Graphing Functions: The Case Of F(x) = (5 - 5x^2) / X^2

Hey guys! Today, we're diving deep into the awesome world of functions and graphing. Specifically, we're going to tackle a question that might seem a bit tricky at first glance: Which graph represents the function $f(x)=\frac{5-5 x^2}{x^2}$? Don't worry, we'll break it down step by step, making sure you understand every little bit. We'll explore the key characteristics of this function, like its domain, asymptotes, intercepts, and end behavior, which are crucial for accurately sketching its graph. By the end of this, you'll be a pro at identifying the correct graphical representation of this type of rational function.

Understanding the Function $f(x)=\frac{5-5 x^2}{x^2}$

Alright, let's get up close and personal with our function, $f(x)=\frac{5-5 x^2}{x^2}$. First things first, understanding the domain is super important. The domain tells us all the possible x-values that the function can take. For rational functions (that's functions with a numerator and a denominator, like ours), the key is to avoid dividing by zero. So, we need to find out where the denominator, $x^2$, equals zero. Setting $x^2 = 0$, we find that $x=0$. This means that $x=0$ is not in the domain of our function. The domain is all real numbers except for 0. This is a critical piece of information because it immediately tells us that there will be a break or a gap in our graph at $x=0$. This break often manifests as a vertical asymptote.

Now, let's talk about asymptotes. These are like invisible lines that the graph approaches but never quite touches. We've already hinted at a vertical asymptote at $x=0$ because of the domain restriction. To confirm this, we can look at the behavior of the function as x gets really close to 0. If the function's value shoots off to positive or negative infinity as x approaches 0 from the left or right, then we definitely have a vertical asymptote there. In our case, as $x \to 0$, the denominator $x^2$ gets extremely small, while the numerator $5-5x^2$ approaches 5. Dividing a number close to 5 by a very, very small positive number ($x^2$ is always positive when $x \neq 0$) results in a very large positive number. So, $f(x) \to \infty$ as $x \to 0$ from both the left and the right. This confirms our vertical asymptote at $x=0$.

Next up are horizontal asymptotes. These describe the behavior of the function as x goes towards positive or negative infinity. To find them for rational functions, we compare the degrees of the polynomial in the numerator and the denominator. Our function is $f(x)=\frac{5-5 x^2}{x^2}$. The degree of the numerator ($5-5x^2$) is 2, and the degree of the denominator ($x^2$) is also 2. When the degrees are the same, the horizontal asymptote is the line $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$. In our case, the leading coefficient of the numerator is -5, and the leading coefficient of the denominator is 1. Therefore, the horizontal asymptote is $y = \frac{-5}{1} = -5$. This tells us that as x gets very large (positive or negative), the graph of our function will get closer and closer to the line $y=-5$.

Let's not forget about x-intercepts and y-intercepts. An x-intercept is a point where the graph crosses the x-axis, meaning $f(x)=0$. To find x-intercepts, we set the numerator equal to zero: $5 - 5x^2 = 0$. Solving for x, we get $5 = 5x^2$, so $x^2 = 1$. This gives us $x = 1$ and $x = -1$. So, our function has x-intercepts at $(1, 0)$ and $(-1, 0)$. A y-intercept is a point where the graph crosses the y-axis, meaning $x=0$. However, we already established that $x=0$ is not in the domain of our function. Therefore, there is no y-intercept. This is consistent with having a vertical asymptote at $x=0$.

Finally, let's consider the end behavior. We already know the horizontal asymptote is $y=-5$. This means as $x \to \infty$, $f(x) \to -5$, and as $x \to -\infty$, $f(x) \to -5$. We can also analyze the behavior of the function around the vertical asymptote. As we saw, $f(x)$ goes to positive infinity as $x$ approaches 0 from either side. This symmetry around the y-axis (because $f(-x) = \frac{5-5(-x)^2}{(-x)^2} = \frac{5-5x^2}{x^2} = f(x)$, meaning the function is even) is a key feature. So, we expect a graph that comes down from the horizontal asymptote, hits the x-axis, goes up towards infinity, and on the other side, comes down from infinity, hits the x-axis, and goes back towards the horizontal asymptote.

