Answer:
B. Discontinuity at (−2, 1), zero at (−3, 0)
Explanation:
The given function is:
![(x^(2)+5x+6)/(x+2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/otxpljpxigjd1b616qets5hmmekg8n4egv.png)
The expression in numerator can be expressed as factors as shown below:
![(x^(2)+5x+6)/(x+2)\\\\ =(x^(2)+2x+3x+6)/(x+2)\\\\ =(x(x+2)+3(x+2))/(x+2)\\\\ =((x+2)(x+3))/(x+2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/aug3ys2vh2giez84c837vsgdsjraq3xx9o.png)
Note that for x = -2, both numerator and denominator will be zero. When both the numerator and denominator of a rational function are zero for a given value of x we get a discontinuity at that point. This discontinuity is known as a hole. This means there is a hole at x = -2
Cancelling the common factor from numerator and denominator we get the expression f(x) = x + 3
Using the value of x = -2 in previous expression we get:
f(x) = -2 + 3 = 1
Thus, there is a discontinuity(hole) at (-2, 1)
For x = -3, the value of the expression is equal to zero. This means x = -3 is a zero or root of the function.
Thus, (-3, 0) is a zero of the function.
Therefore, option B would be the correct answer.