Domain Of F(x) = (-x² + 6x - 5) / (x² - 7x + 12) And Integer Intersections Explained

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Hey guys! Today, we're diving deep into the fascinating world of functions, specifically focusing on how to determine the domain of a rational function and pinpoint its integer intersections. We'll be tackling the function f(x) = (-x² + 6x - 5) / (x² - 7x + 12), a classic example that perfectly illustrates these concepts. So, buckle up and let's get started!

Understanding the Domain of a Function

Let's start with the basics. What exactly is the domain of a function? In simple terms, the domain represents all the possible input values (x-values) that you can plug into the function without causing any mathematical mayhem. Think of it as the function's playground – it's the set of numbers where the function is allowed to play. For most functions, this is pretty straightforward, encompassing all real numbers. However, certain types of functions, like our rational function here, have specific restrictions we need to consider.

Rational functions, which are essentially fractions where both the numerator and denominator are polynomials, introduce a crucial caveat: we cannot divide by zero. Division by zero is a big no-no in the mathematical universe, leading to undefined results. Therefore, when determining the domain of a rational function, our primary mission is to identify any x-values that would make the denominator equal to zero. These values are the troublemakers, the ones we need to exclude from the domain.

To find these problematic x-values for our function, f(x) = (-x² + 6x - 5) / (x² - 7x + 12), we need to focus on the denominator: x² - 7x + 12. Our goal is to solve the equation x² - 7x + 12 = 0. This equation tells us exactly which x-values will cause the denominator to vanish, making the function undefined. We can solve this quadratic equation using several methods, such as factoring, completing the square, or the quadratic formula. Factoring is often the quickest and easiest route if the quadratic expression factors nicely, and in this case, it does!

We need to find two numbers that multiply to 12 and add up to -7. Those numbers are -3 and -4. Therefore, we can factor the denominator as (x - 3)(x - 4) = 0. Setting each factor equal to zero gives us the solutions x = 3 and x = 4. These are the values that make the denominator zero, and consequently, the values that are not in the domain of our function. So, the domain of f(x) is all real numbers except for 3 and 4. We can express this mathematically using interval notation as (-∞, 3) U (3, 4) U (4, ∞). This notation means that the function is defined for all x-values less than 3, between 3 and 4, and greater than 4.

Finding Integer Intersections

Now that we've conquered the domain, let's shift our focus to another important aspect of functions: intersections, specifically integer intersections. Intersections are the points where the graph of our function crosses the x-axis (x-intercepts) or the y-axis (y-intercept). These points provide valuable insights into the function's behavior and are crucial for sketching its graph. We're particularly interested in integer intersections, meaning the points where both the x and y coordinates are integers.

To find the x-intercepts, we need to determine where the function equals zero. In other words, we need to solve the equation f(x) = 0. For a rational function, this boils down to finding the zeros of the numerator, since a fraction is zero only if its numerator is zero (and the denominator is non-zero, of course!). So, we set the numerator of our function, -x² + 6x - 5, equal to zero: -x² + 6x - 5 = 0. To make things a bit easier, we can multiply both sides by -1 to get rid of the negative leading coefficient: x² - 6x + 5 = 0.

This is another quadratic equation, and just like before, we can try factoring it. We're looking for two numbers that multiply to 5 and add up to -6. Those numbers are -1 and -5. So, we can factor the quadratic as (x - 1)(x - 5) = 0. Setting each factor equal to zero gives us the solutions x = 1 and x = 5. These are the x-coordinates of our x-intercepts. To find the corresponding y-coordinates, we plug these x-values back into the original function, f(x). However, since we know the function is zero at these points (that's how we found them!), the y-coordinates are simply 0. Therefore, our x-intercepts are the points (1, 0) and (5, 0). Both of these points have integer coordinates, so they are indeed integer intersections.

Next, let's find the y-intercept. The y-intercept is the point where the graph crosses the y-axis, which occurs when x = 0. To find the y-coordinate of the y-intercept, we simply plug x = 0 into our function: f(0) = (-0² + 6(0) - 5) / (0² - 7(0) + 12) = -5 / 12. So, the y-intercept is the point (0, -5/12). The y-coordinate is a fraction, not an integer, so this is not an integer intersection.

Therefore, the only integer intersections for our function f(x) = (-x² + 6x - 5) / (x² - 7x + 12) are the x-intercepts (1, 0) and (5, 0).

Putting It All Together

Let's recap what we've learned. We successfully determined the domain of the rational function f(x) = (-x² + 6x - 5) / (x² - 7x + 12) by identifying the values that make the denominator zero and excluding them. We found that the domain is all real numbers except for 3 and 4. We then explored integer intersections, finding the x-intercepts by setting the numerator equal to zero and the y-intercept by setting x equal to zero. We discovered that the function has two integer x-intercepts, (1, 0) and (5, 0), but no integer y-intercept.

