Question

What are the discontinuity and zero of the function f(x) = quantity x squared plus 5 x plus 4 end quantity over quantity x plus 4?


a Discontinuity at (−4, −3), zero at (−1, 0)

b Discontinuity at (−4, −3), zero at (1, 0)

c Discontinuity at (4, 3), zero at (−1, 0)

d Discontinuity at (4, 3), zero at (1, 0)

Answer 1

Answer:

Option a -  Discontinuity at (−4, −3), zero at (−1, 0)

Step-by-step explanation:

Given : Function $\dfrac{x^2+5x+4}{x+4}$

To find : What are the discontinuity and zero of the function ?

Solution :

To find the discontinuity for our given function $\dfrac{x^2+5x+4}{x+4}$

We will equate denominator to 0,

$x+4=0$

$x=-4$

Now we simplify the expression,

$f(x)=\dfrac{x^2+5x+4}{x+4}$

$f(x)=\dfrac{(x+4)(x+1)}{(x+4)}$

$f(x)=x+1$

Since the denominator term is cancelled out, so our give function has a removable discontinuity.

Now, we will find value of y by substituting x=-4 in function $f(x)=x+1$.

$f(-4)=-4+1$

$f(-4)=-3$

The discontinuity of given function is at point (-4,-3) .

For zeros of the function put function equate to zero.

$x+1=0$

$x=-1$

The zero is at (-1,0).

Therefore, Option a is correct.