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

tertre
Mar 12, 2017

#1**+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.

x^{2} + 7x - 30 = 0

x^{2} + 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

#2**+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