Graph each inequality. y ≤ log(x - 1) - 2

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Accepted Answer

Hello everybody. I hope you're doing it today today. We're going to look at this question that states draw the graph for the given inequality where the inequality is Y is less than, or equal to log of open parentheses, X minus three, closed parentheses minus four. Now for a question like this, the first thing I'm going to notice is that the inequality that we're working with is a non strict inequality. And this means that we will therefore have a solid line in our graph. Now, non strict inequalities are inequalities that have less than, or equal to or greater than, or equal to if this was just a less than sign or a greater than sign. This, this, this is what would be a strict inequality and instead we would use a dashed line in our graph. So that's the first thing that we have to notice. Now, shifting focus back into solving or inequality for the graph. I'll change that lesson or equal to sign for an equal sign or I'll have log um or Y is equal to of open parentheses, X minus three closed parentheses minus four. Now, the next thing I'll do is try to figure out a parent function for my inequality here. So apparent function, if I look at my Y equals um equation, I know that the minus four and the minus three are just transformations of a graph. So what I wanna do is eliminate those transformations and see what I would be left with. In this case, that parent function, once I do that would be Y equals log of X without the minus three and without the minus four as transformations. So now that I have the parent function, I know different rules and what this parent function will look like. The log of X graph has a vertical Asymptote. And I'll use VA as an abbreviation at X equal zero or the Y axis and an X intercept or an I'll use X dash I N T as an abbreviation at one comma zero. So those, that's something that I know about my paragraph and I can draw a small and quick version of what that could look like. So I draw Y axis and an X axis. I'll put a point at one comma zero to the left of that point. I'll be turning downwards and approaching the Y axis but never touching it. And I, and to the right of that point, I turned up and to the right quickly to have kind of a curved L A curved upside down L. So now that I know these things about my parent graph and my parent function, I can see how I go from my parent function to my current function. So we'll do that by adding transformations little by little. So first, I'm going to add the minus three and I'll have Y equals log of open parentheses X minus three. Now, this means that the log of X graph shifts three units, right? So that's what happens with how, what, that's what happens when we add the minus three, our parent function moves three units to the right now, if we add our other transformation, we have Y is equal to a log of open parentheses, X minus three, closed parentheses minus four. And now that minus four means that the, the graph of log of open parentheses, X minus three closed parentheses shifts down four units. So that minus four at the end shifts are graphed down by four units. So what, what do these two transformations mean? Well, it means that if we had a vertical ascent so at X equals zero, no, it will be at, well, let's see what happens with the transformations. We had it at X equal zero and we move it three units to the right, moving three units down will affect our vertical as to or moving four units that won't accept, won't affect our vertical as so, but moving three units to the right will because it's a vertical line. Now, if we started at X equals zero or the Y axis and we move three units to the right now, we are at X equals three. So our vertical ascent to moves from X equal zero to X equals three and where we had an X intercept at one comma zero. Now that moves three units to the right, meaning our X instead of being one, now we move with three units to the right, meaning that we're at an X of four. And if we move down four units where we were initially at A Y of zero, we moved down four. Now we're at A Y of negative four. So now that X intercept just becomes a regular point in our graph. So we know what happens to the Verla and, and now we have another point now because we have an inequality, we know we're going to have to shade our region in our graph because there won't just be one answer toward an inequality. And inequality has many multiple answers. In this case, it, it could be any Y that is less than, or equal to the log of our equation. So in this case, I'm just going to test the point for comma zero. It's a random point. You can pick any point that you want. And basically, you just want to see if that point fits within your graph and within your inequality, if it does, then you shade in the region with that point. If it doesn't, you shade in the region that does not include that point. So if we test four comma zero, we'll have that zero because zeros are Y is less than, or equal to log of open parentheses four minus three because four is our X close parentheses minus four. And then we have that zero is less than, or equal to, well, four minus three is one and the log of one is zero. So we have zero minus four, which is negative four. So we ask ourselves is zero less than or equal to negative four. No. So this is false. And therefore the 0.4 comma zero will not be in the shaded region. So with all this information, we can go ahead and start graphing. So we draw a Y axis and an X axis and I'm going to put my Y axis closer to the left side of my X axis because I know my graph won't extend to the negative side of the X axis because of a vertical Asymptote. So I'll drop points in our X axis at one, two, three, four and five. And on the Y axis, I'll draw points at negative one, negative two, negative three, negative four and negative five. So now let's start plugging in the things that we know, we know we have a vertical as at X equals three. So we represent that by drawing a dash line going up and down at A Y at an X of three. So we draw a vertical dash line at X equals three. And that is our horizontal as our vertical asym. Now, we know we also have a point at four comma negative four. That is where what happened to our ex intercept once we moved it. So I'll put a point at negative four comma four. And now I just have to look at my paragraph and see how that looked and use that same shape and use it in this graph. So to the left of four comma negative four, I'm turning downwards and approaching the vertical Aceto but never touching it. And to the right of four comma four, I'm turning to the right and a bit upwards and extending towards infinity. So once again, I have the shape of an, of a curved L that has been turned upside down and just like that, we have graphed our inequality. Now, before you finish this off, you have to remember that this isn't inequality. So you have to shade in a region. And as we mentioned earlier, we know that the 0.4 comma zero is not part of that region. So we have to shade in under our graph. So that's exactly what we'll do. And after you stayed in that region, now you are finished with the question. So I hope that was helpful. And until next time

Graph each inequality. y ≤ log(x - 1) - 2

Key Concepts

Logarithmic Functions

Inequalities and Their Graphs

Domain Restrictions

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