Given the graph of f(x)=x², Sal graphs g(x)=(x-2)²-4, which is the graph of f shifted 2 units to the right and 4 units down.
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- Why does Sal move the graph 2 units to the right when it says -2 at1:05?
Shouldn't he move it to the left by 2 units?(12 votes)
- Think about it on a coordinate plane: if I shift 0 two to the right, I get 2. Because you have to think about it in terms of 0, not 2, you subtract 2 instead of adding 2.(19 votes)
- I wish these videos were explained better. I feel like its really brief and I still don't understand how you get to where you are by the end. I wish I could be more specific but I guess I don't understand it. There is no introduction really explaining whats going on either. Also taking the simplest explanation and then jumping to advanced problems there is no bridge.(11 votes)
- My favorite video about this topic is the Shifting Functions Introduction video. I love how Sal explains this concept there.(1 vote)
- Does the fact that the x-2 is squared change the graph? Why not?
Also, is f(x)=x the same thing as f(x)=x^2? Why or why not?(6 votes)
- They are different.
F(x) = X and F(x) = x - 2 are linear functions. When they are graphed, they create straight lines. We can tell they are linear because there are 2 variables: X and Y (remember, F(x) is Y) and both variables have exponents = 1.
F(X) = X^2 and F(X) = (X-2)^2 are quadratic functions or 2nd degree (exponent on X = 2).
When graphed, these create what is called a parabola (looks like a U-shape).
This section on linear vs. non-linear functions might help you: https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/linear-nonlinear-functions-tut/v/recognizing-linear-functions(5 votes)
- Anyone besides me have trouble with the interactive graphs? On the quiz after "Shifting Functions" I bet I took the quiz four times to even get three out of four correct. And each time I know I had the formula for shifting graphs figured out correctly but when I graphed the functions with the interactive graphs I got each one wrong. And I believe it was a result of me not being able to set the graphs exactly right. For example, the last graph I tried the tip of the graph was 8 places to the right and one place up. And when I set the graph to that location the system still counted it wrong. In the answer it said shift the tip of graph four places right and one place up.
Thanks in advance,
Fred Haynes(5 votes)
- Hmm I see your problem. In order to graph a function, you have to have it in vertex form;
a(x-d)² + c <---- Basic Form
Example: (x-3)² + 3
Since there's no a, you don't have to worry about flipping on the x axis and compressing or stretchign the function. Now we look at d. d = -3. In order to find the zeros of the function, x must equal 3. That's why the equation moves to the right when d is negative and when d is positive, the equation moves to the left. So simply to say, (x-d)² = 0 When you know d, then your x will be something that subtracts d to equal 0. So in this example, the function shifts 3 units to the right. Now we look at c. c is 3, which means that the graph moves 3 units up. And voila! There you have it! If you need more clarification, don't hesitate to ask. I hope this helped!(5 votes)
- If you are subtracting a number from x, wouldn't that shift it to the left since it would decrease the value of x?(2 votes)
- Think of the equations: g(x)=x and f(x)=x-1. The same value of g(x) and f(x) will be made by an x value one greater in the function f than the x from the function g . Therefore, the graph will move to the right.(3 votes)
- In the subsequent practice section, the term 'transformation' is used, not shift. I assume every shift is a transformation, but is every transformation a shift?(1 vote)
- at the time of 1.55, why we shift the coordinate (5,9) to the coordinate (5,5)?(2 votes)
- When comparing g(x) with f(x), we need to know not only what happens with the x values (shift 2 units to the right) but we also need to know what happens with the y values.
The constant term in f(x) is zero (in other words, there isn't one), but the constant term in g(x) is - 4. This tells us that the points in g(x) are 4 units lower than in f(x).
That's why (5,9) was moved down 4 units to reach (5,5).
Hope this helps!(2 votes)
- At0:45, This logic about shifting the x to the right when g(x)=(x-2)^2 -4 is on the graph, doesn't make any sense to me?
"Turn RIGHT to go LEFT? yes! Thank you! or should I say no thank you? Cause in opposite world maybe that really means thank you!" - Lightning McQueen
I need an explanation... Thank you! :D(2 votes)
- think about it this way: you start with y=x^2 which goes through (0,0). So if you have y=(x-2)^2, what value of x is needed so that y will be 0 (still the x intercept)? so if you put x=2, you get the point (2,0) which is 2 to the right. If you have y=(x+2)^2, x has to equal -2 to give the point (-2.0) and thus is a shift to the left.(2 votes)
- How does it affect the graph if the (x+#) is squared?(1 vote)
- Is it correct to say the function f(x) = x^2 can be shifted ONLY to the right e.g. g(x) = (x-2)^2, since the square of the calculation x-2 in parentheses will always result in a positive number?
- [Voiceover] We’re told, "the graph of the function f of x "is equal to x-squared." We see it right over here in grey. It’s shown in the grid below. "Graph the function g of x is equal to "x minus two-squared, minus four "in the interactive graph." This is from the shifting functions exercise on Khan Academy, and we can see we can change the graph of g of x. But let’s see, we want to graph it properly, so let’s see how they relate. Well, let’s think about a few things. Let’s first just make g of x completely overlap. Okay, there you go. Now they're completely overlapping. And let’s see how they’re different. Well, g of x, if you look at what's going on here, instead of having an x-squared, we have an x minus two-squared. So, one way to think about it is, when x is zero, you have zero-squared is equal to zero. But how do you get zero here? Well, x has got to be equal to two. Two minus two-squared is zero-squared, if we don’t look at the negative four just yet. And so, we would want to shift this graph over two to the right. This is essentially how much do we shift to the right. It’s sometimes a little bit counterintuitive that we have a negative there, because you might say, well, negative, that makes me think that I want to shift to the left. But you have to remind yourself is like, well okay, for the original graph, when it was just x-squared, to get the zero-squared, I just have to put x equals zero. Now to get a zero-squared, I have to put in a two. So this is actually shifting the graph to the right. And so, what do we do with this negative four? Well, this is a little bit more intuitive, or at least for me when I first learned it. This literally will just shift the graph down. Whatever your value is of x minus two-squared, it's gonna shift it down by four. So what we wanna do is just shift both of these points down by four. So this is gonna go from nine, and this is gonna go from the coordinate five comma nine, to five comma, if we go down four, five comma five. And this is gonna go from two comma zero, to two comma negative four. Two comma negative four. Did I do that right? I think that’s right. What, essentially, what we have going on is, g of x is f of x shifted two to the right and four down. Two to the right and four down. And notice, if you look at the vertex here, we shifted two to the right and four down. And I shifted this one also, this one also I shifted two to the right, and four down. And, there you have it. We have graphed g of x, which is a shifted version of f of x.