 In this video, we provide the solution to question number 12 from Practice Exam 4 for Math 1050, in which case we have to graph the logarithmic function f of x equals the natural log of negative x plus 2 plus 2, and we're going to do that below, and we have to put at least one point, label one point, we have to also label the vertical asymptote. So the thing to remember for logarithmic graphs is that the general formula is the following. f of x is going to equal the natural log of x minus h over b plus k. So some things I can see very quickly is that there's a coefficient attached to the x, we have to factor that thing out. So f of x is going to look like when we factor it, the natural log of negative x minus 2, like so, plus 2, for which times a by negative 1 is the same thing as dividing by negative 1. So I actually can write this as the natural log of x minus 2 over negative 1 plus 2. So we see that h is equal to 2, that'll give us a horizontal shift, k is also 2, and this b value is negative 1. So that means we're going to reflect our logarithm across its vertical asymptote. So since the vertical asymptote, since we shifted the graph I should say to the right by 2, the original asymptote is at the y-axis, so we shifted over by 2. And so we see that x equals 2 is the location of the vertical asymptote. The instructions do say we have to put that in the graph. So we're going to put that in there and we're going to label it x equals 2. Because we reflected across the vertical asymptote, the graph is actually going to be over here, pointing something like that as opposed to the standard logarithm which points to the right. So that's an important observation there. Then let's take our x-intercept, which since also the b value is negative 1, that tells us that the original x-intercept for a graph this distance is the b value. We reflected it over here, but we haven't stretched it at all. So this should still be a factor of 1. But we did shift it up by 2. So shifting it up, we see that the original x-intercept is going to move here. So if we follow that journey, right, the original x-intercept starts at 1, 0. We moved it over by 2, so it's at 3, 0. We reflected it over back to 1, 0. And then we shifted it up by 2. So this what was the original x-intercept now has the coordinates 1, 2. And you can check that with the function, of course. If I plug in 1, you're going to get negative 1 plus 2, which is 1. Natural log of 1 is 0 plus 2 would then be 2. So that's a point in there. And so then we're going to try to connect the dots. It's really steep when you're close to the vertical asymptote. And it'll curve away towards the left side. So don't worry about the x-intercepts or y-intercepts, as this is the natural log. Those are going to be irrational numbers anyways. You only have to label one point. So I'm going to actually take the modified x-intercept, the original x-intercept, as it goes through these transformations. And so this then gives us the graph of this logarithmic function.