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?
a. discontinuity at (−2, 1), zero at (3, 0)
b. discontinuity at (−2, 1), zero at (−3, 0)
c. discontinuity at (2, 5), zero at (3, 0)
d. discontinuity at (2, 5), zero at (−3, 0)

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

Step-by-step explanation:

The given function is:

The expression in numerator can be expressed as factors as shown below:

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.

(x, y) -> (x + 1, y + 1), (x, y) -> (y, x)

Step-by-step explanation:

I'll use point B (-3,0) for this problem.

First, the translation rule tells us to add 1 to the x value and add 1 to the y value. Now we have (-2, 1). Next, the rule tells us to flip the x and y coordinates. Now we have (1, -2). This matches to vertex B''(1, -2).