2. Your friend thinks the function $f(x) = \frac{(x+2)(2x-3)}{(2x-3)^2}$ has an asymptote at $x = \frac{3}{2}$ because $(2x - 3)$ is in the denominator. Your classmate says that there is no asymptote and there is a hole. Who do you agree with? Explain.??
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However, we need to check if this value makes the function undefined. If we substitute x = -2 into the function, we get: f(-2) = (2(-2)-3)/(-2+2) = (-4-3)/0 Since we have a division by zero, the function is undefined at x = -2. This means that x = -2 is not a Show more…
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