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# help

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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
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### 1+0 Answers

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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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