This exercise highlights the importance of understanding the restrictions imposed by rational functions and the techniques for finding key features like intercepts. By carefully analyzing the function's structure, we can gain valuable insights into its behavior and create a more complete picture of its graph. Remember, guys, practice makes perfect! The more you work with these concepts, the more comfortable you'll become in navigating the world of functions.

Diving Deeper into Rational Functions

Now that we've covered the basics of finding the domain and integer intersections, let's delve a little deeper into the fascinating world of rational functions. Understanding these functions is crucial in various areas of mathematics, including calculus and pre-calculus. Rational functions often appear in real-world applications, such as modeling population growth, chemical reactions, and electrical circuits. So, let's explore some more advanced concepts related to our function, f(x) = (-x² + 6x - 5) / (x² - 7x + 12).

Simplifying Rational Functions

One of the first things you might want to consider when dealing with a rational function is whether it can be simplified. Simplification involves factoring both the numerator and the denominator and then canceling out any common factors. This process can make the function easier to work with and provide further insights into its behavior.

Let's apply this to our function. We've already factored the denominator as (x - 3)(x - 4). Now, let's factor the numerator, -x² + 6x - 5. We already factored a slightly modified version of this in the previous section (x² - 6x + 5), which factored as (x - 1)(x - 5). Therefore, -x² + 6x - 5 factors as -(x - 1)(x - 5). So, our function can be rewritten as:

f(x) = -(x - 1)(x - 5) / (x - 3)(x - 4)

In this case, there are no common factors between the numerator and the denominator, so the function is already in its simplest form. However, if we had found a common factor, canceling it would have been an essential step in analyzing the function.

Vertical Asymptotes

We've already established that the values x = 3 and x = 4 are not in the domain of our function because they make the denominator zero. These values are special in another way: they correspond to vertical asymptotes. A vertical asymptote is a vertical line that the graph of the function approaches but never actually touches. They occur at the x-values that make the denominator zero but do not also make the numerator zero (after simplification).

In our case, since the function is already simplified and x = 3 and x = 4 make the denominator zero but not the numerator, these are indeed vertical asymptotes. This means that as x gets closer and closer to 3 or 4, the value of the function will either shoot up towards positive infinity or plummet down towards negative infinity. Vertical asymptotes are crucial for understanding the behavior of a rational function near these excluded x-values.

Horizontal Asymptotes

Besides vertical asymptotes, rational functions can also have horizontal asymptotes. A horizontal asymptote is a horizontal line that the graph of the function approaches as x goes to positive or negative infinity. In simpler terms, it describes the long-term behavior of the function. The existence and location of a horizontal asymptote depend on the degrees of the polynomials in the numerator and denominator.

There are three possible scenarios:

  1. Degree of numerator < Degree of denominator: In this case, the horizontal asymptote is the line y = 0 (the x-axis).
  2. Degree of numerator = Degree of denominator: In this case, the horizontal asymptote is the line y = (leading coefficient of numerator) / (leading coefficient of denominator).
  3. Degree of numerator > Degree of denominator: In this case, there is no horizontal asymptote. Instead, there might be a slant (or oblique) asymptote.

For our function, f(x) = (-x² + 6x - 5) / (x² - 7x + 12), the degree of the numerator (2) is equal to the degree of the denominator (2). Therefore, we fall into the second scenario. The leading coefficient of the numerator is -1, and the leading coefficient of the denominator is 1. So, the horizontal asymptote is the line y = -1 / 1 = -1.

This tells us that as x gets very large (positive or negative), the value of the function will approach -1. This is valuable information for sketching the graph of the function.

Sketching the Graph

By combining our knowledge of the domain, integer intersections, vertical asymptotes, and horizontal asymptotes, we can create a pretty accurate sketch of the graph of f(x) = (-x² + 6x - 5) / (x² - 7x + 12). We know:

  • The domain is all real numbers except 3 and 4.
  • The integer intersections (x-intercepts) are (1, 0) and (5, 0).
  • There are vertical asymptotes at x = 3 and x = 4.
  • There is a horizontal asymptote at y = -1.

With this information, we can sketch the basic shape of the graph. The graph will have three main sections, separated by the vertical asymptotes. We know the graph crosses the x-axis at 1 and 5, approaches the vertical asymptotes at 3 and 4, and approaches the horizontal asymptote at y = -1 as x goes to infinity. A more precise graph could be obtained by plotting additional points, but this gives us a solid foundation.

Further Exploration

We've covered a lot of ground today, guys, from determining the domain and integer intersections to understanding asymptotes and sketching the graph of a rational function. But the world of rational functions is vast and fascinating, and there's always more to explore!

Some topics you might want to investigate further include:

  • Slant asymptotes: What happens when the degree of the numerator is exactly one more than the degree of the denominator?
  • Holes: Sometimes, rational functions have "holes" in their graphs. When do these occur, and how do we find them?
  • Applications of rational functions: How are these functions used to model real-world phenomena?

By continuing to explore these concepts, you'll deepen your understanding of rational functions and their importance in mathematics and beyond. Keep practicing, keep exploring, and most importantly, keep asking questions! You've got this!