question\nfind all vertical asymptotes of the following function.\n f(x)=\frac{4 x^{2}-25}{4 x^{2}-10 x}…

question\nfind all vertical asymptotes of the following function.\n f(x)=\frac{4 x^{2}-25}{4 x^{2}-10 x} \nno vertical asymptotes\none vertical asymptote\ntwo vertical asymptotes\nno vertical asymptotes

question\nfind all vertical asymptotes of the following function.\n f(x)=\frac{4 x^{2}-25}{4 x^{2}-10 x} \nno vertical asymptotes\none vertical asymptote\ntwo vertical asymptotes\nno vertical asymptotes

Answer

Explanation:

Step1: Factor numerator and denominator

Numerator: (4x^{2}-25=(2x + 5)(2x - 5)) (using (a^{2}-b^{2}=(a + b)(a - b)) with (a = 2x), (b=5)). Denominator: (4x^{2}-10x=2x(2x - 5)) (factoring out (2x)). So (f(x)=\frac{(2x + 5)(2x - 5)}{2x(2x - 5)}).

Step2: Simplify the function

Cancel out the common factor ((2x - 5)) (for (x\neq\frac{5}{2})). We get (f(x)=\frac{2x + 5}{2x}), (x\neq\frac{5}{2}).

Step3: Find the vertical asymptote

Set the simplified denominator equal to zero: (2x=0), so (x = 0). Check the original domain. The original function (f(x)=\frac{4x^{2}-25}{4x^{2}-10x}) has domain (x\neq0,x\neq\frac{5}{2}). But after simplification, the non - removable discontinuity is at (x = 0) (since the factor ((2x - 5)) was canceled, so (x=\frac{5}{2}) is a hole, not an asymptote).

Answer:

One Vertical Asymptote ((x = 0))