We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# help

0
328
1

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

Aug 29, 2017

### 1+0 Answers

#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.
 $$3x+2=0$$ $$x+1=0$$

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 $$a+b$$

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

.
Aug 29, 2017