Question 185346
# 1



{{{g(x)=(x^2+x+1)/(5x^2+1)}}} Start with the given function.



{{{5x^2+1=0}}} Set the denominator equal to zero. Remember, vertical asymptotes arise only when there's a division by zero



{{{5x^2=-1}}} Subtract 1 from both sides.



{{{x^2=-1/5}}} Divide both sides by 5



{{{x=sqrt(-1/5)}}} Take the square root of both sides.



Remember, you CANNOT take the square root of a negative number. So this means that the value of "x" will be complex/imaginary. 



In other words, there is NO value of "x" that will make the denominator zero. 



So there are no vertical asymptotes (graph the function to verify)



===============================================================


# 2


{{{(3a^2-6a)/(12-6a)}}} Start with the given expression



{{{(3a(a-2))/(12-6a)}}} Factor the numerator



{{{(3a(a-2))/(-6(a-2))}}} Factor the denominator



{{{-(a(a-2))/(2(a-2))}}} Reduce {{{3/(-6)}}} (the outer coefficients) to get {{{-1/2}}}



{{{-(a*highlight((a-2)))/(2*highlight((a-2)))}}} Highlight the common terms.



{{{-(a*cross((a-2)))/(2*cross((a-2)))}}} Cancel out the common terms.



{{{-a/2}}} Simplify



==============


Answer:


So {{{(3a^2-6a)/(12-6a)}}} simplifies to {{{-a/2}}}



In other words, {{{(3a^2-6a)/(12-6a)=-a/2}}} where {{{a<>2}}}