We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift.
horizontal shift
A horizontal shift occurs when you add or subtract a constant value to the input [latex]x[/latex] of the function [latex]f(x)[/latex].
This shifts the graph of the function horizontally.
Rightward shift: If you subtract a constant [latex]c[/latex] from [latex]x[/latex] before applying the function [latex]f[/latex], the graph of the function shifts to the right by [latex]c[/latex] units.
[latex]g(x) = f(x-c)[/latex]
Leftward shift: If you add a constant [latex]c[/latex] to [latex]x[/latex] before applying the function [latex]f[/latex], the graph of the function shifts to the left by [latex]c[/latex] units.
[latex]h(x) = f(x+c)[/latex]
The image shows the graph of the cube root function [latex]f(x) = \sqrt[3]{x}[/latex] (solid blue line) and its horizontally shifted version [latex]f(x + 1)[/latex] (dashed orange line).Original Function [latex]f(x)[/latex]
The solid blue curve represents the original function [latex]\sqrt[3]{x}[/latex].
The function [latex]f(x)[/latex] passes through the origin [latex](0,0)[/latex] because [latex]\sqrt[3]{0} = 0[/latex].
Horizontally Shifted Function [latex]f(x+1)[/latex]
The dashed orange curve represents the function [latex]f(x+1) = \sqrt[3]{x+1}[/latex].
Each point on the graph of [latex]f(x+1)[/latex] is exactly [latex]1[/latex] unit to the left of the corresponding point on the graph of [latex]f(x)[/latex].
For example:
If [latex]x=0[/latex], then [latex]\sqrt[3]{0+1} = \sqrt[3]{1} = 1[/latex].
If [latex]x=-2[/latex], then [latex]\sqrt[3]{-2+1} = \sqrt[3]{-1} = -1[/latex].
A horizontal shift involves moving the graph of a function left or right without altering its shape. In this case, adding [latex]1[/latex] to the input of the function [latex]f(x) = \sqrt[3]{x}[/latex] results in a horizontal shift of the graph to the left by [latex]1[/latex] unit.
A function [latex]f\left(x\right)[/latex] is given below. Create a table for the function [latex]g\left(x\right)=f\left(x - 3\right)[/latex].
[latex]x[/latex]
2
4
6
8
[latex]f\left(x\right)[/latex]
1
3
7
11
The formula [latex]g\left(x\right)=f\left(x - 3\right)[/latex] tells us that the output values of [latex]g[/latex] are the same as the output value of [latex]f[/latex] when the input value is 3 less than the original value. For example, we know that [latex]f\left(2\right)=1[/latex]. To get the same output from the function [latex]g[/latex], we will need an input value that is 3 larger. We input a value that is 3 larger for [latex]g\left(x\right)[/latex] because the function takes 3 away before evaluating the function [latex]f[/latex].
We continue with the other values to create this table.
[latex]x[/latex]
5
7
9
11
[latex]x - 3[/latex]
2
4
6
8
[latex]f\left(x\right)[/latex]
1
3
7
11
[latex]g\left(x\right)[/latex]
1
3
7
11
The result is that the function [latex]f\left(x\right)[/latex] has been shifted to the right by 3. Notice the output values for [latex]g\left(x\right)[/latex] remain the same as the output values for [latex]f\left(x\right)[/latex], but the corresponding input values, [latex]x[/latex], have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11.
Analysis of the Solution
The graph below represents both of the functions. We can see the horizontal shift in each point.
The graph below represents a transformation of the toolkit function [latex]f\left(x\right)={x}^{2}[/latex]. Relate this new function [latex]g\left(x\right)[/latex] to [latex]f\left(x\right)[/latex], and then find a formula for [latex]g\left(x\right)[/latex].
Notice that the graph is identical in shape to the [latex]f\left(x\right)={x}^{2}[/latex] function, but the [latex]x[/latex]–values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so
Notice how we must input the value [latex]x=2[/latex] to get the output value [latex]y=0[/latex]; the [latex]x[/latex]-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the [latex]f\left(x\right)[/latex] function to write a formula for [latex]g\left(x\right)[/latex] by evaluating [latex]f\left(x - 2\right)[/latex].
To determine whether the shift is [latex]+2[/latex] or [latex]-2[/latex] , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, [latex]f\left(0\right)=0[/latex]. In our shifted function, [latex]g\left(2\right)=0[/latex]. To obtain the output value of 0 from the function [latex]f[/latex], we need to decide whether a plus or a minus sign will work to satisfy [latex]g\left(2\right)=f\left(x - 2\right)=f\left(0\right)=0[/latex]. For this to work, we will need to subtract 2 units from our input values.
The function [latex]G\left(m\right)[/latex] gives the number of gallons of gas required to drive [latex]m[/latex] miles. Interpret [latex]G\left(m\right)+10[/latex] and [latex]G\left(m+10\right)[/latex].
[latex]G\left(m\right)+10[/latex] can be interpreted as adding [latex]10[/latex] to the output, gallons. This is the gas required to drive [latex]m[/latex] miles, plus another [latex]10[/latex] gallons of gas. The graph would indicate a vertical shift.[latex]G\left(m+10\right)[/latex] can be interpreted as adding [latex]10[/latex]to the input, miles. So this is the number of gallons of gas required to drive [latex]10[/latex] miles more than [latex]m[/latex] miles. The graph would indicate a horizontal shift.
Original Function [latex]f(x)[/latex]
