 0 This means x = 11, so the base = 48 and the height = 20. Perimeter = 2(b + h) = 2(68) = 136 cm. 18) Solution: f(63) = g(16) means that the y-value on the line at x = 63 equals the y-value on the parabola when x = 16. Since f (x) = – 2/3 x + 6, f(63) = – 42 + 6 = – 36. We now know that (16, – 36) is a point on the parabola. The vertex is the x-intercept of the line so it is (9, 0). Now we find the rule: g(x) = a (x – 9)² + 0 since (h, k) = (9, 0). Now substitute (16, – 36) to find a: – 36 = a (16 – 9)² The rule of correspondence is g(x) = − 36 49 (x − 9)2 2008 FINAL EXAM QUESTION ON QUADRATIC FUNCTION