question\nfind the equation of all vertical asymptotes of the following function.\n f(x)=\frac{2…

Explanation:

Step1: Find the domain of the function

For the function (f(x)=\frac{2x - 3}{\sqrt{x^{2}-3x - 28}}), the expression under the square - root must be positive, i.e., (x^{2}-3x - 28>0). Factor the quadratic: (x^{2}-3x - 28=(x - 7)(x+4)>0). The solutions of the inequality ((x - 7)(x + 4)>0) are (x<-4) or (x>7) using the sign - chart method (test intervals ((-\infty,-4)), ((-4,7)) and ((7,\infty))).

Step2: Analyze the behavior near the boundary points

Vertical asymptotes occur where the function approaches (\pm\infty). For a rational function (y=\frac{N(x)}{D(x)}) (here (N(x)=2x - 3) and (D(x)=\sqrt{x^{2}-3x - 28})), we consider the limit as (x) approaches the boundary points of the domain. (\lim_{x\rightarrow - 4^{-}}\frac{2x - 3}{\sqrt{x^{2}-3x - 28}}), let (t=x + 4), (x=t - 4). Then (x^{2}-3x - 28=(t - 4)^{2}-3(t - 4)-28=t^{2}-8t + 16-3t + 12-28=t^{2}-11t). As (t\rightarrow0^{-}), (\sqrt{x^{2}-3x - 28}=\sqrt{t(t - 11)}\sim\sqrt{- 11t}) (for (t\rightarrow0^{-})) and (2x-3=2(t - 4)-3=2t-11\sim - 11). So (\lim_{x\rightarrow - 4^{-}}\frac{2x - 3}{\sqrt{x^{2}-3x - 28}}=\lim_{t\rightarrow0^{-}}\frac{2t-11}{\sqrt{t(t - 11)}}) is a finite non - zero value (since the numerator approaches (-11) and the denominator approaches (0) from the positive side as (x\rightarrow - 4^{-}), but the function is not defined on an open interval around (x = - 4) (the domain is (x<-4) or (x>7)). Similarly, for (\lim_{x\rightarrow7^{+}}\frac{2x - 3}{\sqrt{x^{2}-3x - 28}}), let (u=x - 7), (x=u + 7). Then (x^{2}-3x - 28=(u + 7)^{2}-3(u + 7)-28=u^{2}+14u+49-3u - 21-28=u^{2}+11u). As (u\rightarrow0^{+}), (\sqrt{x^{2}-3x - 28}=\sqrt{u(u + 11)}\sim\sqrt{11u}) and (2x-3=2(u + 7)-3=2u + 11\sim11). The function is not defined on an open interval around (x = 7) (the domain is (x<-4) or (x>7)).

Answer:

No Vertical Asymptotes