Finding Equation of Rational Function

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    Function Rational
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To find the equation of a rational function with specific characteristics, it is established that there must be a factor of (x-5) in both the numerator and denominator due to the hole at x=5, and an x in the denominator for the vertical asymptote at x=0. The requirement for a horizontal asymptote at g(x) = 2 indicates that the degrees of the numerator and denominator must be equal, necessitating a leading coefficient of 2 in the numerator. Confusion arises regarding the y-intercept, which is clarified as the provided point (-1.5, 0) is actually an x-intercept, indicating a flaw in the original problem statement. The discussion concludes that the problem needs correction for accurate interpretation.
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Homework Statement


Must find equation meeting these requirements:
1. Hole at x=5
2. Vertical Asymptote at x= 0
3. Horiztonal Asymptote at g(x) = 2
4. Y-int at (-1.5, 0)

Homework Equations

The Attempt at a Solution


I know that there must be an (x-5) in both the numerator and denominator due to the hole in the graph. I know that there is an x in the denominator due to the vertical asymptote. And since the degree's have to be even (from top and bottom numerator and denominator) due to the the horizontal asymptote, there has to be a value of 2 in the numerator. Howevere now I am confused where the y-int comes into th equation?
 
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Veronica_Oles said:

Homework Statement


Must find equation meeting these requirements:
1. Hole at x=5
2. Vertical Asymptote at x= 0
3. Horiztonal Asymptote at g(x) = 2
4. Y-int at (-1.5, 0)
There is either an error in the problem you were given or you have copied it incorrectly. The point (-1.5, 0) is an x-intercept, not a y-intercept.
Veronica_Oles said:

Homework Equations

The Attempt at a Solution


I know that there must be an (x-5) in both the numerator and denominator due to the hole in the graph. I know that there is an x in the denominator due to the vertical asymptote. And since the degree's have to be even (from top and bottom numerator and denominator) due to the the horizontal asymptote, there has to be a value of 2 in the numerator. Howevere now I am confused where the y-int comes into th equation?
 
Mark44 said:
There is either an error in the problem you were given or you have copied it incorrectly. The point (-1.5, 0) is an x-intercept, not a y-intercept.
Okay that's good to know. The problem given was written like how I wrote it, meaning problem is flawed.
 

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