question find all vertical asymptotes of the following function. f(x) = (4x² - 1)/(2x² - x)

Explanation:

Step1: Factor the function

Factor the numerator and denominator. The numerator $4x^{2}-1=(2x + 1)(2x - 1)$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The denominator $2x^{2}-x=x(2x - 1)$. So, $f(x)=\frac{(2x + 1)(2x - 1)}{x(2x - 1)}$.

Step2: Simplify the function

Cancel out the common factor $(2x - 1)$ (for $x\neq\frac{1}{2}$). $f(x)=\frac{2x + 1}{x},x\neq\frac{1}{2}$.

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: $x = 0$. When $x = 0$, the numerator $2x+1=2\times0 + 1=1\neq0$.

Answer:

$x = 0$