 We often have the impression that mathematics is about solving problems and Well, it is but more often than not mathematics is really about taking a problem and Transforming it into a problem. We've already solved and this appears in things like graph transformations So consider a function like f of x equals x squared and with a little bit of effort I can graph y equals f of x and it's gonna look something like that and so the question is Can I use this information that I know how to solve that I have a Solution for graph y equals f of x y equals x squared I have a solution to graph y equals x squared and can I use this information to help us graph seemingly related functions like y equals x minus four squared or Y equals minus x squared or y equals x squared minus five and the idea is that these all have some Passing resemblance to f of x equals x squared and my question is are the graphs of these Going to look something like this And in order to answer this question I want to see what happens when I alter the function slightly and see what happens to its graph So let's take a look at this and so for starters We'll take a look at the comparison between graph y equals f of x and y equals the same function plus some constant and For starters, we'll assume that our constant is positive and well Well, suppose I have a generic graph that shows y equals f of x now It's very important to remember that this graph is supposed to be generic and what that means is we should take no Information whatsoever from any point on the graph. It does not necessarily mean there's a crossing of the axis here That's above the axis any of this information is for argumentative purposes only we can't use it With the exception that if I have a point on the graph The x and y coordinates are going to be based off the function My x coordinate is whatever the x coordinate is my y coordinate. It's the graph of y equals f of x So my y coordinate has to be f of x and so the point of the graph x f of x Now let's consider that other graph y equals f of x plus k Well, I know that a point on this graph is going to have x value Whatever it is the y value is f of x plus k So I know that on the graph y equals f of x plus k. There's going to be a point x f of x plus k So how can I get to that point? Well, I'm at the point x f of x plus k and I could get to this point f f of x plus k by Taking a vertical shift of k units now here We've assumed that k is positive k is negative that plus k is going to drop us down by some amount But I've gone from this point if I move upwards by k units my horizontal displacement Has it changed it's still x but my vertical displacement is f of x which took me to here Plus k which takes me up to there So I can go from a point on my original graph to a point that new graph by going vertically a distance k And I can do that for every point of the graph of y equals f of x and I can produce my Transformed graph and what this tells me is that the graph of y equals f of x plus k is The same as the graph of y equals f of x shifted vertically upwards by K units and if k is positive, that's a vertical shift upwards if k is negative. That's a shift downward So let's take a look at another one y equals f of x and y equals f of x minus h and so I'll compare those two graphs And I'll take our generic graph again completely generic And I'll look at the graph of y equals f of x minus h so again my x and y Coordinates x is whatever it'll be my y coordinate is f of x minus h So I have this point on the graph of y equals f of x minus h So to get to this point Well, let's see What if I started the point x minus h on the graph of y equals f of x well if my x value is x minus h and My y value is f of x then my y value for the point is going to be f of x minus h so I'll find that point and I want to get to from the point on this graph to the point on this graph and so that means my x and y coordinates my point has to go from here To here and I can get there by moving h units horizontally So again this point I've gone over x minus h units I've gotten up f of x minus h units if I go horizontally over h more units that takes me to x coordinate x My vertical displacement hasn't changed so my vertical my y coordinate is still f of x minus h in both cases And again, I can do that for every point of the graph and what that means is that my graph will look something like that And this suggests that the graph of y equals f of x minus h is going to be the graph of y equals f of x shifted horizontally by h units again to the right if h is positive or to the left if h is negative Well, how about the graph of y equals f of x and y equals f of minus x So I'll start with a generic graph and this time I'll be a little bit more generic I'm seeing something like this now for reasons that will become apparent in a second I do need to know where the y axis and I don't need the x axis for this particular one But I do need to know where the axis is located So let's think about that if I have a point on the graph of y equals f of x if I'm over here The coordinates are going to be some negative number x negative x and My y value is f of x. Here's my x coordinate So my y value is going to be f of negative x. So I'm at a point over here someplace Now if I want to get to the point on the graph x My y value is f of negative x f negative x I need to go to the opposite side of the y axis But I'm going to keep the same vertical distance f of negative x is my height f of negative x is my height I'm at the same height, but rather than being back x. I'm forward x So it's the same distance horizontally just on the other side and again I can do that for any other point on the graph and my graph f of negative x is going to look something like this and What that suggests is the graph of y equals f of negative x is going to be the graph of y equals f of x Reflected across the vertical axis Finally, we'll compare the graph of y equals f of x and y equals negative f of x so again, I'll take my generic graph y equals f of x and I have a point on the graph x f of x I also have a point on this graph of x negative f of x whatever x is my y value is Negative f of x. So here's a point on the other graph and again I can get there by going across the x axis so again here my horizontal displacement has a change I'm still back whatever the distance was but instead of going up f of x I'm going down minus f of x and I can do this for every point of the graph and Produce my graph of y equals negative f of x and what that suggests is the graph of y equals negative f of x is going to be the graph of f of x reflected across the horizontal axis And what this allows us to do is to take our nice simple y equals x squared graph and transform it into something else Now if I have the graph of y equals f of x if it looks like that I can sketch the graph of y equals some transformation of f of x Now it's easiest to follow the order of operations and sketch several intermediate graphs Just to get a sense of how these graphs look you might want to sketch them all the same set of axes But we do want to sketch the intermediate graphs The hardest way of doing this problem is to try all the transformations all at once and draw the single graph at the end So let's take this step by step Order of operations tells us what has to be done first So expressions inside parentheses have to be evaluated first and I have to worry about this x minus three so I want to graph f of x minus three and so this is going to be a horizontal translation of three units to the right so my first step I'm going to slide that graph over to the right The next thing Application has to be performed so I want to sketch the graph of y equals negative f of x minus three so that negative f of x is going to flip that graph across the y-axis and there's my reflection The last thing I do is this plus five and so this plus five is going to shift my graph vertically by five units and I'll end up there And there's my final graph of y equals negative f of x minus three