Question

For the function f(x)=2x²/2x²7-9,what are the vertical asymptotes, if any?

A) x=1 and x=−9/2
B) x=3 and x=− 2/1
C) x=2 and x=− 2/9
D) x=−3 and x= 2/1
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Answer 1

The given function appears to have a typo since the corrected intended function does not lead to any of the provided answer options for vertical asymptotes. Assuming a typo and correcting the denominator leads to a single vertical asymptote at x = 7/2, which does not match any choices given.

Step-by-step explanation:

The student has provided a function with an error in its notation. However, assuming the intended function is f(x) = \dfrac{2x^2}{2x^2 - 7} - 9, to find the vertical asymptotes, we need to look for values of x that make the denominator zero since the function will approach infinity or negative infinity at these points. To do this, set the denominator equal to zero:

2x^2 - 7 = 0 \Rightarrow x^2 = \dfrac{7}{2} \Rightarrow x = \pm\sqrt{\dfrac{7}{2}}

However, these values do not match any of the given options. If we consider a typo and correct the function to f(x) = \dfrac{2x^2}{2x - 7} - 9, we get:

2x - 7 = 0 \Rightarrow x = \dfrac{7}{2}

Since this still does not match any of the provided options, it seems there's a misunderstanding either in the function's formula or in the answer options provided by the student. Hence, no choice of A), B), C), or D) is correct based on the function as interpreted here.