Graphing functions Use the guidelines of this section to make ...

Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x² - 6x

Accepted Answer

Welcome back, everyone. In this problem, we want to draw the graph of the function F X equals X2 minus 4 X. And here we have our graph. On which we're going to draw um F of X, OK. Now, if we're going to draw the graph, there are a few things that we need here, OK? Uh, let's try to figure out a few of the properties of the graph, and then we'll be able to draw the graph of the function. First, let's start by trying to figure out the functions domain, OK? And now if we think about it, this is a polynomial function. The domain of a polynomial function is all real numbers. So in interval notation, that means the domain goes from open parentheses negative infinity, infinity closed parentheses. OK. Next, let's try to figure out this graph's symmetry, OK. Is our graph going to be symmetric? Well, there are two things that we can test here. First, we can test if our function is even, and a function is even. If F of negative X is equal to F of X, for all values of X. For function F here, if we put negative X into our function, then it's going to be negative X. Oops, sorry, negative X, that's inside our bracket, negative X squared minus 4 multiplied by negative X. Now we know here negative X2 is going to be X2. And -4 multiplied by negative X, that's going to be + 4 X. And this is not equal to FFX. Therefore, our function is not even, OK? Uh function is not even. Next, we can test it for a function is odd, and a function is odd if F of X, OK? F of negative X sorry, is equal to negative F of X. Now we know so far, let me just fix this here, we know so far that F of negative X is X2 + 4 X. OK? Uh, is this going to be equal to negative FFX? Well, Negative F of X is going to be the negative value of X2 minus 4 X, which would be equal to negative X2 + 4 X. and this is not equal to F of negative X. So what does that tell us? Well, our function is neither even nor is it odd, OK? So it's neither. That means it's not symmetric about the X or Y axis. Now, what else can we do? Third thing is to find the derivatives of our function, because for all derivatives, when the first derivative equals 0, it gives us the critical points of the graph, and when the second derivative equals 0, the X value is that for the inflection points, that of the inflection points, OK. So let's go ahead and try to find the critical points, OK. Were the first derivative of our function. Is equal to 0, OK? Now we know here that for X2 minus 4 X, the first derivative is equal to 2 X minus 4, OK? So at 2 X minus 4 equals 0, OK. Then 2 X will be equal to 4, and thus X will be equal to 2. Thus, this is where Our graph has a critical point. Our graph has one critical point at X equals 2, OK? Next, let's find out if there are any points of inflection. But before we even think about that, we can think about the intervals on which the function is increasing or decreasing based on our first derivative. Now we know then. That our function is increasing, OK, where our first derivative is X is greater than 2, OK, or sorry, when F is, first, when X is greater than 2, our function is increasing. Sorry, let me just make that note here. So when X is greater than 2, F is increasing. And when X is less than 2, F is decreasing. OK. And we also know that the function is concave up. OK, it's always going to be concave because F is always positive, OK. F is always positive. F is always positive to the function is and give up on the entire domain. OK, so on. Negative infinity infinity. Now that we have this information from our first derivative, let's, next, let's try to think about what happens in our second derivative to get our inflection points. And remember, we said that for our inflection points, OK. At A second derivative equals 0. Yeah. All right, let me write this in another way or inflection points. Occur Where our second derivative is equal to 0. We already have our first derivative. So now, our second derivative is going to be equal to 2, OK, because we differentiate to X, we get 2, differentiate -4, we get 0. Now, since the second derivative of the given function is 2, that means there are no inflection points, OK? There are no inflection points. For our our graph and know if we think about what we had before here. Since there are no inflection points, only one critical point, and since the function is decreasing before X equals to and increasing after, the critical point X equals 2 is a local minimum. OK, let me write that in red here. So it's gonna be a minimum point, a local minimum point. And now for our local minimum, means the minimum within the local area that we can see for our graph, to find a minimum value, we can substitute X equals 2 into the function, OK? So, when we do that, 2 squared is gonna be 4 + 4 multiplied by 2 is gonna be 8 and 4 minus 8 equals -4. So since FF2 equals -4, that means the local minimum 2-4 is the local minimum. OK. No, what else can we, what other feature can we think about for our graph? Where we can locate all the asymptotes and determine the end behavior. That's another important thing. So our end behavior. Now for our graph, OK, as X approaches infinity, OK, then we also know that F of X approaches infinity. And as X approaches negative infinity, OK, we know that F of X is going to approach infinity because our leading term X2 dominates, OK? So that's what we know about our functions and behavior. Next, You can find a Y intercept, OK. And for or Y intercept. That occurs when X equals 0 and F of 0 equals 02 minus 4 multiplied by 0, which is 0. So a Y intercept is at 00. While for our X intercepts. It occurs where F of X equals 0 and when F of X equals 0. That means X2 minus 4 X is gonna be equal to 0, thus X multiplied by X minus 4 equals 0. And I should probably put that below it here. And that, if that's the case, then either X equals 0 or X equals 4. So our X intercepts are 00, is also a Y intercept, and 40. 00 and 40 are the X intercepts of our function. So we, we've looked at a few properties, right? Let's just summarize all of what we found. So we found our X intercepts to be 0040, or Y intercept is 00. We know the end behavior of our function. We know where the local minimum is and its behavior as it approaches and leaves the minimum point. And we also know, OK, that it's an, it's, it's not an odd or even function, OK? And we know the domain of our function, OK? So with all of that in mind, we have enough information to plot our graph. So, now we're gonna go to our graph, put our X intercepts, which, OK, X and this is the X and Y intercept, 00 and 40. OK. We're gonna go to our local minimum 2-4. OK. And know that we have that and we know the behavior, you know our graph is going to look like this and give up, OK. So this is going to be the graph of F of X. Thanks a lot for watching everyone. I hope this video helped.

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x² - 6x

Key Concepts

Quadratic Functions

Vertex of a Parabola

Intercepts

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