Question 78340
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Can you tell me if this answer is correct?
Determine the vertical asympotes for the function 

{{{f(x)=(x^2-16)/(3x^2-3)}}}

My answer:

{{{x=0}}}, {{{x=sqrt(1)}}}, {{{x= -sqrt(1)}}}

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<pre><font size = 5><b>
The {{{x=0}}} is incorrect!  The other two are correct
except that you would never leave those answers
unsimplified as {{{x=sqrt(1)}}} and {{{x= -sqrt(1)}}}

Since {{{sqrt(1) = 1}}}, the answers should be 
simplified as:

{{{x = 1}}} and {{{x = -1}}}

You simply set the denominator of

{{{f(x)=(x^2-16)/(3x^2-3)}}}

equal to 0

{{{3x^2-3=0}}}

Divide every term on both sides through by 3

{{{(3x^2)/(3)}}}{{{-3/3=0/3}}}

{{{x^2-1=0}}}

There are two ways to solve that:

1. Factor the left side as the difference
of squares:

{{{(x-1)(x+1)=0}}}

Set each factor = 0

Setting the first factor = 0,

{{{x-1=0}}}

{{{x = 1}}}

Setting the second factor = 0,

{{{x+1=0}}}

{{{x = -1}}} 

2. The way you apparently did it:

{{{x^2-1=0}}}
 
Add +{{{1}}} to both sides

{{{x^2 = 1}}}

Take square roots of both sides

{{{x=sqrt(1)}}} and {{{x= -sqrt(1)}}}

which simplifies to

{{{x=1}}} and {{{x= -1}}}

That is perfectly OK.

[But there is no way you could have 
gotten {{{x=0}}}!]

Let's look at the graph of 

{{{f(x)=(x^2-16)/(3x^2-3)}}}

{{{graph(300,300,-5,5, -30, 50,(sqrt(x-1.01)/sqrt(x-1.01))(x^2-16)/(3x^2-3),(sqrt(.99-x)/sqrt(.99-x))(sqrt(.99-abs(x))/sqrt(.99-abs(x)))(x^2-16)/(3x^2-3),(sqrt(-1.1-x)/sqrt(-1.1-x))(x^2-16)/(3x^2-3))}}}

Now let's draw in the two vertical 
asymptotes at x=-1 and a=1:

{{{graph(300,300,-5,5, -30, 50,(sqrt(x-1.01)/sqrt(x-1.01))(x^2-16)/(3x^2-3),(sqrt(.99-x)/sqrt(.99-x))(sqrt(.99-abs(x))/sqrt(.99-abs(x)))(x^2-16)/(3x^2-3),(sqrt(-1.1-x)/sqrt(-1.1-x))(x^2-16)/(3x^2-3),999(x-1),-999(x+1))}}}

and you can see that the curve approches 
those two asymptotes on both sides. The
curve gets closer and closer to these 
asymptotes but never touches them! 

Edwin</pre>