Learn about Algebra II: Quadratics and Shifts Video

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Learn about Algebra II: Quadratics and Shifts

On problem 32, what are the solutions to the equation, one plus one over x² is equal to three over x. So at first, this looks like a pretty daunting equation you have this x is denominator of an x² denominator but I think we can simplify it if we just get rid of this x² denominator. The easiest way to do this is to multiply everything by x². So let’s multiply both sides of this equation by x² and then we will get see x² times one is x², x² times one over x² oh that just one and then x² times three over x that’s three x² over ex, x² or divided by x is just x so that is equal to three x. We could subtract three x from both sides and you get x² minus three x plus one is equal to zero this is a simple quadratic and it is not obvious that you can factor it, in fact two numbers when you multiply it equal one and then when you add them equal minus three. I’m guessing it might be imaginary when I fight not be imaginary but this is strange number. So let’s use the quadratic equation, when in doubt use the quadratic equation, so minus b this is b right? B is these three right there, the negative three right, b is negative three so minus b is when you be plus three plus or minus square root of b² minus 3² is nine minus four times a which is one times c which is one is minus four. All of that over 2a, a is one so this is over two. See that is equal to three halves plus or minus is it the square root of five over two, I just separate this out because it mean the choice is it seems like they did that so, you’ve could said that’s three halves plus square root of five over two or three halves minus square root five over two. And I just did that because if you’ll look that’s how they write it and that is choice a. Next problem, 33 I think this one actually might be good to copy and paste the problem. Let me see if I can do this, okay there are two numbers with the following properties. Let me write down the properties I want you copy and paste it for you, okay I copied it and I’ll be going here and then I pasted it for you. All right, now so, the second number, so these two numbers on the following purposes. Second number is three more than the first number, so let’s see s for second number and f for first number. So the second number, the second number is three more than the first number. So second number is equal to the first number plus three, that’s from statement one and the product of the two numbers is nine more than the sum. So the product of the numbers that’s s times f is nine more, nine plus their sum plus s plus f. So let’s see if we have two equations in to unknowns. This is non-linear because I’m multiplying this two variables but I think we should be able to solve them one way or the other. So let’s see, we have what s is equal to, so let’s just substitute that back into this equation so let’s say that s is equal to f plus three so, if we substitute for this S’s we get f plus three times f is equal to nine plus f plus three write instead of s, f plus three and then plus f. See if we can supply this, f times f is f² plus three f is equal to nine plus three is 12 plus two f. Let’s see, subtract two f from both sides, you get f² plus f is equal to 12. Subtract 12, you get f² plus f minus 12 is equal to zero. And this one looks factorable, I don’t have to take out the quadratic equation, let’s see this is f plus four times f minus three, right because we multiply those you get negative 12 and when you add those you get plus one so that is equal to zero. So in order for this to be true, one of, one or both of this have to be equal to zero so f plus four is equal to zero that means f is, f could be equal to minus four if f minus three is equal to zero then that’s f could be three. So, f could be minus four or three. Now s is f plus three so if we’re dealing with the minus four scenarios if f is equal to minus four then what is s? Then s is going to be minus four plus three then s is going to be equal to minus one and then f is equal to three then s is equal to six. So let’s see if we see either of these combinations. Minus four minus one, that’s choice b, excellent. All right problem 34 let me see maybe I should copy and paste these word problems so we can see how parse the problems. So if copied it, we go here then pasted it. Jenny is solving the equation x² minus eight x equals nine by completing the square. What numbers should be added to both sides of the equation to complete this squares. So x² minus eight x is equal to nine and I wrote it with space for a reason. When you’re completing the square, you’re trying to turn the left hand sides of this equation into some type of a perfect square right. So if it’s a perfect square I have two numbers and it’s the same number when you one add them together you get minus eight and when you square them, you should get something else right? So what’s half of minus eight? Half of minus eight is minus four and if I add, six so minus four squared is 16. So if I add 16 to both sides I’m all