Functions: True Statements For F(x) = 10/(x+5)

Hey guys! Today, we're diving deep into the world of functions, specifically a rational function, f(x)=rac{10}{x+5}. We've got a multiple-choice question here asking us to identify four correct statements about this function. Let's break it down piece by piece and figure out which of these statements are actually true. Understanding functions like these is super important, not just for passing exams, but for grasping how mathematical relationships work in the real world. So, grab your notebooks, and let's get started on uncovering the truths about f(x)=rac{10}{x+5}. We're going to look at intercepts, range, domain, and end behavior to make sure we nail all four correct answers. It's going to be a fun ride through algebra, I promise!

Understanding the Function f(x)=rac{10}{x+5}

Alright, let's get down to business with our function f(x)=rac{10}{x+5}. This is a classic example of a rational function, which basically means it's a fraction where both the numerator and the denominator are polynomials. In our case, the numerator is a constant, 10, and the denominator is a linear expression, $x+5$. Rational functions have some unique characteristics, like vertical asymptotes and horizontal asymptotes, which really define their behavior. The denominator, $x+5$, is the key here. We know that we can't divide by zero, so the function is undefined when $x+5=0$, which means $x=-5$. This value, $x=-5$, is critical because it tells us there's a vertical asymptote at $x=-5$. This asymptote is a vertical line that the graph of the function approaches but never touches. It's like a boundary line that shapes the graph. Now, let's think about the $y$-intercept. To find the $y$-intercept, we need to see what happens when $x=0$. Plugging in $x=0$ into our function gives us f(0) = rac{10}{0+5} = rac{10}{5} = 2. So, the $y$-intercept is indeed at the point $(0, 2)$. This is a really common way to check our understanding of intercepts – just substitute the appropriate value for $x$ or $f(x)$ and solve. For the $x$-intercept, we need to find the value of $x$ where the function's output, $f(x)$, is zero. For a rational function, this happens when the numerator is zero, provided the denominator is not also zero at that same $x$-value. In our function, the numerator is 10, which is never zero. This means that $f(x)$ will never equal 0. Therefore, there is no $x$-intercept for this function. This is a crucial point to remember about rational functions: they don't always have both $x$ and $y$-intercepts. Sometimes they have neither, or just one. The fact that the numerator is a non-zero constant is the key indicator here. We also need to consider the range of the function. The range is the set of all possible output values ($y$-values) that the function can produce. Since the numerator is a non-zero constant (10), the function $f(x)$ can never be equal to 0. Also, because of the vertical asymptote at $x=-5$, the function's values will shoot off towards positive or negative infinity as $x$ approaches -5. As $x$ gets very, very large (approaching positive or negative infinity), the value of rac{10}{x+5} gets closer and closer to 0. This means there's a horizontal asymptote at $y=0$. A horizontal asymptote describes the behavior of the function as $x$ approaches positive or negative infinity. So, the function can take on any value except for the value of the horizontal asymptote, which is 0. The statement about the range being (-ree, 2) igtriangleup (2, ree) implies that 2 is excluded from the range, which isn't necessarily true based on our analysis of the horizontal asymptote. We'll revisit this when we check the options. Finally, let's think about the end behavior. As $x$ approaches positive infinity (x o ree), $x+5$ also approaches positive infinity, so rac{10}{x+5} approaches 0 from the positive side. As $x$ approaches negative infinity (x o -ree), $x+5$ also approaches negative infinity, so rac{10}{x+5} approaches 0 from the negative side. This confirms our horizontal asymptote at $y=0$. So, we've got a good handle on the key features of this function. Now, let's systematically evaluate each statement provided in the question to find the four correct answers. It's all about carefully checking each property against our findings.

Analyzing the Statements: Step-by-Step

Let's tackle each statement one by one, guys, and see if it holds true for our function f(x)=rac{10}{x+5}. This is where we apply everything we've learned about intercepts, range, and end behavior.

Statement A: The $y$-intercept is $(0, 2)$.

To find the $y$-intercept, we set $x=0$ and evaluate $f(0)$.

$f(0) = \frac{10}{0+5} = \frac{10}{5} = 2$.

