graphing rational functions example #7 end showend show slide #1 nextnext we want to graph this...
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2-3x +x+4f(x)=
x
x
y
Graphing Rational FunctionsExample #7
End Show Slide #1 Next
We want to graph this rational function showing all relevant characteristics.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
x
y
Graphing Rational FunctionsExample #7
Previous Slide #2 Next
First we must factor both numerator and denominator, but don’t reduce the fraction yet.
Numerator: Factor out the negative. Then factor to 2 binomials.Denominator: Prime
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #3 Next
Note the domain restrictions, where the denominator is 0.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #4 Next
Now reduce the fraction. In this case, there isn't a common factor. Thus, it doesn't reduce.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0
x
y
Graphing Rational FunctionsExample #7
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.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles
x
y
Graphing Rational FunctionsExample #7
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.Since this example didn't reduce, it has no holes.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles
x
y
Graphing Rational FunctionsExample #7
Previous Slide #7 Next
Next look at the degrees of both the numerator and the denominator. Because the denominator's degree,1, is exactly 1 less than the
numerator's degree,2, there will be an oblique asymptote, but no horizontal asymptote.
2-3x +f(x)=
x+4x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles
2 4=-3
-3x +x+
x 4x
1++
x
x
y
Graphing Rational FunctionsExample #7
Previous Slide #8 Next
To find the O.A. we must divide out the rational expression. In this case, since the fraction didn't reduce we will use the original form. Also, since the denominator is a monomial, I'll divide each term in the numerator by the x.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
2-3x +x+4 4=-3x+1+
x x
x
y
Graphing Rational FunctionsExample #7
Previous Slide #9 Next
The O.A. will be y=(the polynomial part of the division).
2-3x +f(x)=
x+4x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
No int.w/ O.A.
2
=--3x
3x++x+4x
1+4x
2
=
4-3x+
-3
1+ =-3x+1x
-3x +
4
x+
=0x
4+1
xx
4=0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #10 Next
Now we need to find the intersections between the graph of f(x) and the O.A. The more formal way to find this is to set the function equal to the O.A., but you see this reduces to when the remainder is 0. Since, 4 is
never 0, there are no intersections w/ the O.A.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
No int.w/ O.A. 4
x - int.= , -13
2-3x +x+4 4=-3x+1+
x x
2-3x +x+4
=-3x+1x
4-3x+1+ =-3x+1
x4
=0x4=0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #11 Next
We find the x-intercepts by solving when the function is 0, which would be when the numerator is 0. Thus, when 3x-4=0 and x+1=0.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
No int.w/ O.A. 4
x - int.= , -13
No y- int.
2-3x +x+4 4=-3x+1+
x x
2-3x +x+4
=-3x+1x
4-3x+1+ =-3x+1
x4
=0x4=0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #12 Next
Now find the y-intercept by plugging in 0 for x. In this case, there wouldn't be a y-intercept since the function is undefined at x=0.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
No int.w/ O.A. 4
x - int.= , -13
No y- int.
2-3x +x+4 4=-3x+1+
x x
2-3x +x+4
=-3x+1x
4-3x+1+ =-3x+1
x4
=0x4=0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #13 Next
Plot any additional points needed.In this case we don't need any other points to determine the graph.
Though, you can always plot more points if you want to.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
No int.w/ O.A. 4
x - int.= , -13
No y- int.
2-3x +x+4 4=-3x+1+
x x
2-3x +x+4
=-3x+1x
4-3x+1+ =-3x+1
x4
=0x4=0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #14 Next
Finally draw in the curve.Let's start on the interval for x<0, the graph has to pass through the point (-1,0) and approach both asymptotes without crossing the O.A. or the x-
axis again.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
No int.w/ O.A. 4
x - int.= , -13
No y- int.
2-3x +x+4 4=-3x+1+
x x
2-3x +x+4
=-3x+1x
4-3x+1+ =-3x+1
x4
=0x4=0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #15 Next
For the interval for x>0, the graph has to pass through the point (4/3,0) and approach both asymptotes without crossing the O.A. or the x-axis
again.
2-3x +x+4f(x)=
x
-(3x - 4)(x+1)
=x
; x 0
V.A.:x=0 NoHoles O.A.:y=-3x+1
No int.w/ O.A. 4
x - int.= , -13
No y- int.
2-3x +x+4 4=-3x+1+
x x
2-3x +x+4
=-3x+1x
4-3x+1+ =-3x+1
x4
=0x4=0
x
y
Graphing Rational FunctionsExample #7
Previous Slide #16 End Show
This finishes the graph.