Math 1090
Quiz 4
March 9, 2015
Answer the following questions in the space provided. The value of every
question is indicated at the beginning. Time: 12 minutes.
Name:
UID:
1. (10 points) For this piecewise function

 −x2 + 2, x < −1
|x|,
−1 ≤ x < 2
f (x) =

−x2 + 2, x ≥ 2
(a) Fill in the table
x
−2 −1 0 1 2
f (x) −2 1 0 1 −2
(b) Sketch the graph of f .
2. Consider the function f (x) =
2+x
1−x
(a) What are (if any) the vertical asymptotes?
Solution. The vertical asymptotes are determined by setting the denominator
equal to zero. In this case, we see that there is only one asymptote: the line x = 1.
(b) For every vertical asymptote x = a, find the limits limx→a f (x) and lima←x f (x).
Solution. When x = 1, we have f (1) = 30 , so the function blows up to ±∞ and
we need to determine the sign. Note that
lim
x→1
+
2+x
=
= +∞,
1−x
+
lim
1←x
2+x
+
=
= −∞
1−x
−
What we mean by this is the following: as x approaches 1 from the left (think of
' +2.9
> 0. On the other hand, as x approaches 1 from the right
x = 0.9) then 2+x
1−x
+0.1
2+x
+3.1
(think of x = 1.1) then 1−x = −0.1
< 0.
(c) What are, if any, the horizontal asymptotes?
Solution. The horizontal asymptotes are horizontal lines to which the function
approaches as |x| grows. For this we compute
x
2+x
= lim
= lim −1 = −1
x→±∞ −x
x→±∞
x→±∞ 1 − x
lim
so as |x| grows, the function approaches the horizontal line y = −1.
(d) Find the x-intercepts and the y-intercept.
Solution. The y-intercept is obtained by setting x = 0, namely f (0) = 2+0
= 2, so
1−0
it is the point (0, 2). In order to find the x-intercepts, we need to solve the equation
2+x
= 0 =⇒ x = −2, so there is only one x-intercept (−2, 0).
1−x
(e) Sketch the graph of the function
Solution. Using all the information above we obtain the following graph.
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