Exponential and Logarithmic Functions: Background You’ll Need 2
Change the appearance of functions on a graph by shifting, stretching, or flipping them
Performing a Sequence of Transformations
Combining transformations follows a specific order of operations similar to the mathematical order of operations (PEMDAS: Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). When applying transformations to a function, the sequence ensures that each transformation is applied correctly and the resulting graph reflects the intended changes.
order of transformations
When transforming a function [latex]y = a \cdot f(bx-c)+d[/latex], the general order of transformation is as follows:
Horizontal Shifts by [latex]c[/latex] units.
[latex]f(x)[/latex] is shifted to the right if [latex]c[/latex] is positive
[latex]f(x)[/latex] is shifted to the left if [latex]c[/latex] is negative.
Horizontal Stretches/Compressions by a factor of [latex]\dfrac{1}{b}[/latex].
[latex]f(x)[/latex] is compressed horizontally if [latex]|b|>1[/latex].
[latex]f(x)[/latex] is stretched horizontally if [latex]0<|b|<1[/latex].
Reflections
Reflection across the y-axis if [latex]b[/latex] is negative.
Reflection across the x-axis if [latex]a[/latex] is negative.
Vertical Stretches/Compressions by a factor of [latex]a[/latex].
[latex]f(x)[/latex] is stretched vertically if [latex]|a|>1[/latex].
[latex]f(x)[/latex] is compressed vertically if [latex]0<|a|<1[/latex].
Vertical Shifts by [latex]d[/latex] units.
[latex]f(x)[/latex] is shifted to the upward if [latex]d[/latex] is positive
[latex]f(x)[/latex] is shifted to the downward if [latex]d[/latex] is negative.
When combining vertical transformations written in the form [latex]af\left(x\right)+k[/latex], first vertically stretch by [latex]a[/latex] and then vertically shift by [latex]k[/latex].
[latex]\\[/latex]
When combining horizontal transformations written in the form [latex]f\left(bx-h\right)[/latex], first horizontally shift by [latex]\frac{h}{b}[/latex] and then horizontally stretch by [latex]\frac{1}{b}[/latex].
[latex]\\[/latex]
When combining horizontal transformations written in the form [latex]f\left(b\left(x-h\right)\right)[/latex], first horizontally stretch by [latex]\frac{1}{b}[/latex] and then horizontally shift by [latex]h[/latex].Given [latex]f(x)=|x|[/latex], identify the transformations and graph the transformed function
[latex]h(x)=f(x+1)-3 = |x+1|-3[/latex]
Step-by-Step Transformations
Original Function: The original function is [latex]f(x)=|x|[/latex].
Transformations:
Horizontal Shift: The term [latex]|x+1|[/latex]indicates a horizontal shift to the left by [latex]1[/latex] unit. This is because [latex]x+1 = 0[/latex] when [latex]x=-1[/latex]so the entire graph of i[latex]f(x)=|x|[/latex] is moved [latex]1[/latex] unit to the left.
Vertical Shift: The term [latex]-3[/latex] indicates a vertical shift downward by [latex]3[/latex] units.
Graph
Original Graph
[latex]y=|x|[/latex]
Left by [latex]1[/latex] unit
[latex]y=|x+1|[/latex]
Downward by [latex]3[/latex] units
y = |x+1|-3
[latex](0,0)[/latex]
[latex](-1,0)[/latex]
[latex](-1,-3)[/latex]
[latex](1,1)[/latex]
[latex](0,1)[/latex]
[latex](0,-2)[/latex]
Use the given graph of [latex]f(x)[/latex] to draw the transformed function: [latex]g(x)=f(\frac{1}{2}x+1)-3[/latex].
The original function [latex]f(x)[/latex] has key points: [latex](-2,0)[/latex], [latex](0,2)[/latex], and [latex](2,0)[/latex].