set and why that work when that’s not the perfect square. This is now x minus four squares is equal to nine plus this 25. There not even asking us to solve it, they just wanted to know what we have to add to both sides. So it’s 16 d and remember the whole logic here and I’ve done a few videos in completing square is. What number do I had here to make this look a perfect square. You say okay, I have a minus 8 x so I take half of this number, right. Because it’s the same number added to itself twice is going to become minus eight I take half of that number then I squared. So half of minus eight is minus four and you squared, you get the 16. So I have 16 to both sides, you get this and you could actually solve for this x minus four is plus or minus five and you keep going. And that’s actually, where the quadratic equation comes from anyway next problem. 16 was choice number d all right. So I’m going to copy and paste this entire problem here. Let’s go up here you paste it here. okay which of the following most actually describe the translation of the graph, okay y is equal to x plus three squared minus two to the graph y equals x minus two squared plus two. So this translations and all that, that’s of, so the y translation it tends to be pretty easy to figure out. Because if I have some let me just draw some example graph. So if I have the graph x², the graph x² looks something like this, see if I can draw it the graph. X² looks something like this, right and there is x, when x is equal to zero right are minimum point right? And the other value it increases and need both direction. The graph of x² plus two you shifting up right? This is the graph of x² plus two, you’d shifted up by two and the graph of x² minus two, you would shift it down by two right this would be x² plus two and this would be x² minus two. So the shift and the wide direction is very easy to see, so if we’re going from something minus two, if we’re going from something minus two, two plus two if we’re going from minus two to plus two, we we’re going to shifting it up four right? So that’s always easy one to just eyeball and figure out. So we’re definitely be shifting from minus two to two, so it’s up four. So it’s either going to be choice a or choice d. The left right shift is often a little bit more hard for people to visualize or to at least internalize but let’s would, I give you the attempt. If this is the graph, let’s just go back to this, this is the graph of x², this yellow line right there, right? That’s the graph of x², now let me ask you a question, what is the graph of x minus I don’t know x minus three squared. So this shifted down three to negative direction or three of the depositors. You’re intuition might say I’m subtracting three when I had, when I did minus two I shifted down. But it's actually the opposite here, because you have to think about for what value of x am I going to have as zero squared here, right? And that happens when x is equal to three so you can think of this way. Now, when we’re these points, when x is equal to three it’s the same thing as this point when we have just x² right? Because when you put three in here, this whole expression becomes zero and as you get above three, that’s like going above zero and as you go below three, that’s like going below zero. So this graph would just get shifted to the right by three, right? That’s x minus three shift to the right by three, x plus three would go on the other direction because when x is minus three that’s what equals your imaginary that down. So let’s think about this, we’re going from x plus three, right? So if this is x², x plus three is actually it will look something, let me do in a different color. X plus three is actually shifted to the left, and the way I was thinking about it. there’s twist think about it, but the y shifted intuitive and the x shift might not be if you have a plus three here, you actually shifting in the downward direction and the way it actually think about that, the intuition is when will this whole expression equal zero. This whole expression equal zero when x is equal to minus three, so that’s the point of what you’re getting zero squared. And I actually didn’t-- I’m drawing this graph, so I'm not doing the y shift here, right? So this is going to be shifted to the left three, this is going to be shifted to the right by two. So this is shifted to the left three, this is shifted to the right by two to go from this to this, you’re shifting to the right by five, right? So x plus three squared is here, so this actual graph x plus three squared minus two is going to be here, right? And then to go here, you have a plus two’s here, shifting the graph up by four and then you’re going to x minus two. So this graph right here is going to be up here. So you’re shifting up by four and then you’re shifting to the right by five. Actually if going to confuse with you’re shifting, left or right, you got to say I got a different three-- plus three minus two is five and five is only there and you should hopefully understand the problems a little bit deeper that that. But anyway, I’ll see you in the next video.

