Download Solutions - WSU Department of Mathematics

Solutions
Practice Exam 1
Definitions and Concepts
Directions: Complete the following definitions by inserting the word/s or mathematical text as indicated
by the labeled underlined blanks. If asked, formulate your own definition in your own words (as long as it is
still correct of course).
1. (2pts each) Let X be the set of all ”x” coordinates and Y be the set of all ”y” coordinates defined by a
relation. We say that x is a function of y if for every (a) coordinate there is only (b) associated (c)
coordinate defined by the relation. In this instance if x is a function of y then the domain (set of all variables
we ”plug in”) is (d) and the range (set of all variable we ”get back”) is (e) .
(a) y
(b) one
(c) x
(d) Y
(e) X
2. (2pts each) Given a relation between y and x, we say y is a function of x if every
graph of the relation no more than (b) time.
(a)
line intersects the
(a) vertical
(b) one
3. (2pts each) A function f(x) is a (a) function if in addition to passing the
since f(x) is a function of x, it also passes the (c) line test.
(a) 1-1 or one to one
(b) vertical
(c) horizontal
1
(b)
line test, which is given
Main Problems
4. Given the circle x2 − 6x + y2 + 14y + 43 = 0
(a) Write the equation for the circle in standard form.
Solution. Using the technique of completing the square we proceed as follows:
x2 − 6x + y2 + 14y + 43 = 0
⇒ x2 − 6x + 9 + y2 + 14y + 49 = 9 + 49 − 43
⇒ (x − 3)2 + (y + 7)2 = 15
(b) What is the center and radius?
Solution. The center is (3, −7) and the radius is
√
15
5. Find the equation of the line perpendicular to the line 5y − 2x = −1 passing through the point (−1, 15)
Solution. First we find the slope we want be taking the negative reciprocal of the slope of the line we were
given (since our line is perpendicular) as follows:
5y − 2x = −1
⇒ 5y = 2x − 1
2
1
⇒y= x−
5
5
5
Thus the slope we want is − . Using the point slope form of a line y − y1 = m(x − x1 ) we get
2
5
y − 15 = − (x + 1)
2
2
6. Use the following graph for parts a, b.
(a) Is y a function of x assuming a domain as shown in the graph above? If yes state a theorem to support
your answer. If not give a counterexample.
Solution. Yes, the graph passes the vertical line test.
(b) Is x a function of y assuming a domain as shown in the graph above? If yes state a theorem to support
your answer. If not give a counterexample.
Solution. No, the graph fails the horizontal line test. Take for a counter example the points (−0.5, 0) and
(0.5, 0)
3
7. Define the piece-wise function f(x) by the graph below
(a) Determine the x and y intercepts
Solution. x-intercepts: {(−6, 0), (0, 0)} y-intercept: (0, 0)
(b) Determine f(−5), and f(0).
Solution. f(−5) = 2 and f(0) = 0
(c) Find the domain and range of f.
Solution. The domain is R or (−∞, ∞) and the range also R or (−∞, ∞)
4
8. Define y = 1 −
√
3 − 2x
(a) Find the domain and range of the function
Solution. We cannot take square roots of negative numbers and get back real numbers thus 3 − 2x ≥ 0,
]
(
3
3
which implies x ≤ and a domain of −∞,
. After graphing the function you would see that the
2
2
range is (−∞, 1] .
[
]
1
(b) Find the A.R.C. (average rate of change) of the function over the interval − , 1 .
2
f(x2 ) − f(x1 )
which if you recall is the slope of the secant
x2 − x1
−1
−1
line through the point (x1 , f(x1 )), (x2 , f(x2 )). Since x1 =
and x2 = 1 we have f(x1 ) = f(
=
2
2
√
√
√
1
1 − 3 + 2 = 1 − 3 + 1 = 1 − 2 = −1 and f(x2 ) = 1 − 3 − 2 = 1 − 1 = 0. Thus the ARC of f over
2
]
[
1
− , 1 is
2
2
0+1
=
1
3
1+
2
Solution. The formula for the A.R.C. is
(c) Identify the parent function g(x)
√
Solution. g(x) = x
(d) Rewrite the function into the form y = Ag(B(x − C)) + D
Solution. Rewriting the function explicitly first we have
√
y = 1 − 3 − 2x
√
= − 3 − 2x + 1
√ (
)
3
= − −2 x −
+1
2
now we use the fact that g(x) =
√
x to get
(
(
3
y = −g −2 x −
2
))
+1
(e) Write out the transformations needed to transform the graph of y = g(x) into the graph of y = Ag(B(x−
C)) + D.
3
Solution. It should be clear from part (d) that A = −1, B = −2, C =
and D = 1. Applying the
2
transformations in the order BCAD we have
i. A horizontal flip and stretch over the y axis by a factor of B = −2.
3
ii. A horizontal shift to the right by C =
2
iii. A vertical flip over the x axis with no vertical stretch since A = −1
iv. A vertical shift up by D = 1
(f) Sketch a graph of y = Ag(B(x − C)) + D.
Solution.
5
6