Simplifying the Function for Better Insight

Sometimes, simplifying the function can reveal even more about its behavior. Let's rewrite $f(x)=\frac{5-5 x^2}{x^2}$ by dividing each term in the numerator by the denominator: $f(x) = \frac{5}{x^2} - \frac{5x^2}{x2}$ $f(x) = \frac{5}{x^2} - 5$ This simplified form, $f(x) = \frac{5}{x^2} - 5$, is incredibly useful! It directly shows us that the graph of $f(x)$ is essentially the graph of $y = \frac{5}{x^2}$ shifted down by 5 units. The graph of $y = \frac{1}{x^2}$ is a well-known U-shaped curve that is symmetric about the y-axis, with vertical asymptotes at $x=0$ and horizontal asymptotes at $y=0$. Multiplying by 5, $y = \frac{5}{x^2}$, just stretches this graph vertically, but the asymptotes and symmetry remain the same. Now, when we subtract 5, we shift the entire graph downwards. This means the vertical asymptote remains at $x=0$, but the horizontal asymptote shifts from $y=0$ to $y=-5$. This simplified form makes it much easier to visualize the final graph. We can also see that $f(x) = -5$ when $\frac{5}{x^2} = 0$, which is impossible, but as $x \to \pm \infty$, $\frac{5}{x^2} \to 0$, so $f(x) \to -5$. This confirms our horizontal asymptote. For the x-intercepts, setting $f(x)=0$ gives $\frac{5}{x^2} - 5 = 0$, so $\frac{5}{x^2} = 5$, which means $x^2 = 1$, and thus $x = \pm 1$. This again confirms our x-intercepts at $(1,0)$ and $(-1,0)$. This simplified form is a fantastic tool for understanding the transformations applied to a basic function and predicting the final graph's characteristics.

Identifying the Correct Graph

Now that we've thoroughly analyzed the function $f(x)=\frac{5-5 x^2}{x^2}$, we can use these characteristics to identify the correct graph among a set of options. Remember, guys, the key features we're looking for are:

1. Vertical Asymptote: There must be a vertical asymptote at $x=0$. This means the graph should approach infinity (or negative infinity) on either side of the y-axis and never touch it.
2. Horizontal Asymptote: There must be a horizontal asymptote at $y=-5$. The graph should approach the line $y=-5$ as x goes to positive or negative infinity.
3. X-intercepts: The graph must cross the x-axis at $x=1$ and $x=-1$. These are the points $(1, 0)$ and $(-1, 0)$.
4. No Y-intercept: The graph should not cross the y-axis.
5. Symmetry: The graph should be symmetric with respect to the y-axis (since the function is even).

Let's consider what different graphs might show. If a graph has a vertical asymptote at a different location, or no vertical asymptote at all, it's incorrect. If it has a horizontal asymptote at a different y-value (like $y=0$ or $y=5$), it's also wrong. Crucially, the graph must pass through the points $(1, 0)$ and $(-1, 0)$. If it misses these points or crosses the x-axis elsewhere, it's not our function.

Imagine you're looking at a few potential graphs. One graph might show a U-shape opening upwards with a vertical asymptote at $x=0$, but it might have a horizontal asymptote at $y=0$. This would be the graph of $y = \frac{5}{x^2}$. Another graph might look correct in terms of asymptotes but might not pass through the correct x-intercepts. You need a graph that simultaneously satisfies all the conditions we've identified.

Visualizing the Graph's Shape

Given our analysis, the graph of $f(x)=\frac{5-5 x^2}{x^2}$ will have two distinct branches, separated by the vertical asymptote at $x=0$. For $x > 0$, the graph will approach $y=-5$ as $x \to \infty$, pass through $(1, 0)$, and then rise sharply towards positive infinity as $x \to 0^+$. Similarly, for $x < 0$, the graph will approach $y=-5$ as $x \to -\infty$, pass through $(-1, 0)$, and then rise sharply towards positive infinity as $x \to 0^-$. The overall shape will resemble two separate