y-intercept

Today, we're discussing an essential concept in graphing rational functions: the y-intercept. Can someone tell me what we mean when we say 'y-intercept'?

I think it's the point where the graph meets the y-axis?

Exactly! The y-intercept is where the function crosses the y-axis. Now, how do we find this point?

Do we set x to zero?

Yes! We substitute x = 0 into our function. For example, if we have $f(x) = \frac{x-4}{x+2}$, what would be $f(0)$?

That would be $f(0) = \frac{0-4}{0+2} = \frac{-4}{2} = -2$.

Great job! So the y-intercept is at (0, -2). Let’s remember this process: to find the y-intercept, always substitute x with zero.

Now, let’s consider another function: $f(x) = \frac{x^2 - 1}{x - 1}$. Can anyone tell me how to find the y-intercept?

We just plug in x = 0, right?

Exactly! Let's do it together. What do we get for $f(0)$?

So that would be $f(0) = \frac{0^2 - 1}{0 - 1} = \frac{-1}{-1} = 1$.

Well done! The y-intercept for this function is at (0, 1). Let’s summarize: when evaluating for the y-intercept, always remember to check if the denominator is not zero at that point as well.

Now that we’re comfortable finding the y-intercept, I want to challenge you with a quick practice question! Find the y-intercept for $f(x) = \frac{2x + 3}{x - 2}$. What do you think?

We set x = 0, so it would be $f(0) = \frac{2(0) + 3}{0 - 2} = \frac{3}{-2} = -1.5$.

Excellent! The y-intercept is at (0, -1.5). And why is checking the denominator important?

To make sure it doesn’t equal zero, or else the function would be undefined.

Exactly! Always ensure that the function is defined at the point of evaluation. Let’s keep reinforcing this learning as we move forward.