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9.3 Graphing General Rational Functions

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Factor the following completely:
1. 3x2-8x+4
(3x-2)(x-2)
2. 11x2-99
4. x3+2x2-4x-8
(x-2)(x+2)2
5. 2x2-x-15
11(x+3)(x-3)
(2x+5)(x-3)
3. 16x3+128
6. 10x3-80
16(x+2)(x2-2x+4)
10(x-2)(x2+2x+4)
5.3 Graph General Rational
Functions
p. 319
What information can you get from the
equation of a rational graph?
What is local minimum?
Yesterday, we graphed rational
functions where x was to the first
power only.
What if x is not to the first power?
Such as:
f ( x) пЂЅ
x
x пЂ«1
2
Steps to graph when x is not to the 1st power
1. Find the x-intercepts. (Set numer. =0 and solve)
2. Find vertical asymptote(s). (set denom=0 and solve)
3. Find horizontal asymptote. 3 cases:
a. If degree of top < degree of bottom, y=0
lead. coeff. of top
y
пЂЅ
b. If degrees are =,
lead. coeff. of bottom
c. If degree of top > degree of bottom, no horiz.
asymp, but there will be a slant asymptote.
4. Make a T-chart: choose x-values on either side &
between all vertical asymptotes.
5. Graph asymptotes, pts., and connect with curves.
6. Check solutions on calculator.
Graph y = x2 6+ 1 . State the domain and range.
SOLUTION
The numerator has no zeros,
so there is no x-intercept.The
denominator has no real
zeros, so there is no vertical
asymptote.
The degree of the numerator, 0, is less than
the degree of the denominator, 2. So, the line
y = 0 (the x-axis) is a horizontal asymptote.
The graph passes through the points (–3, 0.6), (–1,
3), (0, 6), (1, 3), and (3, 0.6). The domain is all real
numbers, and the range is 0 < y ≤ 6.
Ex: Graph. State domain & range.
yпЂЅ
x
x пЂ«1
2
1. x-intercepts: x=0
2. vert. asymp.: x2+1=0
x2= -1
No vert asymp x пЂЅ пЂ­ 1
(No real solns.)
4. x
y
-2
-.4
-1
-.5
0
0
1
.5
3. horiz. asymp:
2
1<2
(deg. of top < deg. of bottom)
y=0
.4
Domain: all real #’s except -2 & 2
Range: all real #’s except 0<y<3
2
2x
Graph y = x2– 9 .
SOLUTION
The zero of the numerator 2x2 is 0, so 0 is an xintercept. The zeros of the denominator x2–9 are
+3, so x = 3 and x = –3 are vertical asymptotes.
The numerator and denominator have the same
degree, so the horizontal asymptote is y =am = 2 = 2
1
bn
Plot points between and beyond the vertical
asymptotes.
Ex: Graph, then state the domain and range.
yпЂЅ
3x
2
x пЂ­4
2
1. x-intercepts:
4. x
3x2=0
4
x2=0
x=0
3
2. Vert asymp:
1
x2-4=0
0
x2=4
-1
x=2 & x=-2
-3
3. Horiz asymp:
-4
(degrees are =)
y=3/1 or y=3
y
4
5.4
On right of x=2
asymp.
-1
0
Between the 2
asymp.
-1
5.4
4
On left of x=-2
asymp.
Domain: all real numbers
Range:
пЂ­1
2
п‚Ј yп‚Ј
1
2
2 +3x – 4
x
Graph y = x – 2
.
SOLUTION
The numerator factors as (x + 4)(x –1), so the xintercepts are – 4 and 1. The zero of the
denominator x – 2 is 2, so x = 2 is a vertical
asymptote.
The degree of the numerator, 2, is greater than
the degree of the denominator, 1,so the graph
has no horizontal asymptote. The graph has the
same end behavior as the graph of y = x2–1 = x.
Plot points on each side of the vertical
asymptote.
Ex: Graph, then state the domain & range.
x пЂ­ 3x пЂ­ 4
2
yпЂЅ
1. x-intercepts:
xпЂ­2
x2-3x-4=0
4. x
y
(x-4)(x+1)=0
-1
0
x-4=0 x+1=0
Left of x=2
x=4 x=-1
0
2
asymp.
2. Vert asymp:
1
6
x-2=0
3
-4
Right of
x=2
x=2
asymp.
4
0
3. Horiz asymp: 2>1
(deg. of top > deg. of bottom)
no horizontal asymptotes, but there is a slant!
Slant asymptotes
• Do synthetic division (if possible); if not, do long
division!
• The resulting polynomial (ignoring the
remainder) is the equation of the slant
asymptote.
Ignore the remainder,
In our example:
use what is left for the
2 1 -3 -4
equation of the slant
2 -2
asymptote: y=x-1
1 -1 -6
Domain: all real #’s except 2
Range: all real #’s
Manufacturing
A food manufacturer wants to
find the most efficient
packaging for a can of soup
with a volume of 342 cubic
centimeters. Find the
dimensions of the can that
has this volume and uses the
least amount of material
possible.
SOLUTION
STEP 1
Write an equation that gives the height h of the
soup can in terms of its radius r. Use the
formula for the volume of a cylinder and the
fact that the soup can’s volume is 342 cubic
centimeters.
V = r 2h
342 = r2h
342 = h
r2
Formula for volume of cylinder
Substitute 342 for V.
Solve for h.
STEP 2
Write a function that gives the surface area S
of the soup can in terms of only its radius r.
Formula for surface area of cylinder
342
Substitute ПЂ r2 for h.
Simplify.
STEP 3
Graph the function for the surface area S using
a graphing calculator. Then use the minimum
feature to find the minimum value of S.
You get a minimum value
of about 271,which
occurs when r 3.79 and
h ПЂ 342 2 7.58.
(3.79)
ANSWER
So, the soup can using the least amount of material
has a radius of about 3.79 centimeters and a height of
about 7.58 centimeters. Notice that the height and the
diameter are equal for this can.
• What information can you get from the equation
of a rational graph?
X-intercept from setting the numerator=0
Vertical asymptotes from setting the denominator
=0
If m<n, the horizontal asymptote is y = 0
If m=n, the horizontal asymptote is
y =coefficient of numerator/coefficient of
denominator
If m>n, there is no horizontal asymptote
• What is local minimum?
Tells where the most efficient solution is in
application problems.
5.3 Assignment
p. 322
3-6, 4-19 odd
People with graphing calculators
must show a table and plot the
points in the table.
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