Finding Constants for Rational Functions with Specific Vertical Asymptotes

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To find constants 'a' and 'b' for the rational function y=(ax^2+7)/(9-bx^2) with vertical asymptotes at ±3/5, the denominator must equal zero at these points. This leads to the equation 9-bx^2=0, which simplifies to b=25 when substituting x=±3/5. The value of 'a' does not affect the vertical asymptotes, allowing it to be any non-zero number, although it can influence horizontal asymptotes if the degrees of the numerator and denominator are equal. The discussion emphasizes that the critical factor for vertical asymptotes is ensuring the denominator is zero without the numerator also being zero. Therefore, the correct constants are a=-1 and b=25.
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Homework Statement


Find two constants for 'a' and 'b' such that the verticle asymptote will be \pm \frac{3}{5}

y=\frac{ax^2+7}{9-bx^2}


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

y=\frac{-ax^2-7}{bx^2-9}

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

y=\frac{-x^2-7}{25x^2-9}

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 9-bx^2=0
so that x=\pm\fract{3}{\sqrt{b}}
so then equate that to \pm\frac{3}{5}[/tex] and find b
 
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 9-bx^2=0
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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