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**+2 **

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. | ||

| Subtract the constant term from both sides of the equation. | ||

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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