Which of the following functions has one root, x=3 and one vertical asymptote, x=2?

Question

A) ln(-x)

B) ln(x+1)

C) ln(x+2)

D) ln(x-2)

E) none of these

Answer 1

To determine which function has a root at x=3 and a vertical asymptote at x=2, we need to examine the properties of the natural logarithm functions provided in the options.

1. Root at x=3: A root means that the function equals zero at x=3. For a natural logarithm function ln⁡(f(x)), this means f(x)=1 when x=3 because ln⁡(1)=0.
2. Vertical asymptote at x=2: A vertical asymptote occurs where the argument of the natural logarithm goes to zero (since ln⁡(0) is undefined and tends towards −∞). For a function ln⁡(f(x)), this means f(x)=0 at x=2.

Let's check each option:

A) ln⁡(−x)

1. This function has a vertical asymptote at x=0 because −x=0 when x=0.
2. This function cannot have a root at x=3 since ln⁡(−3) is undefined (the logarithm of a negative number is not defined in the real numbers).

B) ln⁡(x+1)

1. This function has a vertical asymptote at x=−1 because x+1=0 when x=−1.
2. To have a root at x=3, we need x+1=1. Solving gives x=0, which does not match x=3.

C) ln⁡(x+2)

1. This function has a vertical asymptote at x=−2 because x+2=0 when x=−2.
2. To have a root at x=3, we need x+2=1. Solving gives x=-1, which does not match x=3.

D) ln⁡(x−2)

1. This function has a vertical asymptote at x=2 because x−2=0 when x=2.
2. To have a root at x=3, we need x−2=1. Solving gives x=3, which matches the given root.

E) None of these

1. This option is valid if none of the above functions match both conditions.

Based on the analysis, the correct answer is:

D) ln⁡(x−2)

Answer 2

x ln(x–2)

→2 →–∞

3 0

ln(x–2)→–∞ as x→2

ln(x–2) = ln(3–2) = ln(1) = 0 when x = 3

So the answer is

D) ln(x-2)

Answer 3

It would be D. Ln(x-2) creates a zero when ln(1) = 0. To satisfy this we can set x-2 =1 meaning that there is a zero at x=3. Also a vertical asymptote would be present at x=2, at a point where the function is undefined. Similar to a vertical asymptote of a polynomial, we usually solve for the denominator. In this case we know if x=2 this would give us undefined, indicating a vertical asymptote at x = 2 as well. Hope this helped!

Answer 4

D) ln(x-2) has vertical asymptote x=2

root 3 means f(3) = 0 = ln(3-2) = ln1 = 0

it has a root at x=3 too

Answer 5

Let's start off with a dissection of the question being asked and go through each condition one-by-one.

1) the function has a root at x = 3. f(3) = 0

2) there is a vertical asymptote at x = 2. f(x) is undefined at x = 2.

If you notice, all of the answer choices are in the form f(x) = ln(mx+b), where m (m=-1,1) and b (b=0,1,2,-2) are constants. This fact allows us to deduce the function based upon the first two conditions.

First examine each condition:

1) f(3)=0

if f(x) = ln(mx+b), 0 = ln(3m+b)

e^0 = 3m+b

1 = 3m+b

2) f(x) is undefined & has an asymptote at x = 2

If you remember, ln(x) has an asymptote at x = 0 (ln(0) is undefined)

Therefore, it is fair to assume that ln(mx+b) has an asymptote when mx+b = 0

If the asymptote is at x = 2, we are left with the equation 2m+b = 0

Final step:

We are given the set of linear equations

2m+b = 0

3m+b = 1

if we subtract those two equations, we are given

3m+b - (2m+b) = 1 - 0

m = 1

and substituting m=1 into the equation 2m+b=0, we are given

2(1)+b = 0

b = -2

Therefore, our final equation is f(x) = ln(1x-2) or ln(x-2)

Answer choice D