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On problem 32, what are the solutions to the equation, one plus one over x² is equal to three over x. So at first, this looks like a pretty daunting equation you have this x is denominator of an x² denominator but I think we can simplify it if we just get rid of this x² denominator. The easiest way to do this is to multiply everything by x². So let’s multiply both sides of this equation by x² and then we will get see x² times one is x², x² times one over x² oh that just one and then x² times three over x that’s three x² over ex, x² or divided by x is just x so that is equal to three x. We could subtract three x from both sides and you get x² minus three x plus one is equal to zero this is a simple quadratic and it is not obvious that you can factor it, in fact two numbers when you multiply it equal one and then when you add them equal minus three. I’m guessing it might be imaginary when I fight not be imaginary but this is strange number.

So let’s use the quadratic equation, when in doubt use the quadratic equation, so minus b this is b right? B is these three right there, the negative three right, b is negative three so minus b is when you be plus three plus or minus square root of b² minus 3² is nine minus four times a which is one times c which is one is minus four. All of that over 2a, a is one so this is over two. See that is equal to three halves plus or minus is it the square root of five over two, I just separate this out because it mean the choice is it seems like they did that so, you’ve could said that’s three halves plus square root of five over two or three halves minus square root five over two. And I just did that because if you’ll look that’s how they write it and that is choice a.

Next problem, 33 I think this one actually might be good to copy and paste the problem. Let me see if I can do this, okay there are two numbers with the following properties. Let me write down the properties I want you copy and paste it for you, okay I copied it and I’ll be going here and then I pasted it for you. All right, now so, the second number, so these two numbers on the following purposes. Second number is three more than the first number, so let’s see s for second number and f for first number. So the second number, the second number is three more than the first number. So second number is equal to the first number plus three, that’s from statement one and the product of the two numbers is nine more than the sum. So the product of the numbers that’s s times f is nine more, nine plus their sum plus s plus f. So let’s see if we have two equations in to unknowns. This is non-linear because I’m multiplying this two variables but I think we should be able to solve them one way or the other.

So let’s see, we have what s is equal to, so let’s just substitute that back into this equation so let’s say that s is equal to f plus three so, if we substitute for this S’s we get f plus three times f is equal to nine plus f plus three write instead of s, f plus three and then plus f. See if we can supply this, f times f is f² plus three f is equal to nine plus three is 12 plus two f. Let’s see, subtract two f from both sides, you get f² plus f is equal to 12. Subtract 12, you get f² plus f minus 12 is equal to zero. And this one looks factorable, I don’t have to take out the quadratic equation, let’s see this is f plus four times f minus three, right because we multiply those you get negative 12 and when you add those you get plus one so that is equal to zero. So in order for this to be true, one of, one or both of this have to be equal to zero so f plus four is equal to zero that means f is, f could be equal to minus four if f minus three is equal to zero then that’s f could be three. So, f could be minus four or three.

Now s is f plus three so if we’re dealing with the minus four scenarios if f is equal to minus four then what is s? Then s is going to be minus four plus three then s is going to be equal to minus one and then f is equal to three then s is equal to six. So let’s see if we see either of these combinations. Minus four minus one, that’s choice b, excellent.

All right problem 34 let me see maybe I should copy and paste these word problems so we can see how parse the problems. So if copied it, we go here then pasted it. Jenny is solving the equation x² minus eight x equals nine by completing the square. What numbers should be added to both sides of the equation to complete this squares. So x² minus eight x is equal to nine and I wrote it with space for a reason. When you’re completing the square, you’re trying to turn the left hand sides of this equation into some type of a perfect square right. So if it’s a perfect square I have two numbers and it’s the same number when you one add them together you get minus eight and when you square them, you should get something else right? So what’s half of minus eight? Half of minus eight is minus four and if I add, six so minus four squared is 16.

