 In this video, we provide the solution to question number 10 for practice exam 4 for math 1050 in which case we are given some graph F, which we can see right here. We're supposed to find the equation for F. And so let's try to analyze this thing. It looks like a logarithm, the vertical asymptote really gives it away. And so what is the general formula for a logarithmic graph, right? The general formula we're looking for is y equals some log, all the answers look like natural log, so I'm going to write it that way. We have the natural log of x minus h over b plus k. So we have some type of shift in play, maybe a horizontal shift, a vertical shift. We also have maybe a stretch of some kind. Okay, so to find the horizontal shift, you want to look for the asymptote. The asymptote gives you the horizontal shift. Notice that for a standard logarithm, asymptote is the y axis. So we have shifted it to the left by a factor of two. And so since we're at x equals negative two, that's going to give you the h value. So h here is going to be negative two. That's great. What else can we say? The distance between the distance between the asymptote and the x intercept, which I should mention the x intercept has been moved up by two. A shift up by two would suggest that k value is two here. And then the distance between the vertical asymptote and the original x intercept after the shifting, that's going to be your b value, which here we see b is going to equal one, right? We shall also mention that the curvature goes to the right of the asymptote, so there's no reflection in there whatsoever. And so therefore, our formula is going to look something like the following f of x equals the natural log of, well, we had x minus a negative two, so that's going to be x plus two. B value is one, so you don't see a fraction there. And then we shift everything up by two. So we're looking for the answer, the log of x plus two plus two. And so then the correct graph would then be A. I should say the correct equation for the graph would be choice A.