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Vertical stretch or compression

In the equation f ( x ) = m x , the m is acting as the vertical stretch    or compression of the identity function. When m is negative, there is also a vertical reflection of the graph. Notice in [link] that multiplying the equation of f ( x ) = x by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. This means the larger the absolute value of m , the steeper the slope.

This graph shows seven versions of the function, f of x = x on an x, y coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. Seven multi-colored lines run through the point (0, 0). Starting with the lines in the top right quadrant and moving clockwise, the first line is f of x = 3 times x and has a slope of 3, the next line is f of x = 2 times x which has a slope of 2, the next line is f of x = x which has a slope of 1, the next line is f of x = x divided by 2 which has a slope of .5. The last line in this quadrant is f of x = x divided by 3 which has a slope of one third x. In the bottom right quadrant moving clockwise, the first line is f of x = negative x divided by 2, which has a slope of negative one half, the middle line is f of x = negative x which has a slope of negative 1, and the last line is f of x = negative 2 times x which has a slope of  negative 2.
Vertical stretches and compressions and reflections on the function f ( x ) = x

Vertical shift

In f ( x ) = m x + b , the b acts as the vertical shift    , moving the graph up and down without affecting the slope of the line. Notice in [link] that adding a value of b to the equation of f ( x ) = x shifts the graph of f a total of b units up if b is positive and | b | units down if b is negative.

This graph shows six versions of the function, f of x = x, on an x, y coordinate plane. The x-axis runs from negative 8 to 8, and the y axis runs negative 8 to 8. There are five lines parallel to each other. The first line extends from the bottom left quadrant to the upper right quadrant on the coordinate plane. This line shows f of x = x plus 4 which has a slope of 1 and a y-intercept at 4. The next line also extends from the bottom left quadrant to the upper right quadrant and shows f of x = x plus 2 which has a slope of 1 and a y-intercept at 2. The next and middle line, extends from the lower left quadrant, through the center of the graph at point (0, 0) to the upper right quadrant and shows f of x = x. The next line extends from the lower left quadrant, through the lower right quadrant to the upper right quadrant. This line shows f of x = x minus 2 which has a slope of 1 and a y-intercept at -2. The last line extends from the lower left quadrant, through the lower right quadrant to the upper right quadrant.This line shows f of x = x minus 4 which has a slope of 1 and a y-intercept at -4.
This graph illustrates vertical shifts of the function f ( x ) = x .

Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.

Given the equation of a linear function, use transformations to graph the linear function in the form f ( x ) = m x + b .

  1. Graph f ( x ) = x .
  2. Vertically stretch or compress the graph by a factor m .
  3. Shift the graph up or down b units.

Graphing by using transformations

Graph f ( x ) = 1 2 x 3 using transformations.

The equation for the function shows that m = 1 2 so the identity function is vertically compressed by 1 2 . The equation for the function also shows that b = 3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in [link] .

This graph shows two functions on an x, y coordinate plane. One shows an increasing function of y = x divided by 2 that runs through the points (0, 0) and (2, 1). The second shows an increasing function of y = x and runs through the points (0, 0) and (1, 1)).
The function, y = x , compressed by a factor of 1 2

Then show the vertical shift as in [link] .

This graph shows two functions on an x, y coordinate plane. The first is an increasing function of y = x divided by 2 and runs through the points (0, 0) and (2, 1).  The second shows an increasing function of y = x divided by 2 minus 3 and passes through the points (0, 3) and (2, -2).  An arrow pointing downward from the first function  to the second function reveals the vertical shift.
The function y = 1 2 x , shifted down 3 units
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Graph f ( x ) = 4 + 2 x using transformations.

This graph shows three functions on an x, y coordinate plane. One shows an increasing function y = x that passes through points (0, 0) and (2, 2).  A second shows an increasing function y = 2 times x that passes through the points (0, 0) and (2, 4).  The third is an increasing function y = 2 times x plus 4 and passes through the points (0, 4) and (2, 8).
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In [link] , could we have sketched the graph by reversing the order of the transformations?

No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.

f ( 2 ) = 1 2 ( 2 ) 3 = 1 3 = −2

Writing the equation for a function from the graph of a line

Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at [link] . We can see right away that the graph crosses the y -axis at the point ( 0 , 4 ) so this is the y -intercept.

This graph shows the function f of x = 2 times x plus 4 on an x, y coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. This function passes through the points (-2, 0) and (0, 4).

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point ( 2 , 0 ) . To get from this point to the y- intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

m = rise run = 4 2 = 2

Substituting the slope and y- intercept into the slope-intercept form of a line gives

Questions & Answers

Need help solving this problem (2/7)^-2
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-1
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An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
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lim x to infinity e^1-e^-1/log(1+x)
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If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
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how do they get the third part x = (32)5/4
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make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
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20/(×-6^2)
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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