So if I add 16 to both sides I’m all set and why that work when that’s not the perfect square. This is now x minus four squares is equal to nine plus this 25. There not even asking us to solve it, they just wanted to know what we have to add to both sides. So it’s 16 d and remember the whole logic here and I’ve done a few videos in completing square is. What number do I had here to make this look a perfect square. You say okay, I have a minus 8 x so I take half of this number, right. Because it’s the same number added to itself twice is going to become minus eight I take half of that number then I squared. So half of minus eight is minus four and you squared, you get the 16. So I have 16 to both sides, you get this and you could actually solve for this x minus four is plus or minus five and you keep going. And that’s actually, where the quadratic equation comes from anyway next problem.

16 was choice number d all right. So I’m going to copy and paste this entire problem here. Let’s go up here you paste it here. okay which of the following most actually describe the translation of the graph, okay y is equal to x plus three squared minus two to the graph y equals x minus two squared plus two. So this translations and all that, that’s of, so the y translation it tends to be pretty easy to figure out. Because if I have some let me just draw some example graph. So if I have the graph x², the graph x² looks something like this, see if I can draw it the graph. X² looks something like this, right and there is x, when x is equal to zero right are minimum point right? And the other value it increases and need both direction. The graph of x² plus two you shifting up right? This is the graph of x² plus two, you’d shifted up by two and the graph of x² minus two, you would shift it down by two right this would be x² plus two and this would be x² minus two. So the shift and the wide direction is very easy to see, so if we’re going from something minus two, if we’re going from something minus two, two plus two if we’re going from minus two to plus two, we we’re going to shifting it up four right? So that’s always easy one to just eyeball and figure out.

So we’re definitely be shifting from minus two to two, so it’s up four. So it’s either going to be choice a or choice d. The left right shift is often a little bit more hard for people to visualize or to at least internalize but let’s would, I give you the attempt. If this is the graph, let’s just go back to this, this is the graph of x², this yellow line right there, right? That’s the graph of x², now let me ask you a question, what is the graph of x minus I don’t know x minus three squared. So this shifted down three to negative direction or three of the depositors. You’re intuition might say I’m subtracting three when I had, when I did minus two I shifted down. But it's actually the opposite here, because you have to think about for what value of x am I going to have as zero squared here, right? And that happens when x is equal to three so you can think of this way. Now, when we’re these points, when x is equal to three it’s the same thing as this point when we have just x² right? Because when you put three in here, this whole expression becomes zero and as you get above three, that’s like going above zero and as you go below three, that’s like going below zero. So this graph would just get shifted to the right by three, right? That’s x minus three shift to the right by three, x plus three would go on the other direction because when x is minus three that’s what equals your imaginary that down. So let’s think about this, we’re going from x plus three, right?

So if this is x², x plus three is actually it will look something, let me do in a different color. X plus three is actually shifted to the left, and the way I was thinking about it. there’s twist think about it, but the y shifted intuitive and the x shift might not be if you have a plus three here, you actually shifting in the downward direction and the way it actually think about that, the intuition is when will this whole expression equal zero. This whole expression equal zero when x is equal to minus three, so that’s the point of what you’re getting zero squared. And I actually didn’t-- I’m drawing this graph, so I'm not doing the y shift here, right?

So this is going to be shifted to the left three, this is going to be shifted to the right by two. So this is shifted to the left three, this is shifted to the right by two to go from this to this, you’re shifting to the right by five, right? So x plus three squared is here, so this actual graph x plus three squared minus two is going to be here, right? And then to go here, you have a plus two’s here, shifting the graph up by four and then you’re going to x minus two. So this graph right here is going to be up here. So you’re shifting up by four and then you’re shifting to the right by five. Actually if going to confuse with you’re shifting, left or right, you got to say I got a different three-- plus three minus two is five and five is only there and you should hopefully understand the problems a little bit deeper that that. But anyway, I’ll see you in the next video.

Transcription by: Scribe4you Transcription Services