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Math 116.04 - Quiz 4
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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.
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(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 .