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2
2
3x -2x -1f(x)=
2x - x -1
Graphing Rational FunctionsExample #2
END SHOW Slide #1 Next
We want to graph this rational function showing all relevant characteristics.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1)=
(2x+1)(x -1)
Graphing Rational FunctionsExample #2
Previous Slide #2 Next
First we must factor both numerator and denominator, but don’t reduce the fraction yet.
Both factor into 2 binomials.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
Graphing Rational FunctionsExample #2
Previous Slide #3 Next
Note the domain restrictions, where the denominator is 0.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
Graphing Rational FunctionsExample #2
Previous Slide #4 Next
Now reduce the fraction. In this case, we cancel the common factor of(x-1) in both the numerator and the denominator.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
Graphing Rational FunctionsExample #2
Previous Slide #5 Next
Any places where the reduced form is undefined, the denominator is 0, forms a vertical asymptote. Remember to give the V. A. as the full equation
of the line and to graph it as a dashed line.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
Graphing Rational FunctionsExample #2
Previous Slide #6 Next
Any values of x that are not in the domain of the function but are not a V.A. form holes in the graph. In other words, any factor that reduced completely
out of the denominator would create a hole in the graph where it is 0.Thus, there is a hole at 1. From the reduced form, y=(3·1+1)/(2·1+1)=4/3.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
( x +1)3=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
H.A.: y = 32
Graphing Rational FunctionsExample #2
Previous Slide #7 Next
Next look at the degrees of both the numerator and the denominator. Because both the denominator's and the numerator's degrees are the
same, 2, there will be a horizontal asymptote at y=(the ratio of the leading coefficients) and there is no oblique asymptote.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
H.A.: y = 32
(3x +1)(2x+1)
=32
6x +2=6x +3 2=3
Graphing Rational FunctionsExample #2
Previous Slide #8 Next
Next we need to find where the graph of f(x) would intersect the H.A. To do this we set the reduced form equal to the number from the H.A., and solve
for x.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
3H.A.: y =
2
(3x +1) 3=
(2x+1) 2
6x +2=6x +3 2=3
No I nt. w/ H.A.
Graphing Rational FunctionsExample #2
Previous Slide #9 Next
In this case, solving the equation, led to a statement that is always false. Thus, there are no values of x where the 2 graph intersect. Hence, there
are no intersections between the graph of f(x) and the H.A.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
3H.A.: y =
2
(3x +1) 3=
(2x+1) 2
6x +2=6x +3 2=3
No I nt. w/ H.A. 1
x - int=-3
Graphing Rational FunctionsExample #2
Previous Slide #10 Next
We find the x-intercepts by solving when the function is 0, which would be when the numerator is 0. Thus, when 3x+1=0.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
3H.A.: y =
2
(3x +1) 3=
(2x+1) 2
6x +2=6x +3 2=3
No I nt. w/ H.A. 1
x - int=-3
0+0-1y - int= =1
0-1
Graphing Rational FunctionsExample #2
Previous Slide #11 Next
Now find the y-intercept by plugging in 0 for x.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
3H.A.: y =
2
(3x +1) 3=
(2x+1) 2
6x +2=6x +3 2=3
No I nt. w/ H.A. 1
x - int=-3
0+0-1y - int= =1
0-1
x=-1
-3+1 -2
y= = =2-2+1 -2
Graphing Rational FunctionsExample #2
Previous Slide #12 Next
Plot any additional points needed.Here I only plotted one more point at x=-1 since a point hadn't been plotted
to the left of the V.A. You can always choose to plot more points than required to help you find the graph.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
3H.A.: y =
2
(3x +1) 3=
(2x+1) 2
6x +2=6x +3 2=3
No I nt. w/ H.A. 1
x - int=-3
0+0-1y - int= =1
0-1
x=-1
-3+1 -2
y= = =2-2+1 -2
Graphing Rational FunctionsExample #2
Previous Slide #13 Next
Finally draw in the curve.For the part to the right of the V.A., we use that it can't cross the H.A. and it
has to approach the V.A. and the H.A.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
3H.A.: y =
2
(3x +1) 3=
(2x+1) 2
6x +2=6x +3 2=3
No I nt. w/ H.A. 1
x - int=-3
0+0-1y - int= =1
0-1
x=-1
-3+1 -2
y= = =2-2+1 -2
Graphing Rational FunctionsExample #2
Previous Slide #14 Next
For the part to the left of the V.A., we use that it can't cross the H.A. and it has to approach the V.A. and the H.A.
2
2
3x -2x -1f(x)=
2x - x -1
(3x +1)(x -1) 1
= ;x 1,-(2x+1)(x -1) 2
(3x +1)=
(2x+1)
1V.A.: x = -
2
4Hole at 1,
3
3H.A.: y =
2
(3x +1) 3=
(2x+1) 2
6x +2=6x +3 2=3
No I nt. w/ H.A. 1
x - int=-3
0+0-1y - int= =1
0-1
x=-1
-3+1 -2
y= = =2-2+1 -2
Graphing Rational FunctionsExample #2
Previous Slide #15 END SHOW
This finishes the graph.