Algebra II (2018 edition)
Sal is given a drawing with four graphs and four formulas of square-root functions. He uses transformations to match each formula with its appropriate graph. Created by Sal Khan.
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- If there is no sign behind the radical should you always assume that it means the principle root?(6 votes)
- If there is not a - (minus) ,or ± (plus or minus) sign before the radical, you can automatically assume it is looking for the principal root.(12 votes)
- What about CUBE roots. How would I solve a cube root?(3 votes)
- Cube roots is no different from square roots, except for the fact that you're cubing your number. Square roots only have two factors. Cube roots have three. For example, the square root of 64 is 8 because 8X8=64. The cube root of 64 would be 4 because 4X4X4=64. Another example of cube roots could be 27. The Cube root of 27 can be expressed as 3X3X3. Therefore, the cube root of 27 is 3. Got it :)(5 votes)
- Question: Supposing you were to graph one of these functions, how would you graph the slope? Whenever I have radical functions, I plug in points. Is there an easier, less time-consuming way?(2 votes)
- How do you find the point of intersection on a graphing calculator?(1 vote)
- when you are on the graph, hit second, calculate. You will then see an option called intersect. On my calculator, it is number 5. Just select that option, and then put the constraints on opposite sides of the point of intersection.(2 votes)
- Why, when I graph line B does it not intersect with its x-intercept? I find this to almost always be the case with these graphs.(1 vote)
- Is this topic in Algebra I(1 vote)
- No one noes. It has been a mystery for a long time. But there are some tales heard that it lies within Algebra II(1 vote)
- Wait in the video for the square root of -x-1, why wouldn't the vertex be 1.2. In the video, Sal said the vertex was -1,2. I don't get why though.(1 vote)
- The vertex is (-1, 2) because you are supposed to move the initial function y = sqrt(x) to the right 1 unit, up two units (indicated by the -1 in the radical and the +2 outside the radical), then reflect it over the y-axis (by the order of transformations). Since a reflection over the y-axis causes a point's x value to become negative (-x, y), the initial vertex (1, 2) is changed to (-1, 2) from the transformations this radical equation has from y = sqrt(x).(1 vote)
- what happens and how to graph The square root of x-2? Or negative numbers in general cause im confused.(1 vote)
- It's simple. You just need to draw the graph of the square root of x and move 2 to the left. There should be NO points less than 0 (for y).(1 vote)
- Hey I had a question:
Is the endpoint of an even root function (like square root or fourth root) considered a vertex, point of inflection, or an extrema? I know that some in some function groups, their vertex is a point of inflection and an extrema, but what about even roots?(1 vote)
Match each function with its graph. And we have graph D, A, B, and C. And let's just start with the graph of B because, actually, this one looks the closest to the square root of x, which would look something like that. But it's clearly shifted. And it's flipped over the horizontal axis. The fact that it's flipped over the square horizontal axis, that means that we're taking the negative square root. So it's going to be one of these two cases right over here. We're shifting it down by 2. So we should have a negative 2, and both of these cases have a negative 2 here. And we're shifting it relative to the square root of x. We're shifting it to the right by 1. So if we're shifting to the right by 1, what we see under the radical should be x minus 1. And both of these have an x minus 1 on it. So which of these is B? And which of these is C? And I encourage you to think about that. Pause the video if you'd like. Well, the difference between these two is this one has a scaling factor of 2 here while this one doesn't. And of course, this is going to turn into more and more negative values. This is going to be getting negative faster. And you see that graph C here gets negative faster. And you could even try some points. In either case, when x is equal to 1, in either case, what we have under the radical becomes 0. x minus 1 is 0. x minus 1 is 0 when x is equal to 1. And in either case, our y value's negative 2. But then, you see, as x increases, this one right over here gets negative twice as fast. You see that right over here. This one right over here, y has gone from negative 2 to negative 4 here. For graph C, y has gone from negative 2 to negative 6. So it's gone down by 4. This has only gone down by 2. So it's pretty clear that graph B corresponds to the one that doesn't have the 2 out front. So this is graph B right over here. And graph C corresponds to the one that has a negative 2 out front. So let me try that. The negative 2 out front. So now we just have to think about these two equations and match them to these two graphs. And so both of these, they haven't been flipped around the horizontal axis. They've been flipped around the vertical axis. And that's why whatever we have under the radical we've essentially taken the negative of it. And actually, we could figure out which ones these are just by looking at how much they've been shifted in the y direction relative to the square root of x-- which would look something like that. I know I can't draw on this one right over here. Well, this one has been shifted up by 2. D has been shifted up by 2. This one over here, g of x has been shifted up by 2. And you could also see that it's been shifted to the left by 1. If you're shifting to the left by 1, normally, under the radical, you would have x plus 1. But then, we flipped it over around the vertical axis. And so that takes the negative of that. That's why it's negative x minus 1. You could view this as the negative of x plus 1. So either way, that is D. Throw that under the D. And just deductive reasoning tells us A. This is A, and it makes sense. We see it has shifted up by 1. So it's plus 1. And we could view this as the negative of x plus 4. So it's been shifted 4 to the left. And we see that is definitely the case relative to the square root of x. We got it right.