Transcript Document

Exercises
Graph the following functions. Solutions will follow the exercises.
1
1. f ( x ) 
x3
2. f ( x) 
1
x3
x2 1
7. y 
x2
8. p ( x ) 
3
x2  4
x
x2 1
3. y 
4x
2x  1
9. f ( x ) 
4. y 
4x
1  2x
x 2  2x  8
10. g ( x)  2
x  2 x  15
5. g ( x) 
3x  3
x4
x 2  x  20
6. h ( x ) 
x5
11. y 
x 1
x2 1
12. q ( x) 
6x  2
4x  5
Exercise Solutions
The following graph solutions are done in blue.
The asymptotes are done in red.
1
1. f ( x) 
x3
1
2. f ( x) 
x3
Notice the difference in the equations of 1 and 2 and then
compare the difference in the graphs of 1 and 2.
3.
y
4x
2x  1
4x
4. y 
1  2x
Again notice the difference in the equations of 3 and 4 and then
compare the difference in the graphs of 3 and 4.
3x  3
5. g ( x) 
x4
x 2  x  20
6. h( x) 
x5
Complete solution at the end.
(5, 9)
x2 1
7. y  2
x
3
8. p( x)  2
x 4
9.
f ( x) 
x
x2 1
Complete solution at the end.
x 2  2x  8
10. g ( x)  2
x  2 x  15
Complete solution at the end.
11. y  x2  1
x 1
(1, .5)
6x  2
12. q ( x) 
4x  5
Complete solutions for problems 6, 9, and 10,
x 2  x  20
6. h( x) 
x5
h( x ) 
( x  5)( x  4)
x5
h(x)  x  4, x  5
First we need to ask ourselves, “Does
the function reduce?”
Yes, in fact, it reduces to a linear
function which is restricted at x = 5.
This means that when x = 5, there is
a hole in the graph.
x
y
5 9
0
4
hole
x
9. f ( x )  2
x 1
First we need to ask ourselves, “Does the function reduce?”
No!
Next we need to graph the horizontal and vertical asymptotes.
Looking at the denominator, if we set it equal to 0, we get imaginary
values for x. This means that there is no vertical asymptote. We can
also see that since x2 is always positive and adding 1 to x2 is always
positive, the denominator is always positive and can never be equal to
zero.
Since the numerator degree is smaller than the denominator degree,
the horizontal asymptote is y = 0. This line is shown in red.
Sometimes the function
crosses the horizontal
asymptote.
Huh, something new!
So, after asymptotes are
drawn, we need to check
and see if the function
crosses the horizontal
asymptote
The horizontal asymptote is defined by the line y = 0. I will
call the number 0, the “horizontal asymptote value” for this
function.
Any line or curve which crosses this red line will have a y value
of zero at the point of crossing.
All we need to do is find the x value that goes with this y value of
0. If we set the function equal to 0 and solve for x, we will have
the x value which corresponds to the y value of 0.
x
Multiplying both sides by the LCD,

0
we get x = 0
x2 1
Therefore, the point, (0, 0) is on the function and is also on the horizontal
asymptote. Hence the function crosses the asymptote at (0, 0).
When a function does not cross the horizontal asymptote, the above
process will yield a false statement, meaning there is no
solution…there is no x value that corresponds to a y value of 0.
The point of crossing has
been emphasized with the
red dot.
Since there are no vertical
asymptotes, we will find a
few points to the left of this
point of crossing and a few
points to the right: (-3, -.3),
(-2, -.4), (-1, -.5), (-.5, -.4),
(.5, .4), (1, .5), (2, .4), (3, .3).
Since y = 0 is an asymptote, the function must approach it and the
only point of crossing was found to be (0,0). Connecting the dots
and approaching the asymptote on both ends gives us the graph of
the function.
x 2  2x  8
10. g ( x)  2
x  2 x  15
g ( x) 
( x  4)(x  2)
( x  5)(x  3)
Vertical Asymptotes: x = -5
and x = 3
Horizontal Asymptote: y = 1
Does the function cross the
horizontal asymptote?
To find out, set the function
equal to the “horizontal
asymptote value”.
Does the function reduce?
No
x 2  2x  8
1
2
x  2 x  15
Multiply both sides by the LCD and solve.
x2 – 2x – 8 = x2 + 2x – 15
-4x = -7
The function crosses the “H. A.” at
(1.75, 1).
x = 1.75
Now we must plot at least
one point on each side of
both Vertical Asymptotes and
one point on each side of the
point of crossing.
 10,1.75
2.5,1.9
 3,  .5
These points show us which
quadrant the function lies for
each section of the graph
defined by the “V. A.s”
3.9, 0
With our knowledge of how functions behave close to their
asymptotes and the fact that the function crosses the horizontal
asymptote in only one point, we can now draw the function. If
you are uncomfortable doing this with only one point in each
section, you can, of course, plot more points.
Note: Functions never
cross vertical asymptotes.