Answer:
The function is not continuous at x= -1 and x= 8
Step-by-step explanation:
The given function is,
\(y=\dfrac{x-5}{x^2-7x-8}\)
Factoring the denominator,
\(=x^2-7x-8\)
\(=x^2-8x+x-8\)
\(=x(x-8)+1(x-8)\)
\(=(x+1)(x-8)\)
Now the function becomes,
\(y=\dfrac{x-5}{(x+1)(x-8)}\)
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.
