What are the discontinuity and zero of the function f(x) = x^2 + 8x + 7 / x + 1

Answer: - Discontinuity at (-1,6)  - The zero is at (-7,0)Step-by-step explanation: Given the function [tex]f(x)=\frac{x^2 + 8x + 7}{x+1}[/tex], you need to factor the numerator. Find two number whose sum be 8 and whose product be 7. These are 1 and 7, then: [tex]f(x)=\frac{(x+1)(x+7)}{(x+1)}[/tex] Then, the denominator is zero when [tex]x=-1[/tex] Therefore, [tex]x=-1[/tex] does not belong to the Domain of the function. Then, (-1,6) is a discontinuity point. Simplifying, you get: [tex]f(x)=x+7[/tex] You can observe that a linear function is obtained. This function is  equal to zero when [tex]x=-7[/tex], therefore the zero of the function is at (-7,0).