Answer:
The function is not continuous at x= -1 and x= 8
Step-by-step explanation:
The given function is,
[tex]y=\dfrac{x-5}{x^2-7x-8}[/tex]
Factoring the denominator,
[tex]=x^2-7x-8[/tex]
[tex]=x^2-8x+x-8[/tex]
[tex]=x(x-8)+1(x-8)[/tex]
[tex]=(x+1)(x-8)[/tex]
Now the function becomes,
[tex]y=\dfrac{x-5}{(x+1)(x-8)}[/tex]
The function is not defined (it does not exist) for x= -1 and for x= 8, because the denominator is zero for those values of x.
To be continuous, the function has to be defined. So at those points the function is not continuous.
*At x= -1 and at x= 8 , the function has a vertical asymptote.
