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MATH 123, EXAM #1TR
Total Points: 100
Show all work for full credit
2-8-2001
CIRCLE FINAL ANSWERS (except graphs). Do all work on blank paper. Do not write anything on this
page except your full name and the last 4 digits of your ID number. Staple this page on top when handing
in.
1. (5 points each) For f (x) = 2x – 1 and g (x) = x 2 – 2x, find:
a) f (0)  g (4)
b)
f (3)
g (2)
= (-1)(8) = -8
is undefined since denominator is 0
2. (6 points for part-a, 5 points each, for the other parts) Write the equation of the line through the two
indicated points. Write the final answer in the form Ax + By = C, with A, B and C as integers.
a) (-2, 6), (4, -3)
3x + 2y = 6
b) (-5, 9), (-5, -5)
x = -5
slope is –1.5
slope is undefined (hence vertical line)
c) Now graph the line from part (a), labeling the coordinates of both intercepts.
See last page
3. (24 points) For the function m(x) = 0.1x 2 – 1.2x – 1 find: (round values to 2 decimal places)
a) (5) The vertex, then write the equation in the form m(x) = a(x – h) 2 + k
Vertex is (6, -4.6)
b) (4) The maximum or minimum value of the function (state which it is)
Min. value is -4.6
c) (4) The range (of the function)
All real greater than or equal to -4.6
d) (6) All intercepts (label each as x-int. or y-int.)
y-int: (0, -1)
x-ints: (-.78,0) &
(12.78,0)
e) (5) Now graph the function, labeling the vertex and all intercepts (with their appropriate coordinates).
See last page
MATH 123, EXAM #1TR
Total Points: 100
Show all work for full credit
4. (5 points each) For the function f ( x) 
a) The domain
5x
find:
x  x  12
2
All reals except -4 and 3 (These values cause denominator to be 0)
Only intercept is (0,0)
b) All intercepts (label each as x-int. or y-int.)
c) All asymptotes (label each as horizontal or vertical)
Vertical: x = -4 and x = 3
Horizontal: y = 0
d) Now sketch a graph of the function, labeling all intercepts and asymptotes.
5. (5 points each) Solve for x. (Must show work for full credit. Trial and error gets very little credit!)
a) 2 ln( x  1)  ln( x 2  5)
b) 9 x 1  31 x
c) 2 x 2 e x  3xe x
d) log x 8  3
e) 28  7(5 x )
(x – 1) 2 = x 2 – 5
Rewrite as (3 2)(x – 1) = 3 1+x
x=3
2(x – 1) = 1 + x or x = 3
Since e x  0, divide both sides by e x and solve to get:
x = 0 or x = 1.5
Rewrite as x –3 = 8. And since (1/2) –3 = 2 3 = 8, x = 0.5
4 = 5x
x=
log 4
log 5
= 0.8614
MATH 123, EXAM #1TR
Total Points: 100
Show all work for full credit
f) ln( x  3)  ln x  2 ln 2
(x + 3)/x = 2 2
x + 3 = 4x
x=1
2c)
Graph of 3x + 2y = 6
Note: Intercepts are (2,0) and (0,3)
3e)
Graph of m(x) = 0.1x 2 – 1.2x – 1
See first page for coordinates of vertex
and intercepts