Asked by James
                State the equation of a rational function if the vertical asymptote is x = 5, the horizontal asymptote is y = 2, the x-intercept is -1/2 and the y-intercept is -1/5.
            
            
        Answers
                    Answered by
            mathhelper
            
    could be as simple as y = 2x/(x-5) - 1/5
    
                    Answered by
            oobleck
            
    not quite. The above has an x-intercept of x=0
y = (2x+1)/(x-5)
    
y = (2x+1)/(x-5)
                    Answered by
            mathhelper
            
    Thanks oobleck.
Guilty of not reading the whole question.
Totally missed the x-intercept part
    
Guilty of not reading the whole question.
Totally missed the x-intercept part
                    Answered by
            Bosnian
            
    Detailed explanation.
Rational function can be written in the form:
f(x) = P(x) / Q(x)
Vertical Asymptote is point where denominator is equal zero.
Vertical asymptote at x = 5
This requires a factor of ( x - 5 ) in the denominator because:
For x = 5
x - 5 = 0
Horizontlal Asymptote is point where f(x) = 0
This requires a factor of 2 x in the numerator because:
For x = 2 , f(x) = 0
2 / ( x - 5 )
The x-intercept is the point at which the graph crosses the x-axis.
At this point, the f(x) is zero.
The x-intercept is - 1 / 2
This requires a factor of ( x + 1 / 2 ) in the numerator because:
For x = - 1 / 2 , ( x + 1 / 2 ) = 0
2 ( x + 1 / 2 ) / ( x - 5 )
( 2 x + 1 ) / ( x - 5 )
The degrees of the numerator and denominator are the same at this point.
You just need to add constant factors to make the ratio of the leading coefficients equal to 2 / 1.
The ratio of leading coefficients is already 2 / 1
So:
f(x) = ( 2 x + 1 ) / ( x - 5 )
    
Rational function can be written in the form:
f(x) = P(x) / Q(x)
Vertical Asymptote is point where denominator is equal zero.
Vertical asymptote at x = 5
This requires a factor of ( x - 5 ) in the denominator because:
For x = 5
x - 5 = 0
Horizontlal Asymptote is point where f(x) = 0
This requires a factor of 2 x in the numerator because:
For x = 2 , f(x) = 0
2 / ( x - 5 )
The x-intercept is the point at which the graph crosses the x-axis.
At this point, the f(x) is zero.
The x-intercept is - 1 / 2
This requires a factor of ( x + 1 / 2 ) in the numerator because:
For x = - 1 / 2 , ( x + 1 / 2 ) = 0
2 ( x + 1 / 2 ) / ( x - 5 )
( 2 x + 1 ) / ( x - 5 )
The degrees of the numerator and denominator are the same at this point.
You just need to add constant factors to make the ratio of the leading coefficients equal to 2 / 1.
The ratio of leading coefficients is already 2 / 1
So:
f(x) = ( 2 x + 1 ) / ( x - 5 )
                    Answered by
            oobleck
            
    Extra credit: Luckily, this simple function has a y-intercept of -1/5 as required.
What could you do to change the y-intercept to, say, y = -1?
    
What could you do to change the y-intercept to, say, y = -1?
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.