- #1
Sorry!
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Homework Statement
Find two constants for 'a' and 'b' such that the verticle 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