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## College Algebra

### Course: College Algebra>Unit 12

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.
• 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)
• 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.
• 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.
• 1.. what do we call functions in the form of x^2 and 1/x and √x?
2.. I didn'nt understan the explanation of ""why does it make a sense"".
• x^2 is a quadratic function, 1/x is a rational function, and √x is a radical function.
(1 vote)
• Do you have to use DESMOS throughout this section? or could you work it without it?
• If you understand all the things that cause shifts, it is easy to do most functions without needing a crutch such as DESMOS to graph the shift. I want students to use the calculator as a tool, not a crutch to give them answers.
• I am very frustrated. I have a homework problem with a chart. x values on the top and F(x) values on the bottom and a multiple choice answer asking to find F(0), F(2), and all of the values of x for which F(x)=0
I am very confused
• 𝐹 is a function that takes an input value 𝑥 and returns an output value 𝐹(𝑥).

Each column of the chart consists of an input value (top) and its corresponding output value (bottom).

If the input value is 0, then the output value is 𝐹(0).