Question

question
find all vertical asymptotes of the following function.
f(x)=\frac{4 x^{2}-25}{4 x^{2}-10 x}
no vertical asymptotes
one vertical asymptote
two vertical asymptotes
no vertical asymptotes

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 \(x
eq\frac{5}{2}\)).
We get \(f(x)=\frac{2x + 5}{2x}\), \(x
eq\frac{5}{2}\).

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 \(x
eq0,x
eq\frac{5}{2}\). 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 (\(x = 0\))