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Algebra 1 (Eureka Math/EngageNY)
Course: Algebra 1 (Eureka Math/EngageNY) > Unit 4
Lesson 13: Topic C: Lessons 18-19: Translating graphs of functionsShifting parabolas
CCSS.Math:
The graph of y=(x-k)²+h is the resulting of shifting (or translating) the graph of y=x², k units to the right and h units up. For example, y=(x-3)²-4 is the result of shifting y=x² 3 units to the right and -4 units up, which is the same as 4 units down.
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- For instance y=(x±9)^-1
Why does the sign before 9 have a counter effect on the parabola?(6 votes)- wait, do you mean y=(x±9)^2 - 1? There is no squared value in the original question, just ^-1. Anyways, assuming that you mean y=(x±9)^2 - 1, then I would be happy to tell you how I think of the counter effect, as you put it. I pretend that I am trying to balance something on a seesaw, where the 0 is the pivot point. Imagine that you had a friend who weighed 9 kilos more than you. Your friend is x+9, and you are x. If you and your friend want to balance, you must shift the seesaw in your direction, or the heavier friend will tip it over. Basically, +9 means that it is 9 points too heavy on the positive side, so if the positive side is too heavy, what do you have to do? You have to shift the whole system to the left, so it can still balance. I hope this helps! I also hope that people still know what a seesaw is, even though people don't seem to play outside anymore.(18 votes)
- would it be right to write it down like this?: g(x)+4=(x-3)squared(3 votes)
- Function notation always has the function name by itself.
g(x) = (x-3)^2 - 4 would be the correct form.(2 votes)
- How would you do this? Shifting f(x) 1 unit right then 2 units down. The equation is f(x)=x^2-2x-1.
My book says this
to shift 1 unit right its
g(x)=(x-1)^2-2(x-1)-1
=x^2-4x+2
for shift 2 units down
h(x)=x^2-2x-1-2
=x^2-2x-3
I need some help understanding this please(1 vote)- One issue is that your original question appears to be combining both shifts to the original equation whereas they are showing shifts independently of each other. While it is easier to shift in vertex form it can be done in any form. Shifting to right affects the x such that you subtract (because 1-1=0), so they correctly shift x by substituting x-1 everywhere you see x. With (x-1)^2=x^2-2x+1 and -2(x-1) = -2x + 2, adding all the parts gives x^2 -2x + 1 - 2x + 2 - 1, which when combining like terms gives x^2 - 4x + 2. IF you wanted to combine both shifts in one equation, a shift down would add the -2 to the end to get x^2 - 4x. However, they appear to be doing two independent shifts, so starting from x^2 -2x - 1 and adding -2 to end gives their result.
So either the problem is incorrect (when you have shifting f(x) 1 unit right then 2 units down)or there is more to their solution.(1 vote)
- y=(x-h)^2+k How do negative values of h represent leftward shifts? I cannot get this one, Sal in the video explained that when we shift h units to the right we substract h units from the function.(1 vote)
- This is going to be true for all functions, so lets start with a linear equation y = x + 3. the y intercept is 3 (set x=0) and the x intercept is -3 (set y = 0). If we keep it as a change in y, we have y = x + 3, so it is easy to see the y intercept. By "making it a change in x" instead, we show it as y = (x + 3) + 0. So if we put in a negative 3 for x, we get y = 0 which gives us the correct x intercept. So we had to have the opposite sign for a change in x.
Similarly for the quadratic function such as y = (x + 3)^2 + 5, we would have to set x = -3 in order to make what is inside the parentheses to be 0, we have to change the sign.
SO a change in y follows the sign, a change in x has to be the opposite sign.(4 votes)
- this was a lot more helpfull(2 votes)
- How is y=f(x-3) and y=(x-3)^2 the same?
At aboutminutes into the video. 1:50(2 votes) - Does it matter if we write
y + 4 = (x-3)^2
instead ofy = (x-3)^2 - 4
? Should be the same but is one preferred over the other? Last video he used the former format, where the4
was being added to they
.(1 vote)- If you are asked to write the equation in vertex form, then use y = (x-3)^2 - 4.
