Finding Equation of Rational Function

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SUMMARY

The discussion focuses on finding the equation of a rational function with specific characteristics: a hole at x=5, a vertical asymptote at x=0, a horizontal asymptote at g(x)=2, and a y-intercept at (-1.5, 0). Participants clarify that the point (-1.5, 0) is an x-intercept rather than a y-intercept, indicating a flaw in the problem statement. The correct formulation involves including (x-5) in both the numerator and denominator, an x in the denominator for the vertical asymptote, and a numerator value of 2 to satisfy the horizontal asymptote condition.

PREREQUISITES
  • Understanding of rational functions and their properties
  • Knowledge of asymptotes and intercepts in graphing
  • Familiarity with polynomial degrees and their implications
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the characteristics of rational functions in detail
  • Learn how to identify and graph vertical and horizontal asymptotes
  • Explore the concept of holes in rational functions
  • Practice solving problems involving intercepts and asymptotes
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Students studying algebra, particularly those focusing on rational functions, as well as educators looking to clarify concepts related to asymptotes and intercepts.

Veronica_Oles
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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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