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The graph of $y=\frac{5x^2-9}{3x^2+5x+2}$ has vertical asymptotes at $x = a$ and $x = b$. Find $a + b$.

Guest Aug 29, 2017
 #1
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To find the vertical asymptotes, we must figure out when the denominator of \(\frac{5x^2-9}{3x^2+5x+2}\) equals 0. This will give us the vertical asymptotes. Let's do that!

 

\(3x^2+5x+2=0\) I will use the AC method to factor out this. What number multiplies to get 6 and adds to get 5? 3 and 2, of course! Break up the b-term.
\(3x^2+3x+2x+2=0\) Let's solve this by grouping. I'll use parentheses to make it easier to follow, I hope.
\((3x^2+3x)+(2x+2)=0\) Factor out the GCF of both expressions inside of the parentheses.
\(3x(x+1)+2(x+1)=0\) Now, use the reverse-distributive-property to combine 3x and 2. In other words, \(ac+bc=(a+b)c\)
\((3x+2)(x+1)=0\) Set both factors equal to 0 and solve both independently.
\(3x+2=0\) \(x+1=0\)

 

Subtract the constant term from both sides of the equation.
\(3x=-2\) \(x=-1\)

 

 
\(x_1=-\frac{2}{3}\) \(x_2=-1\)

 

 
   

 

x=-(2/3) and x=-1 are the location of both vertical asymptotes on the graph. 

 

Now, add both answers together to get the result of \(a+b\)

 

\(-1-\frac{2}{3}=\frac{-3}{3}-\frac{2}{3}=-\frac{5}{3}=-1.\overline{66}\)

TheXSquaredFactor  Aug 29, 2017

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