Identifying x-intercepts and vertical asymptotes of a graph

**vaironxxrd** · March 11th 2012, 06:45 AM

I have the following function

$y = \frac{2x^2-7x-4}{x+5}$

X intercepts =
$0 = 2x^2-7x-4$

$(x-4)(2x+1)$
x intercepts are 4, and $-\frac{1}{2}$

Vertical asymptotes =

$0 = x+5$

$x = -5$

Vertical asymptote is were x= -5

Is this correct?

**FernandoRevilla** · March 11th 2012, 07:04 AM

Originally Posted by vaironxxrd

Is this correct?

Yes, it is correct. Only one thing, $x=-5$ is a vertical asymptote not only because $x+5=0$ , we need to prove $\lim_{x\to -5}y=\infty$ (very easy in this case). For example, $x=-5$ is not a vertical asymptote of $y=\dfrac{x^2-25}{x+5}$ although $x+5=0$ .

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