Graphing Rational Functions

[xxMsJojoxx]

How do I find the equation for this rational function?


I know the x intercept is at 2 and -3, so that should make the numerator (x-2)(+3).

Also, the vertical asymptote is 1, so maybe the denominator is (x-1)

So that would make the equation

But the answer is , so I"m not sure how to interpret 1/3 coefficient and the 3x-1 denominator part from the graph.

 

[skeeter]

you have the graph (small enough?) ... I assume you're looking for an equation

x-intercepts look like x = 2 and x = -3, vertical asymptote at x = 1 ...

[MATH]y = k \cdot \dfrac{(x+3)(x-2)}{x-1}[/MATH]
y-intercept looks like y = 2

[MATH]2 = k \cdot \dfrac{(0+3)(0-2)}{0-1} \implies 2 = k(6) \implies k = \dfrac{1}{3}[/MATH]
[MATH]y = \dfrac{1}{3} \cdot \dfrac{(x+3)(x-2)}{x-1}[/MATH]
I don't agree with the denominator of the given answer ... if true, the vertical asymptote would be x = 1/3

 

[Steven G]

1st of all you asked how to graph the function. Obviously that is not what you meant to ask as the graph was given to you. Based on your work and supplied solution you want to find the equation of the given graph.

The way you got the numerator was fine. Your logic for finding the denominator was also correct. I too think that the vertical asymptote is at x=1. The author for some reason thinks that the asymptote is at x=1/3. So x-1/3 can be a factor or multiplying by 3 you can also get (3x-1) as a factor for the denominator. That is what the author used. Again, it should be x-1 !

Now where did the 1/3 in front come from. Well you did not take into account for the y-intercept. The y-intercept for your function is 6 when it should be 2. So how do you handle that?

 

[xxMsJojoxx]

you have the graph (small enough?) ... I assume you're looking for an equation

x-intercepts look like x = 2 and x = -3, vertical asymptote at x = 1 ...

[MATH]y = k \cdot \dfrac{(x+3)(x-2)}{x-1}[/MATH]
y-intercept looks like y = 2

[MATH]2 = k \cdot \dfrac{(0+3)(0-2)}{0-1} \implies 2 = k(6) \implies k = \dfrac{1}{3}[/MATH]
[MATH]y = \dfrac{1}{3} \cdot \dfrac{(x+3)(x-2)}{x-1}[/MATH]
I don't agree with the denominator of the given answer ... if true, the vertical asymptote would be x = 1/3

Thank you for showing me how standard formula with k as coefficient, and showing me how to identify the y-intercept on the equation

you have the graph (small enough?) ... I assume you're looking for an equation

x-intercepts look like x = 2 and x = -3, vertical asymptote at x = 1 ...

[MATH]y = k \cdot \dfrac{(x+3)(x-2)}{x-1}[/MATH]
y-intercept looks like y = 2

[MATH]2 = k \cdot \dfrac{(0+3)(0-2)}{0-1} \implies 2 = k(6) \implies k = \dfrac{1}{3}[/MATH]
[MATH]y = \dfrac{1}{3} \cdot \dfrac{(x+3)(x-2)}{x-1}[/MATH]
I don't agree with the denominator of the given answer ... if true, the vertical asymptote would be x = 1/3

Hi Skeeter. Thanks for showing me the standard formula with k as the cofficient, and explaining how to use the y-intercept to identify the formula!