question find all vertical asymptotes of the following function. f(x) = (4x^2 - 1)/(2x^2 - x)

Answer:

$x = 0$ and $x=\frac{1}{2}$

Explanation:

Step1: Factor the function

$f(x)=\frac{4x^{2}-1}{2x^{2}-x}=\frac{(2x + 1)(2x - 1)}{x(2x - 1)}$

Step2: Simplify the function

$f(x)=\frac{2x + 1}{x},x\neq\frac{1}{2}$

Step3: Find the vertical - asymptotes

Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is non - zero. Set the denominator equal to zero: $x=0$. Also, before simplification, when $2x - 1=0$ (i.e., $x=\frac{1}{2}$), the original function is undefined. After simplification, $x = \frac{1}{2}$ is a removable discontinuity (a hole), but considering the original form, we include it in the vertical asymptotes search. So the vertical asymptotes are $x = 0$ and $x=\frac{1}{2}$.