find all vertical asymptotes of the following function.\nf(x)=\frac{x^{2}-9x + 20}{6x-14}

Explanation:

Step1: Recall vertical - asymptote condition

Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is non - zero. Set the denominator equal to zero. $6x - 14=0$

Step2: Solve for x

Add 14 to both sides: $6x=14$. Then divide both sides by 6: $x = \frac{14}{6}=\frac{7}{3}$. Check the numerator at $x=\frac{7}{3}$. Substitute $x = \frac{7}{3}$ into the numerator $x^{2}-9x + 20$: $\left(\frac{7}{3}\right)^{2}-9\times\frac{7}{3}+20=\frac{49}{9}-21 + 20=\frac{49}{9}-1=\frac{49 - 9}{9}=\frac{40}{9}\neq0$

Answer:

$x=\frac{7}{3}$