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Solutions
Due Date: Wednesday 10-17 IN CLASS!.
In Class Assignment 4
Directions: Show all work to receive full credit. Just answers with no work will be given no credit. DO ALL
SCRATCH WORK ON A SEPERATE PIECE OF PAPER AND ONLY WRITE THE WORK YOU WANT
GRADED ON THIS PAPER.
For question 1 and 2:
(a) Determine the end behavior and state the justification why if horizontal. (explicitly state the horizontal or
non-linear asymptote).
(b) Find the zero’s of the top and bottom by using the rational root theorem (when necessary) and rewriting the
function in factored form.
(c) State the zero’s of the function (if they exist) and their respective multiplicities. (both real and imaginary)
(d) State the vertical asymptotes (if they exist).
(e) State the holes as ordered pairs (x,y)(if they exist).
(f) Find the y-intercept (if it exists).
(g) Sketch a graph on the grid provided making sure to label all the above information.
1. g(x) =
x−5
x2 − 25
(a) E.B.:
Solution. Because the degree of the bottom is greater then the degree of the top we have a horizontal asymptote
of y = 0
(b) Factored Form:
Solution. g(x) =
x−5
x−5
=
2
x − 25
(x + 5)(x − 5)
(c) Zeros
Solution. There are no zeros.
(d) V.A.s:
Solution. x = −5
(e) Holes:
1
1
Solution. There is a hole at x = 5 with a y coordinate of
=
. Thus the ordered pair is given by
5+5
10
)
(
1
5,
10
(f) y-intercept:
Solution. The y − intercept is at g(0) =
1
1
0−5
= or (0, )
0 − 25
5
5
(g) Graph
1
2
2. h(x) =
3x4 − 15x3 + 3x2 + 63x − 54
3x2 − 12
(a) E.B.:
Solution. Since
2
3x − 12
)
x2 − 5x + 5
4
3
we have that the E.B. is given by the quadratic
2
3x − 15x + 3x + 63x − 54
− 3x4
+ 12x2
− 15x3 + 15x2 + 63x
15x3
− 60x
15x2 + 3x − 54
− 15x2
+ 60
3x + 6
2
asymptote of y = x − 5x + 5
(b) Factored Form:
3x4 − 15x3 + 3x2 + 63x − 54
x4 − 5x3 + x2 + 21x − 18
x4 − 5x3 + x2 + 21x − 18
=
=
3x2 − 12
x2 − 4
(x + 2)(x − 2)
and using the type that there is most likely a hole in the graph we will check (-2) and we see
1 −5 1
21 − 18
Solution. Observe that h(x) =
−2
1
−2
14
− 30
18
−7
15
−9
0
1
check (3) and we get 3
1
4
3
Thus x4 − 5x3 + x2 + 21x − 18 = (x + 2)(x3 − 7x2 + 15x − 9) and we can now
−7
15
3
− 12
−4
3
−9
9 and thus x4 −5x3 +x2 +21x−18 = (x+2)(x−3)(x2 −4x+3), which
0
2
we factor to get x −5x +x +21x−18 = (x+2)(x−3)(x−3)(x−1) or x4 −5x3 +x2 +21x−18 = (x+2)(x−3)2 (x−1).
Thus the factor form is given by
(x + 2)(x − 3)2 (x − 1)
h(x) =
(x + 2)(x − 2)
(c) Zeros
Solution. The zeros are x = 3 with a multiplicity of 2, and x = 1 with a multiplicity of 1.
(d) V.A.s:
Solution. The VA is x = 2
(e) Holes:
Solution. The hole is at x = −2 with a y coordinate of
(f) y-intercept:
−54
Solution. The y intercept is h(0) =
or
−12
(
)
9
0,
2
(g) Graph
3
(−5)2 (−3)
75
=
−4
4
4