Welcome back, we're moving fast through this course. This week, we've explored rational functions. We learned about how to sketch their graphs, find their domain and range, and any vertical and horizontal asymptotes. In our investigations, we also encountered slant asymptotes. Finally, we began looking at graphing and answer questions about piecewise function. As usual, we have a few problems from the recent work to summarize. Let's look at those now.
To start our review, we'll look at the graph with the function f of x equals 1 over x squared minus 7x plus 12. This graph has lots of interesting features, but we'll start with some you're familiar with from earlier units just to make sure you haven't forgotten about those. We always want to know the x and y-intercepts. Does it have x and y-intercepts? If so, what are they? If not, type none. Separate multiple answers with semi-colons as necessary.
As you know from earlier in the course, you can find the y-intercept by substituting in x equals 0. So, the y-intercept is simply f of 0 which as you can see gives us 1 over 12. Are there any x-intercepts? We have to solve f of x equals 0 as you can see here and when we solve that we get 1 is equal to 0 which is not possible, so that there are no x-intercepts.
Now, for the asymptotes. Let's check for vertical, horizontal and oblique, otherwise known as slant, asymptotes. If it is not of a particular type, write none in the box. Separate multiple answers with semi-colons. Remember that slant and oblique are referring to the same type of asymptote. They may be used interchangeably.
Vertical assymptotes occur when the denominator is 0. This means you have to solve x squared minus 7x plus 12 equals 0. So, we can factor this to x minus 4 and x minus 3 equals 0, which gives us the vertical asymptotes of x equaling 3 and x equals 4. To see if we have horizontal and oblique or slant asymptotes, we need to look at the degrees of the polynomials in the numerator and the denominator. As the numerator is just one, it has a lower degree than the denominator, which means that we have a horizontal asymptote. So, what is this asymptote? The horizontal asymptote tells me roughly where the graph will go when x is really, really big. And we can see that in this case, when the x gets bigger, that the function will move closer and closer to 0. So, the asymptote is y equals 0. And for e, we have no slant or oblique asymptote.
Let's have a look at the graph with the intercepts and asymptotes you've just found marked on it. State if the function has a maximum or a minimum, and write the point in the box below.
You can see from the graph, there is a maximum point here at 3.5 comma negative maximum. As all of these points over to the left and to the right pass the asymptotes, are above it. That means it is a relative maximum.
Describe the behavior of the graph within the following intervals. Negative infinity to 3, 3 to 3.5, 3.5 to 4, and 4 to infinity. You may select from the following terms. Increasing, decreasing, or constant. Please put one of those in the boxes to the right.
You may have noticed that the symmetric, so what happens on the left is reversed on the right. In the first region, we see the graph starts off at the horizontal asymptote y equals 0, and rises to the vertical asymptote of x equals 3. We can also say it is increasing at the y-value of each point, as we move right, is greater than the y-value of the point before. In the last region, it's the opposite. It starts at the vertical asymptote and falls to the horizontal one, this means it's decreasing. At x equals 3, we have an asymptote. And the curve to the right of that rises from negative infinity up to the maximum that you found in the question before at 3.5 comma negative 4, before falling again towards negative infinity as it approaches the asymptote x equals 4.
In the previous example, we didn't have any slant or oblique asymptotes, so let's take a look at those now. For the function f of x equals 3x squared plus
We can find the slant asymptotes by using long division, and dividing the above function. 3x squared plus 2 divided by x. When we divide that, we get 3x plus 2 over x. As x gets larger, the term becomes infinitely smaller, and we move toward the line y equals 3x, which is our slant asymptote.
Finally, we're going to review these really cool functions which put together, are called Piecewise Functions. Let's take a look at a graph of the following function. Match the pieces of the function with the region of the graph. You can choose from x squared, x plus 2, and 2, and please be sure to fill in the corresponding regions on the right-hand side. Fill in the blue boxes with the appropriate symbols and the green boxes with the appropriate numbers.
For f of x equals 2. Our region is x is less than negative 2. The next part of the graph is our f of x equals x squared, and that is between the region of negative 2 and 2. And it's inclusive of that region as you can see by the less than or equal to signs. The third part is, f of x is equal to x plus 2 for all values greater than 2.