Find Domain, Intercepts & Graph Rational Functions

Hey guys! Today, we're diving deep into the wild world of rational functions. Specifically, we're going to tackle a juicy problem: finding the domain, the x-and y-intercepts, and then graphing a function, making sure to label all those pesky asymptotes. The function we're dissecting is f(x)=rac{3 x^2-8 x-3}{x-3}. Now, before we get our hands dirty with the graphing, the first big question on everyone's mind is: What is the domain of the function? This is crucial, folks, because the domain tells us all the possible x-values that our function can actually handle. Think of it as the VIP list for our function – only certain x-values get to party.

So, how do we figure out the domain? For rational functions, which are basically fractions where the numerator and denominator are polynomials, the golden rule is: never divide by zero! Our function f(x)=rac{3 x^2-8 x-3}{x-3} has a denominator of $(x-3)$. If we set this denominator to zero, we get $x-3=0$, which means $x=3$. So, our function throws a massive tantrum if we try to plug in $x=3$. It's undefined at that point. Therefore, the domain of our function is all real numbers except for $x=3$. We can write this in interval notation as $(-\infty, 3) \cup (3, \infty)$. This means our function is happy to take any x-value from negative infinity up to, but not including, 3, and then any x-value from just after 3 all the way to positive infinity. It's like a road with a bridge out at $x=3$ – you can drive on either side, but you can't cross at the bridge.

Finding the Domain: The Foundation

The domain of a function is a fundamental concept in mathematics, representing the set of all possible input values (usually 'x' values) for which the function is defined and produces a real output. For rational functions, which are ratios of two polynomials, this definition takes on a specific importance. We're dealing with expressions in the form of $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials. The critical condition that determines the domain of such a function is that the denominator, $Q(x)$, cannot be equal to zero. Division by zero is mathematically undefined, and thus, any value of 'x' that makes the denominator zero must be excluded from the function's domain. In our specific case, f(x)=rac{3 x^2-8 x-3}{x-3}, the denominator is $Q(x) = x-3$. To find the values of 'x' that are not in the domain, we set the denominator equal to zero: $x-3 = 0$. Solving this simple linear equation gives us $x = 3$. This tells us that $x=3$ is the only value that makes our function undefined. Consequently, the domain encompasses all other real numbers. We express this exclusion using interval notation, which is a standard way to describe sets of numbers on the number line. The domain is written as $(-\infty, 3) \cup (3, \infty)$. This notation signifies two separate intervals: all real numbers less than 3 (from negative infinity up to, but not including, 3) and all real numbers greater than 3 (from just above 3 up to positive infinity). The 'union' symbol ($\cup$) indicates that the domain includes all numbers in both of these intervals. Understanding the domain is the very first step because it dictates the boundaries within which we can analyze the function's behavior, including where it might have vertical asymptotes or holes.

Unearthing the Intercepts: Where the Graph Meets the Axes

Now that we've locked down the domain, let's talk intercepts! These are super important because they tell us where our graph crosses the x-axis (x-intercepts) and the y-axis (y-intercept). They give us key reference points on our graph.

To find the y-intercept, we simply plug in $x=0$ into our function. Remember, the y-intercept is the point where the graph crosses the y-axis, and any point on the y-axis has an x-coordinate of 0. So, let's do that:

$f(0) = \frac{3(0)^2 - 8(0) - 3}{(0)-3} = \frac{0 - 0 - 3}{0 - 3} = \frac{-3}{-3} = 1$

Boom! The y-intercept is at the point (0, 1). Easy peasy, right?

Next up, the x-intercepts. These are the points where the graph crosses the x-axis, meaning the y-value (or $f(x)$) is zero. To find these, we set the numerator of our rational function equal to zero, provided that the value of x doesn't also make the denominator zero (because we already know those values are off-limits). So, we set the numerator $3x^2 - 8x - 3$ equal to 0:

$3x^2 - 8x - 3 = 0$

This is a quadratic equation, and we can solve it by factoring or using the quadratic formula. Let's try factoring. We're looking for two numbers that multiply to $(3)(-3) = -9$ and add up to $-8$. Those numbers are $-9$ and $1$. So, we can rewrite the middle term:

$3x^2 - 9x + 1x - 3 = 0$

Now, we can factor by grouping:

$3x(x - 3) + 1(x - 3) = 0$

$(3x + 1)(x - 3) = 0$

This gives us two potential solutions: $3x + 1 = 0$ or $x - 3 = 0$. Solving these, we get $x = -1/3$ and $x = 3$.

Hold up! Remember our domain? We established that $x=3$ is not in the domain because it makes the denominator zero. So, even though setting the numerator to zero gave us $x=3$, it's not a valid x-intercept. This is a crucial point in rational functions – sometimes a value that makes the numerator zero also makes the denominator zero. When that happens, it usually indicates a hole in the graph, not an x-intercept.

Therefore, the only valid x-intercept comes from $x = -1/3$. So, the x-intercept is at the point (-1/3, 0).

Analyzing Asymptotes: The Invisible Boundaries

Asymptotes are like invisible lines that our function's graph gets really, really close to but never actually touches. They help us understand the function's behavior as x gets very large or very small, or when x approaches a value that makes the function undefined.

There are three main types of asymptotes: vertical, horizontal, and slant (or oblique). For our function f(x)=rac{3 x^2-8 x-3}{x-3}, let's break them down.

