# 4.1 Linear functions  (Page 9/27)

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## Vertical stretch or compression

In the equation $\text{\hspace{0.17em}}f\left(x\right)=mx,$ the $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ is acting as the vertical stretch    or compression of the identity function. When $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ is negative, there is also a vertical reflection of the graph. Notice in [link] that multiplying the equation of $\text{\hspace{0.17em}}f\left(x\right)=x\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ stretches the graph of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ by a factor of $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}m>\text{1}\text{\hspace{0.17em}}$ and compresses the graph of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ by a factor of $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}0 This means the larger the absolute value of $\text{\hspace{0.17em}}m,\text{\hspace{0.17em}}$ the steeper the slope.

## Vertical shift

In $\text{\hspace{0.17em}}f\left(x\right)=mx+b,$ the $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ to the equation of $\text{\hspace{0.17em}}f\left(x\right)=x\text{\hspace{0.17em}}$ shifts the graph of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ a total of $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ units up if $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is positive and $\text{\hspace{0.17em}}|b|\text{\hspace{0.17em}}$ units down if $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is negative.

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 $\text{\hspace{0.17em}}f\left(x\right)=mx+b.$

1. Graph $\text{\hspace{0.17em}}f\left(x\right)=x.$
2. Vertically stretch or compress the graph by a factor $\text{\hspace{0.17em}}m.$
3. Shift the graph up or down $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ units.

## Graphing by using transformations

Graph $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}x-3\text{\hspace{0.17em}}$ using transformations.

The equation for the function shows that $\text{\hspace{0.17em}}m=\frac{1}{2}\text{\hspace{0.17em}}$ so the identity function is vertically compressed by $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$ The equation for the function also shows that $\text{\hspace{0.17em}}b=-3\text{\hspace{0.17em}}$ so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in [link] .

Then show the vertical shift as in [link] .

Graph $\text{\hspace{0.17em}}f\left(x\right)=4+2x\text{\hspace{0.17em}}$ using transformations.

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.

$\begin{array}{ccc}\hfill f\left(2\right)& =& \frac{1}{2}\left(2\right)-3\hfill \\ & =& 1-3\hfill \\ & =& -2\hfill \end{array}$

## 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 $\text{\hspace{0.17em}}\left(0,\text{4}\right)\text{\hspace{0.17em}}$ so this is the y -intercept.

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 $\text{\hspace{0.17em}}\left(–2,0\right).\text{\hspace{0.17em}}$ 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

$m=\frac{\text{rise}}{\text{run}}=\frac{4}{2}=2$

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

Need help solving this problem (2/7)^-2
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
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?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
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Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
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salma
Commplementary angles
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salma
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Ali
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Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_