[latex]\\[/latex] Step-by-Step Transformations
Horizontal Shift Left by [latex]1[/latex] unit
Horizontal Stretch by a factor of [latex]2[/latex] (Note: this impact the [latex]x[/latex]-values)
Vertical Shift Down by [latex]3[/latex] units
Original Point
Left by [latex]1[/latex] unit
Horizontal Stretch by a factor of [latex]2[/latex]
Down by [latex]3[/latex] units
[latex](-2,0)[/latex]
[latex](-3,0)[/latex]
[latex](-6,0)[/latex]
[latex](-6,-3)[/latex]
[latex](0,2)[/latex]
[latex](-1,2)[/latex]
[latex](-2,2)[/latex]
[latex](-2,-1)[/latex]
[latex](2,0)[/latex]
[latex](1,0)[/latex]
[latex](2,0)[/latex]
[latex](2,-3)[/latex]
Write a formula for the graph shown below, which is a transformation of the toolkit square root function.
The graph of the toolkit function starts at the origin, so this graph has been shifted [latex]1[/latex] to the right and up [latex]2[/latex]. In function notation, we could write that as
Using the formula for the square root function, we can write
[latex]h\left(x\right)=\sqrt{x - 1}+2[/latex]
Analysis of the Solution
Note that this transformation has changed the domain and range of the function. This new graph has domain [latex]\left[1,\infty \right)[/latex] and range [latex]\left[2,\infty \right)[/latex].
A common model for learning has an equation similar to [latex]k\left(t\right)=-{2}^{-t}+1[/latex], where [latex]k[/latex] is the percentage of mastery that can be achieved after [latex]t[/latex] practice sessions. This is a transformation of the function [latex]f\left(t\right)={2}^{t}[/latex] shown below.
[latex]\\[/latex]
Sketch a graph of [latex]k\left(t\right)[/latex].
This equation combines three transformations into one equation.
A horizontal reflection: [latex]f\left(-t\right)={2}^{-t}[/latex]
A vertical reflection: [latex]-f\left(-t\right)=-{2}^{-t}[/latex]
A vertical shift: [latex]-f\left(-t\right)+1=-{2}^{-t}+1[/latex]
We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each of the three transformations. We will choose the points [latex](0, 1)[/latex] and [latex](1, 2)[/latex].
First, we apply a horizontal reflection: [latex](0, 1) (–1, 2)[/latex].
Then, we apply a vertical reflection: [latex](0, −1) (1, –2)[/latex].
Finally, we apply a vertical shift: [latex](0, 0) (1, 1)[/latex].
This means that the original points, [latex](0,1)[/latex] and [latex](1,2)[/latex] become [latex](0,0)[/latex] and [latex](1,1)[/latex] after we apply the transformations.
In the graphs below, the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit.
Analysis of the Solution
As a model for learning, this function would be limited to a domain of [latex]t\ge 0[/latex], with corresponding range [latex]\left[0,1\right)[/latex].
Given the table below for the function [latex]f\left(x\right)[/latex], create a table of values for the function [latex]g\left(x\right)=2f\left(3x\right)+1[/latex].
[latex]x[/latex]
[latex]6[/latex]
[latex]12[/latex]
[latex]18[/latex]
[latex]24[/latex]
[latex]f\left(x\right)[/latex]
[latex]10[/latex]
[latex]14[/latex]
[latex]15[/latex]
[latex]17[/latex]
There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal transformations, [latex]f\left(3x\right)[/latex] is a horizontal compression by [latex]\frac{1}{3}[/latex], which means we multiply each [latex]x\text{-}[/latex] value by [latex]\frac{1}{3}[/latex].
[latex]x[/latex]
[latex]2[/latex]
[latex]4[/latex]
[latex]6[/latex]
[latex]8[/latex]
[latex]f\left(3x\right)[/latex]
[latex]10[/latex]
[latex]14[/latex]
[latex]15[/latex]
[latex]17[/latex]
Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation.
[latex]x[/latex]
[latex]2[/latex]
[latex]4[/latex]
[latex]6[/latex]
[latex]8[/latex]
[latex]2f\left(3x\right)[/latex]
[latex]20[/latex]
[latex]28[/latex]
[latex]30[/latex]
[latex]34[/latex]
Finally, we can apply the vertical shift, which will add 1 to all the output values.
Use the given graph of [latex]f(x)[/latex] to draw the transformed function: [latex]g(x)=f(\frac{1}{2}x+1)-3[/latex].