Vertical Asymptotes: Where the Function Goes Wild

Vertical asymptotes typically occur at the x-values that make the denominator of a rational function equal to zero, after we've simplified the function. We already found that $x=3$ makes the denominator zero. Let's quickly check if this factor cancels out with anything in the numerator. We factored the numerator earlier as $(3x+1)(x-3)$. So, our function can be written as:

$f(x) = \frac{(3x+1)(x-3)}{(x-3)}$

Notice that the $(x-3)$ term appears in both the numerator and the denominator. This means we can cancel it out, but only for $x \neq 3$. After canceling, the simplified function is:

$f(x) = 3x + 1$, for $x \neq 3$

This simplification is HUGE! It tells us that our graph looks exactly like the line $y = 3x + 1$, except at $x=3$. Because the $(x-3)$ factor was canceled out, $x=3$ doesn't result in a vertical asymptote. Instead, it creates a hole in the graph at $x=3$. To find the y-coordinate of this hole, we plug $x=3$ into the simplified function: $y = 3(3) + 1 = 9 + 1 = 10$. So, there's a hole at the point (3, 10).

This is a common pitfall, guys! Always simplify your rational function first. If a factor cancels, it's a hole, not a vertical asymptote. If a factor in the denominator doesn't cancel, that's where you'll find a vertical asymptote.

In this particular function, since all factors canceled out, there are no vertical asymptotes. This is a key takeaway from simplifying rational functions!

Horizontal Asymptotes: The Long-Term Trend

Horizontal asymptotes describe the behavior of the function as $x$ approaches positive or negative infinity ($x \to \pm\infty$). To find them, we compare the degrees of the numerator and the denominator polynomials.

Our original function is f(x)=rac{3 x^2-8 x-3}{x-3}.

• Degree of the numerator (highest power of x): 2
• Degree of the denominator (highest power of x): 1

There are three rules for horizontal asymptotes based on these degrees:

1. If degree of numerator < degree of denominator: The horizontal asymptote is $y=0$.
2. If degree of numerator = degree of denominator: The horizontal asymptote is $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$.
3. If degree of numerator > degree of denominator: There is no horizontal asymptote. Instead, there might be a slant (oblique) asymptote.

In our case, the degree of the numerator (2) is greater than the degree of the denominator (1). Therefore, according to rule #3, there is no horizontal asymptote.

Slant (Oblique) Asymptotes: When the Graph Gets Linear

Since the degree of the numerator is exactly one greater than the degree of the denominator, our function has a slant asymptote. A slant asymptote is a linear line ($y = mx + b$) that the graph of the function approaches as $x \to \pm\infty$. We find the equation of the slant asymptote by performing polynomial long division or synthetic division of the numerator by the denominator.

Let's use polynomial long division for f(x)=rac{3 x^2-8 x-3}{x-3}:

        3x   + 1
      ____________
x - 3 | 3x^2 - 8x - 3
      -(3x^2 - 9x)
      ____________
            x  - 3
           -(x  - 3)
           ________
                0

The division results in $3x + 1$ with a remainder of $0$. This means that $f(x) = 3x + 1$ for all $x \neq 3$. The slant asymptote is the quotient part of the division before the remainder. In this specific case, because the remainder is $0$, the function is the line $y = 3x + 1$. However, we must remember the hole at $x=3$. The slant asymptote is the line that the graph approaches. In cases where the remainder is not zero, the slant asymptote is $y = \text{quotient}$. Here, the slant asymptote is the line $y = 3x + 1$. It's important to note that the function will not cross its slant asymptote if the remainder is not zero. In this special case, the function actually is the line $y=3x+1$, with a hole removed.

Graphing the Function: Putting It All Together

Now for the grand finale – sketching the graph! We've gathered all the essential pieces:

• Domain: $(-\infty, 3) \cup (3, \infty)$
• y-intercept: (0, 1)
• x-intercept: (-1/3, 0)
• Hole: at (3, 10)
• No Vertical Asymptotes
• No Horizontal Asymptotes
• Slant Asymptote: $y = 3x + 1$

Given that our simplified function is $f(x) = 3x + 1$ (for $x \neq 3$), the graph is essentially the straight line $y = 3x + 1$. This is a linear function with a y-intercept of 1 and a slope of 3.

To graph this line:

1. Plot the y-intercept: Mark the point (0, 1) on your graph.
2. Use the slope: From (0, 1), go up 3 units and right 1 unit to find another point (1, 4). Or, go down 3 units and left 1 unit to find (-1, -2). Draw a straight line through these points.
3. Mark the hole: Now, the crucial part for rational functions. Locate the point on the line where $x=3$. Using the simplified function, $y = 3(3) + 1 = 10$. So, the hole is at (3, 10). Draw an open circle at (3, 10) on the line to indicate that this point is not included in the graph of the function.

So, the graph is a straight line $y = 3x + 1$, but with a tiny gap, a hole, at the point (3, 10). It doesn't have any funny bends or curves because the only factor that could have caused a vertical asymptote or a hole that wasn't removable (like a simple pole) was canceled out. The slant asymptote $y=3x+1$ is actually the line itself, which is a special case.

Key Takeaways:

• Simplify first! Always cancel common factors in the numerator and denominator. If a factor cancels, it's a hole. If a factor in the denominator remains, it's a vertical asymptote.
• Domain restrictions persist: Even after canceling, the original restriction ($x \neq 3$ in our case) still applies, leading to a hole.
• Compare degrees for H.A.: The relationship between the degrees of the numerator and denominator dictates horizontal or slant asymptotes.

Keep practicing these steps, guys, and you'll be graphing rational functions like a pro in no time! Let me know if you have any questions!