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# Shifting parabolas

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?
• 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.
• this was a lot more helpfull
• No. The reason is due to fact that the inverse of the function being considered would result in an input being mapped to distinct values which would contradict the definition of a function
• would it be right to write it down like this?: g(x)+4=(x-3)squared
• Function notation always has the function name by itself.
g(x) = (x-3)^2 - 4 would be the correct form.
• 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.
• 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.
• Why is it that whenever you shift to the right, you subtract, and when you shift to the left, you add? Why isn't it the other way?

Thanks.
- AzielKatz1026
• How is y=f(x-3) and y=(x-3)^2 the same?
At about minutes into the video.
• 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
• Yep! And that works with any function. the trick is just internalizing what is inside and what is outside the function.
• Does it matter if we write `y + 4 = (x-3)^2` instead of `y = (x-3)^2 - 4` ? Should be the same but is one preferred over the other? Last video he used the former format, where the `4` was being added to the `y`.
(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).