So, the $y$-intercept is indeed at the point $(0, 2)$. This statement is TRUE. We've already confirmed this in our initial analysis, and it's a solid point for one of our four correct answers.

Statement B: The range of $f(x)$ is (-ree, 2) igtriangleup (2, ree).

The range is the set of all possible $y$-values. We found that there's a horizontal asymptote at $y=0$. This means that as $x$ gets very large (positive or negative), $f(x)$ gets very close to 0, but it never actually equals 0 because the numerator is 10. So, 0 is not in the range. Now, let's think about the values around the vertical asymptote $x=-5$. As $x$ approaches -5 from the right (values slightly greater than -5), $x+5$ is a small positive number, so f(x) = rac{10}{ ext{small positive}} becomes a large positive number. As $x$ approaches -5 from the left (values slightly less than -5), $x+5$ is a small negative number, so f(x) = rac{10}{ ext{small negative}} becomes a large negative number. This means the function can take on any positive value and any negative value. The only value that seems potentially excluded is related to the horizontal asymptote. However, the statement claims the range excludes 2. Let's check if $f(x)$ can ever equal 2. We set rac{10}{x+5} = 2. Multiplying both sides by $x+5$, we get $10 = 2(x+5)$. Distributing the 2, we have $10 = 2x + 10$. Subtracting 10 from both sides gives $0 = 2x$, which means $x=0$. We already know that when $x=0$, $f(x)=2$. So, the function does output the value 2. This means 2 is in the range. Therefore, the statement that the range is (-ree, 2) igtriangleup (2, ree) is FALSE. The correct range, considering the horizontal asymptote at $y=0$, is actually (-ree, 0) igtriangleup (0, ree).

Statement C: The $x$-intercept is at $(-5,0)$.

To find the $x$-intercept, we set $f(x) = 0$ and solve for $x$. So we want to solve rac{10}{x+5} = 0. For a fraction to be equal to zero, the numerator must be zero, and the denominator must be non-zero. In this case, the numerator is 10, which is never zero. Therefore, there is no value of $x$ for which $f(x)=0$. This means there is no $x$-intercept. The statement claims the $x$-intercept is at $(-5,0)$. However, $x=-5$ is where the function is undefined due to division by zero (the vertical asymptote). So, this statement is FALSE. It's important to distinguish between where a function is undefined and where it crosses the $x$-axis.

Statement D: The end behavior is x o ree, f(x) o 0; x o -ree, f(x) o 0.

End behavior describes what happens to the function's output ($f(x)$) as the input ($x$) gets extremely large in either the positive or negative direction. Let's analyze this:

• As x o ree (as $x$ approaches positive infinity): The term $x+5$ also becomes a very large positive number. When you divide a constant (10) by a very large positive number, the result gets closer and closer to zero. So, $f(x) o 0$.
• As x o -ree (as $x$ approaches negative infinity): The term $x+5$ also becomes a very large negative number. When you divide a positive constant (10) by a very large negative number, the result gets closer and closer to zero (from the negative side). So, $f(x) o 0$.

This means the function has a horizontal asymptote at $y=0$. The statement accurately describes this behavior. Thus, this statement is TRUE. This is another correct answer!

Statement E: The function has a vertical asymptote at $x=-5$.

Vertical asymptotes occur where the denominator of a rational function is zero, and the numerator is non-zero at that point. For f(x)=rac{10}{x+5}, the denominator is $x+5$. Setting the denominator to zero, we get $x+5=0$, which gives $x=-5$. At $x=-5$, the numerator is 10, which is not zero. Therefore, there is indeed a vertical asymptote at $x=-5$. This statement is TRUE. This makes it our third correct answer!

Statement F: The function has a horizontal asymptote at $y=0$.

To find the horizontal asymptote for a rational function of the form f(x) = rac{a_n x^n + ext{...}}{b_m x^m + ext{...}}, we compare the degrees of the numerator ($n$) and the denominator ($m$).

In our function, f(x)=rac{10}{x+5}, the numerator is a constant (degree $n=0$), and the denominator is a linear term (degree $m=1$). Since the degree of the denominator ($m=1$) is greater than the degree of the numerator ($n=0$), the horizontal asymptote is always at $y=0$.

This confirms that the statement