Question

What are the discontinuity and zero of the function f(x) = quantity x squared plus 5 x plus 6 end quantity over quantity x plus 2? plz answer fast
Discontinuity at (−2, 1), zero at (3, 0)
Discontinuity at (−2, 1), zero at (−3, 0)
Discontinuity at (2, 5), zero at (3, 0)
Discontinuity at (2, 5), zero at (−3, 0)

Answer 1

Answer:

discontinuity = (-2,1)

zero (-3,0)

Step-by-step explanation:

f(x) = (x^2 +5x+6)

       ------------------

      ( x+2)

First factor the numerator

f(x) = (x+3) (x+2)

       ------------------  

      ( x+2)

The discontinuity of the function is where the denominator goes to zero

x+2 = 0   x=-2

The zero is where the numerator goes to zero  (we do not include the discontinuity)

x+3 =0   x=-3

Canceling the common factor from the numerator and the denominator

f(x) = x+3

f(-2) = -2+3 = 1

f(-3) = -3+3 = 0

Therefore

discontinuity = (-2,1)

zero (-3,0)

Answer 2

Answer:

Discontinuity at (−2, 1), zero at (−3, 0)

Step-by-step explanation:

(x² + 5x + 6)/(x + 2)

(x² + 3x + 2x + 6)/(x + 2)

[x(x + 3) + 2(x + 3)]/(x + 2)

(x + 2)(x + 3)/(x + 2)

Zero: when numerator becomes 0, while the denominator is non-zero.

x + 3 = 0

x = -3

Discontinuity: when both, numerator and denominator become 0.

x + 2 = 0

x = -2