Were on problem 36, it says which of the following sentences is true about the graphs of Y=3×x-5²+1 and Y=3×x+5²+1? So lets just do very something similar in what we did in the past and if you think about it. Both of these equations, Y is going to be one or greater. Let me just you know, let just analyze this a little bit right. This term right here, let me, this term right here since were squaring is always going to be positive, right even if what’s inside the parenthesis becomes negative if you know X-10. Inside the parenthesis, this becomes negative but when you square it. It always becomes positive and you’re going to multiply 3× a positive number. So you’re getting it a positive number.
So the lowest value of this could be a zero. So the lowest value that Y could be is actually one and same thing here. The lowest value that, you know this number can become very negative but when you square it. This going to become positive, so this expression with the squared here is going to be positive when you multiply 3 is going to be positive. So the lowest value here is always going to be zero when you include this whole term. So similarly the lowest value Y can be as one. I just want to think about it a little bit just to give you a little bit of intuition and let’s think of this in the context of what we’ve talked and learn last time with the shifting. So let me draw it a color that you can see. So if that is the Y axis and I'll just draw mainly the positive area.
So let see, so if I were to just draw Y=x²+1. It would look like this where this is 1, well that’s Y=1 and the graph will look something of along this. Y=x²+1, oh that’s a horrible drawing. Normally I wouldn’t do redo it but that was just atrocious. Y=x²+1, look something like that its symmetric you know, you get the idea. You have seen this problem before. This is Y=x²+1. Now if we were to do X-5²+1, what happens to it? Well let me think, what is 3x²+1? Well then it just increases a little bit faster. So if I were to say Y=3x²+1, it might look something like this. It will just increase a little bit faster, 3 times as fast actually. So that would be 3x²+1, right.
The rate of increase in both directions just goes faster because you have this constant term three out there multiplying the numbers. Okay, now what happens when you shift it? When you shift this once to x-5, so where x=0 was the minimum point before. Now if we substitute the five here that will be our minimum point, right because then that whole term becomes zero. So this vertex will now be shifted to the right. We do in another color. So if this is the point 5, now this would be the graph. You just took this graph and you shifted it over to the right by five. I won’t draw the whole thing.
That graph right there would be 3×x-5²+1 and remember the Y shift is always intuitive, if you add one. You’re shifting it up if you subtract one you’re shifting it down. The X shift isn’t. We subtracted 5, x-5. We replaced X with x-5 but we shifted to the right and the intuition is there because now +5 make this expression zero. So that’s 3x-5² and then the same logic, 3×x+5² is going to be in here +1. So if we shift it that’s going to be shifted to the, let me pick a good color, to the left. So it’s going to look something like this, going to be this blue graph shifted to the left. So this is minus five.
So this is the graph right here of 3×x+5²+1. Now hopefully you have an intuition, so let’s read their statements and see which one makes sense. Which of the following is true? There vertices are maximum, no that’s not true of any of these because the vertices is that point right there and there actually at some minimum point, right. A maximum point will look something like that and we know that because you just go positive. This term can only be positive; if this was a -3 then will would flip it over. I guess its not choice A.
The graphs have the same shape with different vertices. Yeah, both of these graphs have the shape of 3x² but they have one vertices just 10 to the left of the other one. So I think B is our choice, so let’s read the other one. The graphs have different shapes with different vertices. No they have the same shape. They definitely have the same shape. I mean they both have this 3x² shape. One graph has a vertex that has a maximum while the other has graph, no that’s not right. They both are upward facing, so they both have minimum points, so its choice B.
Next problem, problem 37, let me see what it says. What are the X intercepts? Let me copy and paste that, okay and I'll paste it there. What are the X intercepts of the graph of that? Well the X intercepts whatever this graph looks like. I don’t know exactly what it looks like. I mean I don’t know what this graphs looks like something like this. Actually I have no idea what it looks like until I solve it. It’s going to look like something like this. When they X intercepts they’re like where does it intersect the X axis? So that’s like there and there. I don’t know of those are the actual points, right and to do that. We set the function equal to zero because this is the point Y=0. You’re essentially saying, when does this function equal zero because that’s the X axis. When Y=0, so you set Y=0 and you get 0=12x²-5x-2.
And whatever I have a coefficient larger than one in front of the x² term. I find that very hard to just eyeball and factor, so I use a quadratic equation. So -B, this is B. B is minus 5, so --5 is plus 5, right. -B plus or minus, the square root of B², -5² is 25-4×A which is 12×C which is ,minus 2. So let just make that ×+2 and put the plus out there, right minus times minus is a plus. All of that over 2A, all of that over 24, 2×A, so that is equal to, let me see, 5+ or minus the square root. Let see 25+4×12×2, right because that was a -2 when we have a minus there before, so 8×12=96, 96 all of that over 24.
What’s 25+96=121, right? This is 121 which is 11², so this becomes 5+ or minus 11/24. So and remember these are the points where these are X values where that original function will equal zero. So it’s always important to remember what were even doing. So lets see, so if x=5+11/24 that is equal to 16/24 which is equal to 2/3 that’s one potential intercept. So you know maybe that’s right here, right that x=2/3 and y=0 and the other value is x=5-11/24 and that’s what -6/24 which is equal to -¼ which could be this point. I actually drew graph that far off of what it could be, so this would be x=-¼ and those are the X intercepts of that graph. So let see 2/3 and -¼ is choice C on the test.
We have time for at least one more. Let see where, oh boy they drew us all these graphs, so which is the graph? Let me shrink it. I want to be able to fit all the graphs, so let me copy and paste their graphs. So this is one where the clipboard is definitely going to come in useful. Okay that’s good enough. Okay so they want to know which is the graph of, so let me--. I’ve never done something this graphical, let see. So that the graph they say is Y=-2×x-1²+1, so that’s what we have to find the graph off. So immediately when you look at it, so you say okay this is like the same thing as Y=-2x²+1 but they shifted the X right. They shifted the X to the right by one. I know it says a -1 but think about it. When X=+1, this is equal to zero.
So it’s going to be shifted to the right by one, right plus one. We know that, we know that’s going to be shifted up by one; right so up plus one and then we have to think. Is it going to be opening upwards or downwards? You think of this way, if this was Y=2x²+1. Then this term would always be positive and it will just come more and more positive as you get further and further away from zero. So it would open up but if you put a negative number there if you say Y=-2x²+1. Then you’re going to open downward. You’re just going to get more and more negative as you get away from your vertex right.
So were shifted to the right by one, were shifted up by one and were going to be opening downwards. So if we look at our choices, only these two are opening downwards and both of them are shifted up by one. There vertex is at Y=1 but this is shifted one to the right and this is shifted one to the left. And remember we said it was x-1², so the vertex happens when this whole expression is equal to zero and this whole expression is equal to zero when x-1, when x=+1. So that’s right here, so its actually choice C. and that’s probably when you’re shifting graph because that’s can be one of them kind of hardest things to engrail but I just really encourage you to explore graphs, practice with graphs with a graphing calculator and really try to plot points, and try to get a really good graph of Y when you go from -2x²+1 to -2×x-1²or when you’re placing X with an X-1 while this shifts the graph to the right by one.