+0  
 
0
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avatar+2765 

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

tertre  Mar 12, 2017

Best Answer 

 #3
avatar+2765 
+10

Thanks so much guys!

tertre  Mar 12, 2017
 #1
avatar+7153 
+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
avatar+12560 
+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
avatar+2765 
+10
Best Answer

Thanks so much guys!

tertre  Mar 12, 2017

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