Find Vertical & Horizontal Asymptotes Of F(x) = (x^2+x-6)/(x^3-1)

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Hey guys! Let's dive into a super important topic in calculus and precalculus: asymptotes. Specifically, we're going to figure out how to find the vertical and horizontal asymptotes of the function f(x) = (x^2 + x - 6) / (x^3 - 1). Asymptotes might sound intimidating, but they're actually pretty cool once you understand the key concepts. Think of them as invisible lines that the graph of a function approaches but never quite touches or crosses.

Understanding Asymptotes

Before we jump into solving our specific problem, let's make sure we're all on the same page about what asymptotes are. There are basically two types we're focusing on today: vertical and horizontal asymptotes. Think of vertical asymptotes as the "no-go zones" for the x-values of our function. These occur where the function becomes infinitely large (positive or negative), which often happens when the denominator of a rational function equals zero. Basically, if plugging in a certain x-value makes the bottom of our fraction zero, there's a good chance we've found a vertical asymptote. Then, horizontal asymptotes tell us what happens to the function as x gets really, really big (approaches positive infinity) or really, really small (approaches negative infinity). They describe the end behavior of our function, showing us if the function levels out to a certain y-value as we move far to the left or right on the graph. This often involves looking at the degrees of the polynomials in the numerator and denominator.

Step-by-Step: Finding the Asymptotes

Okay, enough theory! Let's get our hands dirty and find the asymptotes for f(x) = (x^2 + x - 6) / (x^3 - 1). We'll tackle vertical asymptotes first, then move on to the horizontal ones.

Finding Vertical Asymptotes

The golden rule for vertical asymptotes: find where the denominator equals zero. Remember, we can't divide by zero in math (it's a big no-no!), so any x-values that make the denominator zero are potential locations for vertical asymptotes. So, for our function, we need to solve the equation x^3 - 1 = 0. This looks like a cubic equation, but it's a special one that we can factor. This can be factored using the difference of cubes formula: a^3 - b^3 = (a - b)(a^2 + ab + b^2). In our case, a = x and b = 1. Applying the formula, we get: x^3 - 1 = (x - 1)(x^2 + x + 1). Now we have (x - 1)(x^2 + x + 1) = 0. To solve this, we set each factor equal to zero. So, the first factor gives us x - 1 = 0, which means x = 1. The second factor, x^2 + x + 1 = 0, is a quadratic equation. We could try to factor it, but it doesn't factor nicely. So, let's use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Here, a = 1, b = 1, and c = 1. Plugging those values in, we get x = (-1 ± √(1^2 - 4 * 1 * 1)) / (2 * 1) = (-1 ± √(-3)) / 2. Notice that we have a negative number under the square root. That means the solutions are complex numbers (involving i), not real numbers. Vertical asymptotes happen at real x-values, so these complex solutions don't give us any vertical asymptotes. Thus, we have only one vertical asymptote at x = 1. But before we declare victory, we need to do one more check. We need to make sure that this value doesn't also make the numerator zero. If it does, we might have a hole in the graph instead of a vertical asymptote. Let's keep going!

Checking for Holes

Okay, we found a potential vertical asymptote at x = 1, but we need to be sure it's not a hole. Holes happen when a factor cancels out from both the numerator and denominator. So, let's factor the numerator of our function: x^2 + x - 6. We're looking for two numbers that multiply to -6 and add up to 1. Those numbers are 3 and -2. So, we can factor the numerator as (x + 3)(x - 2). Now, our function looks like this: f(x) = ((x + 3)(x - 2)) / ((x - 1)(x^2 + x + 1)). Do you see any factors that cancel out? Nope! The x - 1 factor in the denominator doesn't appear in the numerator. This means we definitely have a vertical asymptote at x = 1.

Finding Horizontal Asymptotes

Now, let's hunt down those horizontal asymptotes! Remember, horizontal asymptotes describe what happens to f(x) as x approaches positive or negative infinity. There's a handy rule for rational functions like ours that makes this easier. The rule is about comparing the degrees (the highest power of x) of the numerator and denominator.

  • Case 1: Degree of numerator < Degree of denominator: If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is always y = 0.
  • Case 2: Degree of numerator = Degree of denominator: If the degrees are equal, the horizontal asymptote is y = (leading coefficient of numerator) / (leading coefficient of denominator).
  • Case 3: Degree of numerator > Degree of denominator: If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (but there might be a slant or oblique asymptote, which is a topic for another day!).

So, let's apply this to our function, f(x) = (x^2 + x - 6) / (x^3 - 1). The degree of the numerator (x^2 + x - 6) is 2. The degree of the denominator (x^3 - 1) is 3. Since 2 is less than 3 (numerator degree < denominator degree), we're in Case 1! This means our horizontal asymptote is y = 0. Easy peasy!

The Final Answer

Alright, we've done it! We found both the vertical and horizontal asymptotes for f(x) = (x^2 + x - 6) / (x^3 - 1). The vertical asymptote is x = 1, and the horizontal asymptote is y = 0. So, the correct answer from your options is B. vertical asymptote: x=1, horizontal asymptote: y=0.

Graphing to Visualize (Optional)

It's always a good idea to visualize what we've found. If you have access to a graphing calculator or a website like Desmos or GeoGebra, try plotting the function f(x) = (x^2 + x - 6) / (x^3 - 1). You'll see how the graph gets super close to the lines x = 1 (vertically) and y = 0 (horizontally) but never actually touches them. This visual confirmation helps solidify your understanding of asymptotes. If you want to dive deeper, you can graph x = 1 and y = 0 as dashed lines, which will emphasize the asymptotes even more. Remember, graphing is a powerful tool for understanding functions!

Key Takeaways

  • Vertical Asymptotes: Find where the denominator equals zero (and make sure it's not a hole!).
  • Horizontal Asymptotes: Compare the degrees of the numerator and denominator.

Practice Makes Perfect

Finding asymptotes is a skill that gets better with practice. Try working through more examples with different rational functions. Vary the complexity of the polynomials in the numerator and denominator to challenge yourself. You'll start to recognize patterns and become a pro at finding asymptotes in no time!

Final Thoughts

So there you have it! Finding vertical and horizontal asymptotes might seem tricky at first, but by following these steps and understanding the core concepts, you can master them. Remember to factor, compare degrees, and visualize with graphs whenever possible. Keep practicing, and you'll be an asymptote expert in no time. Happy graphing, everyone!