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Math 116.04 - Quiz 4 Name: No notes. Please show enough work to demonstrate your reasoning. (1) Consider the quadratic function f (x) = 2x2 − 12x − 6. (a) Find the vertex of the function by completing the square. (Just using the vertex formula will not receive full credit.) f (x) = 2x2 − 12x − 6 = 2(x2 − 6x − 3) = 2(x2 − 6x + 9 − 9 − 3) (recall that we add and subtract 9 because (−6/2)2 = 9,) = 2((x − 3)2 − 12) = 2(x − 3)2 − 24. So the vertex of the quadratic is at (3, −24). (b) Is the vertex of this function a minimum or a maximum? Explain how you know your answer is correct. Since the coefficient on the x2 term is positive, this quadratic opens up. So the vertex is a minimum. (MORE ON BACK) 1 2 (2) Solve the system of linear equations 2x − 3y + 4z −x + 2y −3x + y = 12 =4 = 2. Taking just the last two equations, we have a system with two variables. We can solve it for x and y, then use the first equation to get z. So start by solving the second equation for x: x = 2y − 4. Now substitute this into the third equation to eliminate x, and solve for y: −3x + y = 2 −3(2y − 4) + y −6y + 12 + y −5y y =2 =2 = −10 = 2. Now that we know y, we can find x using the equation above: x = 2y − 4 = 2(2) − 4 = 0 And finally we can use x and y to find z: 2x − 3y + 4z = 12 2(0) − 3(2) + 4z = 12 −6 + 4z = 12 4z = 18 9 z= . 2 So there is a single solution to the system, x = 0, y = 2, z = 92 .