Graphing Rational Functions Example #7 End ShowEnd Show Slide #1 NextNext We want to graph this rational function showing all relevant characteristics

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


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



Note the domain restrictions, where the denominator is 0.

2-3x +x+4f(x)=

x

-(3x - 4)(x+1)

=x

; x 0

x

y



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



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



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



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



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



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


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



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


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



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


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



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


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



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


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



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


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



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


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


Previous Slide #16 End Show

This finishes the graph.

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Graphing Rational Functions Example #7 End ShowEnd Show Slide #1 NextNext We want to graph this rational function showing all relevant characteristics