There is no benefit to using y + 4 = (x-3)^2. The risk of using this version is that you may mistakenly think the vertex is at (3,+4), when it is at (3,-4).(2 votes)
- So just to be clear:
Shifting to the right by 3 is (x-3)
Shifting to the left by 3 is (x+3)
Shifting down by 3 is (x)-3
Shifting up by 3 is (x)+3(1 vote)- Yep! And that works with any function. the trick is just internalizing what is inside and what is outside the function.(2 votes)
- What is the rule if the equation within the brackets were 1-x?
For example:
y=(1-x)^2 and y=(x-1)^2 and y=1-(1-x)
In these three, which equations shift the shape in which direction?(1 vote) - If shifting the vertex horizontally produces the expression (x-3), then why doesn't the equation become (x-3)+4 when the vertex is shifted down by four units? If the expression inside the parentheses were 1-x, what would be the corresponding rule? Would it be correct to express it as g(x)+4=(1-x)^2?(1 vote)
- Your alternative form does not make sense to me, but you are onto something. A -x inside causes a reflection about the y axis. While vertex form is y=(x-h)^2+k, if you move the constant to the form you are talking about, y-k=(x-h)^2. If the shift is with the variable, you always have to change the sign (both with x and y). In both cases, in order to "act" like 0, you need the opposite sign to make them 0.(1 vote)
Video transcript
- [Instructor] Function g can
be thought of as a translated or shifted version of f of
x is equal to x squared. Write the equation for g of x. Now, pause this video, and see if you can work
this out on your own. All right, so whenever I think
about shifting a function, and in this case, we're
shifting a parabola, I like to look for a distinctive point. And on a parabola, the vertex is going to be our most distinctive point. And if I focus on the vertex of f, it looks like if I shift that to the right by three, and then if I were to shift that down by four, at least our vertices would overlap. I would be able to shift the vertex to where the vertex of g is. And it does look, and we'll validate this, at
least visually, in a little bit, so I'm gonna go minus four
in the vertical direction, that not only would it
make the vertices overlap, but it would make the
entire curve overlap. So we're going to make,
we're gonna first shift to the right by three. And we're gonna think about how
would we change our equation so it shifts f to the right by three, and then we're gonna shift down by four. Shift down by four. Now, some of you might
already be familiar with this, and I go into the intuition in a lot more depth in other videos. But in general, when you shift to the right by some value, in this case, we're shifting
to the right by three, you would replace x with x minus three. So one way to think about this
would be y is equal to f of x minus three, or y is equal to, instead
of it being x squared, you would replace x with x minus three. So it'd be x minus three squared. Now, when I first learned this,
this was counterintuitive. I'm shifting to the right by three. The x-coordinate of my vertex
is increasing by three, but I'm replacing x with x minus three. Why does this make sense? Well, let's graph the shifted version, just to get a little
bit more intuition here. Once again, I go into much more
depth in other videos here. This is more of a worked example. So this is what the shifted
curve is gonna look like. Think about the behavior that we want, right over here, at x equals three. We want the same value
that we used to have when x equals zero. When x equals zero for the original f, zero squared was zero. Y equals zero. We still want y equals zero. Well, the way that we can do that is if we are squaring zero, and the way that we're gonna square zero is if we subtract three from x. And you can validate that at other points. Think about what happens
now, when x equals four. Four minus three is one squared. It does indeed equal one. The same behavior that you used to get at x is equal to one. So it does look like we have
indeed shifted to the right by three when we replace
x with x minus three. If you replaced x with x plus three, it would have had the opposite effect. You would have shifted
to the left by three, and I encourage to think about why that actually makes sense. So now that we've shifted
to the right by three, the next step is to shift down by four, and this one is little bit more intuitive. So let's start with our
shifted to the right. So that's y is equal to
x minus three squared. But now, whatever y value we were getting, we want to get four less than that. So when x equals three, instead
of getting y equals zero, we want to get y equals
four less, or negative four. When x equals four,
instead of getting one, we want to get y is
equal to negative three. So whatever y value we were getting, we want to now get four less than that. So the shifting in the vertical direction is a little bit more intuitive. If we shift down, we subtract that amount. If we shift up, we add that amount. So this, right over here,
is the equation for g of x. G of x is going to be equal
to x minus three squared minus four. And once again, just to review, replacing the x with x
minus three, on f of x, that's what shifted, shifted right by three, by three. And then, subtracting the four, that shifted us down by four, shifted down by four, to give us this next graph. And you can visualize, or
you can verify visually, that if you shift each of these
points exactly down by four, we are, we are indeed going to overlap on top of g of x.