Question

What are the discontinuity and zero of the function f(x) = quantity x squared plus 6 x plus 8 end quantity over quantity x plus 4
Discontinuity at (4, 6), zero at (−2, 0)

Discontinuity at (4, 6), zero at (2, 0)

Discontinuity at (−4, −2), zero at (−2, 0)

Discontinuity at (−4, −2), zero at (2, 0)

Answer 1

Answer:

Discontinuity at (-4,-2), zero at (-2,0).

Step-by-step explanation:

We are given that a function

$f(x)=\frac{x^2+6x+8}{x+4}$

We have to find the discontinuity and zero of the given function.

Discontinuity: It is that point where the function is not defined.

It makes the function infinite.

$f(x)=\frac{x^2+4x+2x+8}{x+4}$

$f(x)=\frac{(x+4)(x+2)}{x+4}$

When x=-4 then

$f(-4)=\frac{0}{0}$ It is indeterminate form

Function is not defined

After cancel out x+4 in numerator and denominator  then we get

$f(x)=x+2$

Substitute x=-4

$f(-4)=-4+2=-2$

Therefore, the point of discontinuity is (-4,-2).

Zero: The zero of the function is that number when substitute it in the given function then the function becomes zero.

When substitute x=-2

Then , $f(0)=-2+2=0$

The function is zero at (-2,0).

Hence, option C is true.