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### Course: Algebra 2>Unit 9

Lesson 1: Shifting functions

# Shifting functions introduction

The graph of y=f(x)+k (where k is a real number) is the same as the graph of y=f(x) only it's shifted up (when k>0) or down (when k<0). Similarly, the graph of y=f(x-h) (where h is a real number) is the same as the graph of y=f(x) only it's shifted to the right (when h>0) or to the left (when h<0).

## Want to join the conversation?

• At , I am still confused about the explanation of why the graph shifted to right when we subtracted from x. I know this rule, but I still am not sure why this occurs! Like, is there a reason why it goes the opposite way?
• When f(x)=y is defined as x^2 then for each x-value f will be its square but when we subtract 1 from x and then square it, then for each x value the y-value will be (x-1)^2.... in a simple manner, when y=x^2, y=0 when x=0 and y=1 when x=1, but when y=(x-1)^2, y=0 when x=1 and y=1 when x=2......therefore the graph appears to shift that many units added to the left... to shift a function up or down it should be of the form: f(x)+h where h is an integer. But when to shift a function towards left or right is should be of the form f(x+h) when h>0 the function shifts towards the left and when h<0 the function shifts towards the left.
• Hello every one, still now i can't understand that the graph shifted to right when we subtracted from x,is there a reason why it goes the opposite way?

can any one explain with an example?
• Suppose we have a graph of a function f(x) that passes through the point (2, 9), so f(2) = 9. We then shift this graph 3 units to the right to form the graph of a new function g(x). This new graph passes through the point (5, 9), so g(5) = 9. Because f(2) = 9, we need to compensate for adding the 3 by defining g(x) = f(x-3), so that g(5) = f(2) = 9. Note that if we had instead used g(x) = f(x+3), then g(5) would equal f(8), which may or may not equal 9.

In short: because we shifted 3 units to the right, we need to subtract 3 from the new x-coordinate in order to achieve the same y-coordinate.
• So in general, this is just vertex form?
• Basically yeah
• When you have a negative value for x, the graph moves to the right and vice versa, but why does this not apply to the vertical direction?
• You should really take a look at some of the answers to similar questions here, they can really help.
Basically, the reason we have to write the reverse for x-transformations but write the positive for up and negative for down in the vertical direction is because we express functions in terms of y.
If you want to increase y by 1 (move the function up by 1), all you have to do is add 1 to every value of the function (tacking a +1 onto the end of the right side of the equation). If you want to transform horizontally, you can't directly just add a +1 to the other side of x as we don't have the function in terms of x. You'll have to manipulate it a bit to get it in terms of x - see this example.
y = 2x^2
x = +/- sqrt(y/2)
Now that we have our function, to move it right 1 we just add 1 to the right side, but then we have to make this equation in terms of y again:
x = +/- sqrt(y/2) + 1
(x - 1)^2 = y/2
y = 2(x - 1)^2
As you can see, trying to shift the function to the right by 1 means that in the y= form, we do the opposite and subtract from x.
• How do i type an absolute value in desmos?
• I figured it out. You have to type abs(what you want to have for absolute value)
• Just addressing two questions that appeared a lot.

Some people are confused why the graph shifted to the right when you subtracted from x (@ ).

So we have (x-1)^2 + k.
What value of x will make (x-1)^2 equal to zero?
In other words, what value of x will make the expression equal to k?
When x = 1, (x-1)^2 is zero, or (x-1)^2 + k is equal to 'k'.
The answer, 1, is positive, so the graph shifted to the right instead of the left.

Likewise, if you have (x+1)^2 + k, the value of 'x' would be -1. Since the answer (-1) is negative, the graph would shift to the left.

Another question I noticed was: Why does the graph go up when k is positive
(@ )?
When k is a positive number, it is being added, so the graph moves up. When k is a negative number, it is being subtracted, therefore the graph shifts down.

Hope this helps.
• suppose f(x) = mx + c
given b > 0
f(x + b) = m(x+b) + c = mx + mb +c
i do not see how it shifts f(x+b) left but not upward.
• Your function is a positively sloped line, so shifting up and shifting left will look the same.
• how are linear functions shifted if there's no vertex? or even any non-quadratic function.
• You can still shift the (0,0) point with transformations. If you have y=x+5, that shifts the parent function up 5. If you have y=-3x-4, it shifts down 4 with the same slope. For any function, you end up shifting point by point, so any one can be shifted.
• What is the reason for the graph being curved or with an angle? And do the graphs shown here go to infinity?
• They do go to infinity, as an infinite number of inputs will give you an infinity amount of outputs. But, note that unlike some other curves, this curve is defined for all values of the input (whatever value you plug in, you're bound to get a finite value)

There's also a different kind of "curve going to infinity". For example, if the function is y = 1/x, this isn't defined for x =0, as you'll get 1/0. So, if you graph it out, you'll see that here too, it'll go onto infinity. But unlike the previous example (where every point had an output), 1/x doesn't have an output at x = 0 and hence, there exists an *asymptote* at x = 0.
• No, `k` is not related to Y. `k` is just an independent variable that you could change, which affects the equation's location.