Sorry!
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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