Question

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

Answer 1

Answer:

- Discontinuity at (-1,6)

 - The zero is at (-7,0)

Step-by-step explanation:

Given the function $f(x)=\frac{x^2 + 8x + 7}{x+1}$, you need to factor the numerator. Find two number whose sum be 8 and whose product be 7. These are 1 and 7, then:

$f(x)=\frac{(x+1)(x+7)}{(x+1)}$

Then, the denominator is zero when $x=-1$

Therefore, $x=-1$ does not belong to the Domain of the function. Then, (-1,6) is a discontinuity point.

Simplifying, you get:

$f(x)=x+7$

You can observe that a linear function is obtained.

This function is  equal to zero when $x=-7$, therefore the zero of the function is at (-7,0).

Answer 2

ANSWER

Discontinuity:

$x = - 1$

Zero:

$x = - 7$

EXPLANATION

The given rational function is

$f(x) = \frac{ {x}^{2} + 8x + 7 }{x + 1}$

We factor function to obtain:

$f(x) = \frac{(x + 1)(x + 7)}{x + 1}$

This function is not continuous when

$(x + 1) = 0$

The function is not continuous at

$x = - 1$

When we simplify the function, we get;

$f(x) = x + 7$

The zero(s) occur at

$x + 7 = 0$

$x = - 7$