Finding Constants for Rational Functions with Specific Vertical Asymptotes

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Homework Help Overview

The discussion revolves around finding constants 'a' and 'b' in the rational function \( y=\frac{ax^2+7}{9-bx^2} \) such that the vertical asymptotes are located at \( \pm \frac{3}{5} \). Participants explore the implications of the function's denominator and how it relates to the specified asymptotes.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the requirement for the denominator to equal zero at the specified asymptotes and explore how to derive 'b' from this condition. There are attempts to rearrange the function and substitute values to find constants. Some express confusion about the implications of changing signs in the equation and the role of 'a' in relation to vertical asymptotes.

Discussion Status

The discussion is active, with various interpretations and approaches being explored. Some participants provide guidance on how to find 'b' by substituting the asymptote values, while others question the role of 'a' in affecting the vertical asymptotes. There is no explicit consensus, but productive dialogue is ongoing.

Contextual Notes

Participants note that the value of 'a' does not affect the vertical asymptotes directly, but there are discussions about its potential impact on horizontal asymptotes under certain conditions. There is also mention of the need for 'a' to be non-zero.

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Homework Statement


Find two constants for 'a' and 'b' such that the vertical asymptote will be [tex]\pm[/tex] [tex]\frac{3}{5}[/tex]

y=[tex]\frac{ax^2+7}{9-bx^2}[/tex]


I rearranged so that it becomes [tex]-bx^2+8[/tex] in the denominator since i know that there are two roots that are [tex]\pm[/tex] it must be a square and since 3 is the numerator of the root it must -9 ... so i rearranged again to get

y=[tex]\frac{-ax^2-7}{bx^2-9}[/tex]

in which case i found the constant for a (-1) and [tex]5^2[/tex] is 25 so i found b as well so the equation would be

y=[tex]\frac{-x^2-7}{25x^2-9}[/tex]

is this right? I have no way to check my answer so i just want to make sure :D
 
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well for a vertical asymptote...the denominator of the function should be zero
in your case [itex]9-bx^2=0[/itex]
so that [itex]x=\pm\fract{3}{\sqrt{b}}[/itex]
so then equate that to [itex]\pm\frac{3}{5}[/tex] and find b[/itex]
 
not going to lie i don't get it...
would 25x^2-9 give u two values of x that equate to 0?
(5x-3)(5x+3).

I just don't understand what your doing there lol.
 
Another try: you want [itex]9-bx^2=0[/itex]
when x = +3/5 and x = -3/5; so if you want to do this in a crude way, just find the expression for b. This is b=(-9)/(x^2). So what is x ? You were already given the x values, since you want the vertical asymtotes at x=+3/5 and x=-3/5. Find b for both of these values by substituting. ...
b=25.

I see no particular big restrictions on a, except that a is not equal to zero.
 
Can't it be for all values of a? as the value of 'a' doesn't affect the vertical asymptotes
 
rock.freak667 said:
Can't it be for all values of a? as the value of 'a' doesn't affect the vertical asymptotes

yah all values can be 'a'. i just made it -1 so that the denominator would have better form... i think i did that right i wasn't EXACTLY sure if i could change all the signs in the equation by multiplying thru by -1 to move the negative to the top ? :S
 
rock.freak667 said:
Can't it be for all values of a? as the value of 'a' doesn't affect the vertical asymptotes

The value of 'a' can affect the asymptotes if it happens to be -7*5^2/3^3.
 
Dick said:
The value of 'a' can affect the asymptotes if it happens to be -7*5^2/3^3.

the 'a' value affects horizontal asymptote if the powers are the same. The only factor for the vertical asymptotes is that it makes the denominator 0 without making the numerator 0.
 

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