Summary

Good morning, class! Today, we will start by exploring what a rational function is. Can anyone tell me how a rational function is defined?

Is it something like a fraction with polynomials?

Exactly! A rational function is represented as $\frac{P(x)}{Q(x)}$, where both $P(x)$ and $Q(x)$ are polynomials, and importantly, $Q(x)$ must not be zero. This ensures that we avoid any undefined situations.

Why is it important that the Q does not equal zero?

Great question! Division by zero is undefined in mathematics, which is why those values must be excluded from the domain. Remember this, it's a key idea in understanding rational functions!

Can you give an example?

Certainly! An example of a rational function is $f(x) = \frac{2x+1}{x-3}$. Here, the denominator is $x-3$, which is zero when x equals 3, so x cannot be 3 in the domain.

So the domain excludes x = 3, right?

That's right! The domain would be all real numbers except 3. Let’s remember it by thinking of the acronym 'R-E-A-L', where the 'D' stands for Domain excluding invalid values.

Moving on, let’s determine the domain of some rational functions. Please give me an example function.

How about $f(x) = \frac{1}{x - 5}$?

Excellent choice! What do we do first to find the domain?

Set the denominator equal to zero! So, $x - 5 = 0$.

Correct! What does that solve to?

x equals 5!

Right again! The domain is all real numbers except for 5. Thus, it's expressed as $\mathbb{R} \backslash \{5\}$. Remember, every time we find a restriction, it’s key to note it down.

Can we try another example?

Of course! Let’s try $f(x) = \frac{x^2 - 4}{x + 2}$. Can anyone find the domain here?

Next, we will look at vertical and horizontal asymptotes. Who can remind me what vertical asymptotes are?

Those happen when the denominator equals zero after simplification?

Absolutely! They provide vital insights into the function’s behavior. For instance, if you have the function $g(x) = \frac{1}{x-2}$, where would you find the vertical asymptote?

At x = 2, since that's where the denominator is zero?

Well done! Now, let’s shift focus to horizontal asymptotes. How can we find them?

It depends on the degrees of the polynomials in the numerator and denominator, right?

Exactly! If the degree of the numerator is less than that of the denominator, we have a horizontal asymptote at $y = 0$.

And if they are equal?

Good question! Then the horizontal asymptote would be at $y = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}}$.

What about if the degree of the numerator is greater?

In that case, we might not find a horizontal asymptote! There could be an oblique asymptote instead.

Now, let’s look at intercepts. Who remembers how to find the x-intercept?

We set f(x) to zero and solve for P(x)!

Correct! And how about the y-intercept?

We set x to zero and evaluate f(0).

That's right! Let’s check an example: For the function $h(x) = \frac{x - 4}{x + 2}$, find the x-intercept.

Setting the numerator to zero gives me x = 4!

Great! Now what about the y-intercept?

If I plug in x = 0, I get $h(0) = \frac{-4}{2} = -2$.

Wonderful! Thus, the x-intercept is at (4, 0) and the y-intercept is at (0, -2). Always remember that intercepts help us understand how the graph interacts with the axes!

Finally, let’s put all these concepts together in graphing rational functions. What is our first step?

Identify the domain and vertical asymptotes?

Exactly! After that, what’s next?

We determine any horizontal or oblique asymptotes.

Correct! Then we find our intercepts. So let’s graph the function $f(x) = \frac{1}{x - 1}$. What do we identify first?

The vertical asymptote at x = 1!

Right! And what about horizontal asymptotes?

There’s a horizontal asymptote at y = 0 since the degree of the numerator is less than the denominator.

Fantastic! Lastly, let's plot the intercepts. Can we conclude how the graph behaves near our asymptotes?

The graph will approach the asymptotes but never touch them!

Correct! Fantastic teamwork today, class. Don't forget—understanding these concepts is crucial for solving real-world problems.