Graph the rational function. f ( x ) = x + 3 x 2 + 5 x + 6 f\left(x\right)=\frac{x+3}{x^2+5x+6} f ( x ) = x 2 + 5 x + 6 x + 3

In this problem, we're asked to graph the rational function x+3overxsquared+5x+6. So let's go ahead and get started with step 1 in factor and find our domain. So in order to factor here, we need to look at our numerator and our denominator. In my numerator, I can't factor that anymore. It stays as x plus 3, but my denominator can be factored into x+3 timesx+2. Now, to find our domain, we wanna set our denominator equal to 0. So I'm gonna take both of those factors, x +3 and x+2 and set them both equal to 0. Now subtracting 3 from both sides here, I end up with x is equal to negative 3. And then subtracting 2 from both sides, I end up with x is equal to negative 2. And that makes my domain restriction x such that x cannot be equal to negative 3 or negative 2. Okay. Let's move on to finding the wholes of our function. So we want to set any common factors equal to 0 here, and it looks like we do have a common factor. We have x plus 3, so I'm going to set x plus 3 equal to 0. Since we just did that up here, I know that this is going to end up giving me x is equal to negative 3, so that tells me I have a hole at negative 3. Now we're gonna keep that in mind as we continue to graph our function. Now we wanna put our function in lowest terms and continue doing every other calculation from our function in lowest terms. So canceling out that common factor, I'm going to end up with the function f of x is equal to 1 over x plus 2 and this is my function in lowest terms. We'll use this for all future calculations. Alright, let's move on to step 3 and find our x intercepts. We wanna set our numerator equal to 0 here, but if I look at my function in lowest terms, I would be setting 1 equal to 0 here, which is not true. This is not a true statement. So I actually don't have an x intercept at all. Whenever you get a false statement in trying to solve for something, it means that you just don't have one. So here we don't have an x intercept. Let's move on to finding our y intercept. Now to find our y intercept, we want to compute f of 0 by simply plugging 0 into our function. So taking our function 1 over x plus 2, plugging 0 in for x here. So one over 0+2 gives me 1 half. So my y intercept is at 1 half. Let's go ahead and plot that on our graph. Right here at 1 half. Okay, now we can go ahead and move on to our next step in finding our asymptotes. So starting with step number 5, our vertical asymptote, asymptote, we want to set our denominator equal to 0. Now, remember, we're working with our function in lowest terms here, so I'm going to use this function here, 1 over x+2, in order to find my vertical asymptote here. So my denominator is x+2, and that's equal to 0. So subtracting 2 from both sides, I end up with x is equal to negative 2. Now this tells me that I need to plot my vertical asymptote negative 2. So let's go ahead and do that on our graph here using a dash line at negative 2. Now we want to find our horizontal asymptotes. So let's go ahead and do that as well by looking at the degree of our numerator and our denominator. Now the degree of my numerator is 0 because it's just a one there, and then the degree of my denominator is 1. So that tells me that the degree of my numerator is less than the degree of my denominator, which means that I simply have a horizontal asymptote at y equals 0, which we can go ahead and plot on our graph. So plotting my horizontal asymptote right on top of that x axis. It might be a little bit hard to see, but it's right there on my x axis, and we're done finding our asymptotes. Now we just want to determine our intervals and plot a point in each of them. So let's go ahead and take a look at our graph and determine our intervals. Remember, we're looking at our graph from left to right, and we're finding intervals based on our vertical asymptotes and x intercepts. So going from this side from negative infinity until I reach the first known thing, The first thing I know is this vertical asymptote here. So my very first interval is from negative infinity to negative 2. Then from negative 2 until my next known thing, I actually don't know anything else here. So this is going to be my other interval and it's my only other interval. So from negative two to infinity. Now remember that you can always choose to plot extra points if you feel like you don't have a great picture of what's going on. Feel free to always choose extra points if you need to. But with these two intervals in mind, let's choose our values for x within each of these intervals. So from negative infinity to negative 2, I know that my hole is at x equals negative 3, so I actually figure out exactly where that point is. So I'm gonna go ahead and choose negative 3 inside of that interval. Then from negative 2 to infinity, I'm gonna go ahead and choose the point negative one. And then if you want to, you can always choose some extra points, even more than just one. So now that we have chosen these values for x, let's go ahead and plug them back into our function. So starting with f of negative 3, plugging that into 1 over negative 3 plus 2, this ends up giving me 1 over negative 1, which is simply negative one. So that's my first ordered pair negative 3 negative 1, and then plugging a negative one into my function f of negative one, this ends up giving me a one over negative one plus 2, which is one over 1, which is going to give me positive one. So this other ordered pair is negative 1, positive 1. Let's go ahead and plot those on our graph here starting with negative 3, negative 1. So negative three, negative one. I know that that's a hole there, actually, so I'm gonna go ahead and plot it as an open circle. Then for my other point, I have negative 1, positive 1. So coming back up to my graph, plotting that point here, negative 1, positive one right here. Okay. Now that we have these extra points, we can go ahead and proceed to our last step here, which is simply to connect everything and draw approaching our asymptotes. So let's go ahead and come back up to our graph and complete graphing our rational function here. So on this side, I want to approach my asymptote here and then connect with that y intercept approaching my asymptote on the other side. Then for my hole here, I know that I need to approach on this side, and then I approach on the other side. Now if you didn't choose to graph exactly where your hole was in those intervals, that's okay because you would have graphed another point and you can always plot the hole afterwards just with an open circle. Make sure you indicate your hole with an open circle no matter where it is. Okay, now we have our final graph of this irrational function and we're all done. Thanks for watching. I'll see you in the next video.

5. Rational Functions

Graphing Rational Functions

College Algebra

5. Rational Functions

Graphing Rational Functions

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Multiple Choice

Graph the rational function.
$f\left(x\right)=\frac{x+3}{x^2+5x+6}$

Graph of f(x) = (x + 3) / (x^2 + 5x + 6) showing vertical and horizontal asymptotes.

Graph of f(x) = (x + 3) / (x^2 + 5x + 6) with two vertical asymptotes and a horizontal asymptote.

Graph of f(x) = (x + 3) / (x^2 + 5x + 6) showing a vertical asymptote at x = -3.

Graph of f(x) = (x + 3) / (x^2 + 5x + 6) with a vertical asymptote at x = -3 and a horizontal asymptote.

Verified step by step guidance

Identify the rational function: $ f(x) = \frac{x+3}{x^2+5x+6} $.

Factor the denominator: $ x^2 + 5x + 6 = (x+2)(x+3) $.

Determine the vertical asymptotes by setting the denominator equal to zero: $ x+2=0 $ and $ x+3=0 $, giving $ x = -2 $ and $ x = -3 $.

Find the horizontal asymptote by comparing the degrees of the numerator and denominator. Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is $ y = 0 $.

Identify any holes in the graph by checking for common factors in the numerator and denominator. Since $ x+3 $ is a common factor, there is a hole at $ x = -3 $.