Finding The Right Function: Domain & Range Mastery

Hey Plastik Magazine readers! Let's dive into a fun math problem today. We're on a quest to find the function that fits a specific domain and range criteria. Basically, we're searching for a function where x can't be a certain number, and y can't be another. Sound like a challenge? Don't worry, we'll break it down step by step and make it super clear. This is all about understanding domain and range, so if you're ready to flex those math muscles, let's get started!

Unpacking Domain and Range: The Basics

Alright, before we jump into the options, let's refresh our memories on what domain and range actually mean. Think of it like this: the domain is all the possible x-values a function can take. It's the set of all the inputs. The range, on the other hand, is all the possible y-values the function can produce. It's the set of all the outputs. Got it? Great!

Now, when we say x ≠ 3, we're saying that the function can't accept 3 as an input. It's like a forbidden number. And when we say y ≠ 2, we're saying that the function can never output 2. This helps us understand what to look for when we're given the options. This means we'll be looking for a function that has a vertical asymptote at x = 3 and a horizontal asymptote at y = 2. Let's see how this works with the options.

To really get this concept down, visualize a graph. The domain is all the x-values you can plot on the graph. The range is all the y-values you can plot. So if there's a hole or a break in the graph, that tells you something about the domain and range. Easy peasy!

Analyzing the Answer Choices: Eliminating the Imposters

Now, let's put our domain and range knowledge to the test by examining each function. We'll methodically check each option to see if it meets our criteria: a domain where x ≠ 3 and a range where y ≠ 2. We'll be using our understanding of rational functions – those with x in the denominator – to figure this out. Remember, we need to find the function that has a vertical asymptote at x = 3 and a horizontal asymptote at y = 2. We are going to analyze each equation to find the function we are looking for.

Option A: $f(x) = \frac{(x-5)}{(x+3)}$

Let's start with option A. Here, the function is $f(x) = \frac{(x-5)}{(x+3)}$. To find the value that is not in the domain, we need to see where the denominator is zero, because that's when the function is undefined. So, we solve x + 3 = 0, which gives us x = -3. So, the domain here is x ≠ -3. This means that option A doesn't meet our domain requirements, so we can discard this answer.

Next, let's look for the horizontal asymptote. To find this, we examine the behavior of the function as x approaches infinity. For rational functions where the degree of the numerator and denominator are the same, the horizontal asymptote is the ratio of the leading coefficients. In this case, the leading coefficients are both 1, so the horizontal asymptote is y = 1. This doesn't match our required range of y ≠ 2.

Option B: $f(x) = \frac{2(x+5)}{(x+3)}$

Moving on to option B, we have $f(x) = \frac{2(x+5)}{(x+3)}$. Again, let's find the value that is not in the domain. We set the denominator to zero, x + 3 = 0, so x = -3. This gives us a domain of x ≠ -3. This doesn't quite work. Our desired domain is x ≠ 3.

Now, for the horizontal asymptote, we'll look at the ratio of the leading coefficients. The numerator's leading coefficient (after expanding) is 2, and the denominator's is 1. Thus, the horizontal asymptote is y = 2. This means option B has a range where y ≠ 2, but the domain does not match our requirements. Keep this one in mind, as it gets close!

Option C: $f(x) = \frac{2(x+5)}{(x-3)}$

Option C presents us with $f(x) = \frac{2(x+5)}{(x-3)}$. Let's examine this one. We set the denominator to zero: x - 3 = 0, so x = 3. Great! This function has a domain of x ≠ 3. This is exactly what we need for the domain. We're getting closer!

Now for the range. We look at the horizontal asymptote. The leading coefficient in the numerator is 2, and the leading coefficient in the denominator is 1, so the horizontal asymptote is y = 2. This gives us a range where y ≠ 2. Boom! We have a winner! This matches both our domain and range criteria perfectly.

Option D: $f(x) = \frac{(x+5)}{(x-3)}$

Finally, let's look at option D. We have $f(x) = \frac{(x+5)}{(x-3)}$. Setting the denominator to zero gives us x - 3 = 0, so x = 3. The domain is x ≠ 3. This matches our domain requirement.

For the horizontal asymptote, we see the ratio of the leading coefficients is 1/1, which is 1. Therefore, the horizontal asymptote is y = 1, which means the range is y ≠ 1. This does not meet our required range of y ≠ 2.

The Verdict: The Chosen Function!

After a thorough investigation, we can confidently say that Option C: $f(x) = \frac{2(x+5)}{(x-3)}$ is the function with a domain of x ≠ 3 and a range of y ≠ 2. Congratulations to everyone who followed along! Understanding domain and range is a fundamental skill in math, and with a bit of practice, you'll be identifying these functions like a pro.

This problem perfectly illustrates how to find the domain and range of a function. You need to remember that domain is the set of all x-values, and range is the set of all y-values. You can usually find the domain by setting the denominator to zero and the range by identifying the horizontal asymptote. Keep practicing, and you'll do great! And that's all, folks! Hope you enjoyed this math adventure. Keep an eye out for more fun content from Plastik Magazine. Until next time!