Question 207647
To find the vertical asymptotes, simply set the denominator equal to zero to get {{{6x^2-x-2=0}}}. From there, solve for 'x' to find the x values that make the denominator equal to zero. These x values will be the equations of the vertical asymptotes.









If you don't know how to solve, then....




{{{6x^2-x-2=0}}} Start with the given equation.



Notice that the quadratic {{{6x^2-x-2}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=6}}}, {{{B=-1}}}, and {{{C=-2}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-1) +- sqrt( (-1)^2-4(6)(-2) ))/(2(6))}}} Plug in  {{{A=6}}}, {{{B=-1}}}, and {{{C=-2}}}



{{{x = (1 +- sqrt( (-1)^2-4(6)(-2) ))/(2(6))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(6)(-2) ))/(2(6))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--48 ))/(2(6))}}} Multiply {{{4(6)(-2)}}} to get {{{-48}}}



{{{x = (1 +- sqrt( 1+48 ))/(2(6))}}} Rewrite {{{sqrt(1--48)}}} as {{{sqrt(1+48)}}}



{{{x = (1 +- sqrt( 49 ))/(2(6))}}} Add {{{1}}} to {{{48}}} to get {{{49}}}



{{{x = (1 +- sqrt( 49 ))/(12)}}} Multiply {{{2}}} and {{{6}}} to get {{{12}}}. 



{{{x = (1 +- 7)/(12)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{x = (1 + 7)/(12)}}} or {{{x = (1 - 7)/(12)}}} Break up the expression. 



{{{x = (8)/(12)}}} or {{{x =  (-6)/(12)}}} Combine like terms. 



{{{x = 2/3}}} or {{{x = -1/2}}} Simplify. 



So the solutions are {{{x = 2/3}}} or {{{x = -1/2}}} 




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Answer:



This means that the vertical asymptotes of {{{f(x)=(3x+5)/(6x^2-x-2)}}} are {{{x = 2/3}}} and {{{x = -1/2}}}