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Unformatted text preview: Solutions to Quiz 15A
www.math.uﬂ.edu/˜harringt
November 17, 2007
1. Let f (x) = −x2 + x + 6
(a) Find the axis of symmetry.
x= −1
1
−b
=
=
2a
2(−1)
2 Note: This is a vertical line x = 1/2.
(b) Find the range of f . (Use interval notation.)
We will ﬁrst observe that the function has a local max since the
leading term of f is −x2 .
Secondly, we need to ﬁnd the largest value of f i.e. we need to
look at the y value of the vertex. So we will consider f (1/2).
12 1
1 2 24
25
1
=−
=
+ +6=− + +
2
2
2
44
4
4
Thus, the interval is (−∞, 25 ].
4
f Notice that we include the point 25
4 in the interval. (c) Find the roots of f .
Here we want to solve f (x) = 0.
−x2 + x + 6 = 0
x2 − x − 6 = 0
(x − 3)(x + 2) = 0 So x = 3 and x = −2.
1 2. Determine validity of the following statements.
(a) The graph of f (x) = 4 − (x − 2)2 has a local max. This statement is
TRUE. Consider rewriting the function as f (x) = −(x − 2)2 + 4.
So the leading term is −x2 and thus it has a local max.
(b) If the discriminant b2 − 4ac = 0, the graph of f (x) = ax2 + bx + c,
(a = 0) will touch the xaxis at its vertex. This statement is
TRUE.
Let’s see why. First, we will compute the xintercept (or commonly
called the root, or zero).
√
−b ± b2 − 4ac
x=
2
√a
−b ± 0
x=
(Since the discriminant is zero)
2a
−b
x=
2a
b
So the xint is ( −a , 0). So the vertex touches the x axis.
2 3. Find the end behavior of (x + 1)2 (x − 1)3 (3 − x)4 (2 − x)2 .
We will ﬁrst need to compute the leading term. The leading term
is:
(x)2 (x)3 (−x)4 (−x)2 = x2 x3 x4 x2 = x11
Note: The negatives cancel each other since the power of x is even.
So our end behaviors look like: Disclaimer for number 3: The arrow lines are striaght and together.
Please note that this is NOT correct. This is just to give an idea
of where the arrows should point. This is a limitation to typing the
solutions. 2 ...
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This note was uploaded on 07/12/2011 for the course MAC 1140 taught by Professor Williamson during the Fall '08 term at University of Florida.
 Fall '08
 WILLIAMSON
 Math, Calculus, Algebra

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