Like

Report

Find the domain and sketch the graph of the function.

$ g(t) = \dfrac{t^2 - 1}{t + 1} $

$(-\infty,-1) \cup(-1, \infty)$

You must be signed in to discuss.

Johns Hopkins University

Campbell University

Oregon State University

University of Nottingham

here we have a rational function. Let's find the domain and sketch the graph. So if you see a rational function, what you want to focus on for the domain is that you can't have zero in the denominator because that is undefined. And we're only looking for values of T that would make that would give us real number outputs. And so that means that T cannot be negative one So we can express our domain like this. Negative Infinity to negative one union, negative one to infinity. That shows all real numbers except negative one. Okay, Now for the graph, what I'm going to do is modify how this function looks a little bit. I'm going to factor the numerator into T plus one times T minus one. Now I'm going to cancel the T plus one from the top of the bottom, but that doesn't mean it's completely gone. That means, actually that there is a hole in the graph. If you've studied rational functions, you've probably seen this before. There's a hole in the graph there, and the whole is at negative one. That's a value that made that canceled factor. Zero. What's the Y? Coordinate of the whole. If you take negative one and substituted into what we have left in the function the T minus one, you end up with negative, too. So there's a hole in the graph at negative one. Negative, too. So now that we've acknowledged that what we have left will look like the graph of G F T equals T minus one. That's what we have remaining after we cancel the T plus one. And that would be a line with a Y intercept of negative one and a slope of one so we can draw that line. We just have to make it have a hole in it at the point. Negative one negative, too. So there's a whole on. Then there's the rest of the line.