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# Question5

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How many vertical asymptotes does the graph of $$y=\frac{x-3}{x^2+7x-30}$$  have?

tertre  Mar 12, 2017

#3
+1844
+10

Thanks so much guys!

tertre  Mar 12, 2017
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#1
+6536
+6

Vertical asymptotes occur when y is undefined.

Y is undefined when the denominator is = 0.

So set the denominator = 0 to find which x values make y undefined.

x2 + 7x - 30 = 0

x2 + 7x + 49/4 = 169/4

x + 7/2 = ±13/2 - 7/2

x = 3 and x = -10

SO there are 2 values that make y undefined.

But from looking at a graph (or by taking the limit as x approaches 3) you can see that at x = 3 there is only a point discontinuity, not an asymptote.

So there is only 1 veertical asymptote at x = -10.

hectictar  Mar 12, 2017
edited by hectictar  Mar 12, 2017
#2
+12119
+5

(x-3)                                    (x-3)/[(x-3)(x+10)]     =  1 /(x+10)

_____                  =

x^2 +7x-30

So as x approaches -10  you will have an asymtope there at -10      One asymtope at x = -10

ElectricPavlov  Mar 12, 2017
#3
+1844
+10