Which Parent Function is Represented by the Graph

Which Parent Function is Represented by the Graph

For Absolute Value Transformations, see theAbsolute Value Transformations section. Here are links to
Parent Part Transformations
in other sections:
Transformations of Quadratic Functions
(quick and piece of cake way);Transformations of Radical Functions;Transformations of Rational Functions;
Transformations of Exponential Functions;Transformations of Logarithmic Functions;
Transformations of Piecewise Functions;
Transformations of Trigonometric Functions;
Transformations of Inverse Trigonometric Functions

You may non be familiar with all the functions and characteristics in the tables; here are some topics to review:

  • Whether functions are
    fifty-fifty
    ,
    odd, or
    neither, discussed hither in the
    Advanced Functions: Compositions, Even and Odd, and Extrema
    .
  • End behavior
    and
    asymptotes
    , discussed in the
    Asymptotes and Graphing Rational Functions
    and
    Graphing Polynomials

    sections
  • Exponential
    and
    Logarithmic
    Functions
  • Trigonometric
    Functions

Basic Parent Functions

Y’all’ll probably written report some “pop”
parent functions
and work with these to learn how to
transform functions

– how to move and/or resize them. We telephone call these basic functions “parent” functions since they are the simplest form of that type of office, meaning they are as close as they tin can get to the
origin
\(\left( {0,0} \correct)\).

The chart below provides some basic parent functions that yous should be familiar with. I’ve too included the
significant
points, or
critical points, the points with which to graph the parent function. I likewise sometimes telephone call these the “reference points” or “ballast points”.

Know the shapes of these parent functions well! Fifty-fifty when using


t

-charts, you must know the general shape of the parent functions in lodge to know how to transform them correctly!

Parent Function Graph Parent Function Graph

\(y=x\)
Linear, Odd

Domain: \(\left( {-\infty ,\infty } \correct)\)
Range: \(\left( {-\infty ,\infty } \right)\)

Terminate Beliefs**:
\(\begin{array}{50}x\to -\infty \text{, }\,y\to -\infty \\10\to \infty \text{, }\,\,\,y\to \infty \end{array}\)

Disquisitional points: \(\displaystyle \left( {-1,-i} \correct),\,\left( {0,0} \right),\,\left( {1,1} \right)\)

\(y=\left| x \right|\)
Absolute Value, Fifty-fifty

Domain: \(\left( {-\infty ,\infty } \correct)\)
Range: \(\left[ {0,\infty } \correct)\)

End Beliefs:
\(\begin{assortment}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}\)

Disquisitional points: \(\displaystyle \left( {-one,ane} \right),\,\left( {0,0} \right),\,\left( {i,1} \right)\)

\(y={{x}^{2}}\)
Quadratic, Even

Domain: \(\left( {-\infty ,\infty } \right)\)
Range: \(\left[ {0,\infty } \right)\)

End Behavior:
\(\begin{assortment}{50}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}\)

Critical points:

\(\displaystyle \left( {-one,i} \right),\,\left( {0,0} \correct),\,\left( {ane,i} \correct)\)

\(y=\sqrt{x}\)
Radical
(Square Root), Neither

Domain: \(\left[ {0,\infty } \right)\)
Range: \(\left[ {0,\infty } \right)\)

Finish Beliefs:

\(\displaystyle \begin{assortment}{50}ten\to 0,\,\,\,\,y\to 0\\10\to \infty \text{,}\,\,y\to \infty \end{assortment}\)

Critical points:

\(\displaystyle \left( {0,0} \right),\,\left( {1,1} \correct),\,\left( {four,2} \right)\)

\(y={{x}^{3}}\)
Cubic, Odd

Domain: \(\left( {-\infty ,\infty } \right)\)
Range: \(\left( {-\infty ,\infty } \correct)\)

Finish Behavior:
\(\begin{assortment}{l}10\to -\infty \text{, }\,y\to -\infty \\10\to \infty \text{, }\,\,\,y\to \infty \finish{array}\)

Critical points:

\(\displaystyle \left( {-ane,-ane} \right),\,\left( {0,0} \right),\,\left( {1,one} \right)\)

\(y=\sqrt[3]{10}\)
Cube Root, Odd

Domain: \(\left( {-\infty ,\infty } \right)\)
Range: \(\left( {-\infty ,\infty } \correct)\)

End Behavior:
\(\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \cease{array}\)

Critical points:

\(\displaystyle \left( {-i,-1} \right),\,\left( {0,0} \right),\,\left( {i,i} \right)\)

\(\begin{array}{c}y={{b}^{ten}},\,\,\,b>ane\,\\(\text{Example:}\,\,y={{two}^{x}})\stop{array}\)

Exponential, Neither

Domain: \(\left( {-\infty ,\infty } \right)\)
Range: \(\left( {0,\infty } \right)\)

Stop Beliefs:
\(\begin{array}{l}x\to -\infty \text{, }\,y\to 0\\ten\to \infty \text{, }\,\,\,y\to \infty \stop{array}\)

Disquisitional points:

\(\displaystyle \left( {-1,\frac{1}{b}} \right),\,\left( {0,1} \right),\,\left( {1,b} \right)\)

Asymptote:  \(y=0\)

\(\begin{assortment}{c}y={{\log }_{b}}\left( ten \right),\,\,b>1\,\,\,\\(\text{Example:}\,\,y={{\log }_{2}}x)\end{array}\)

Log, Neither

Domain: \(\left( {0,\infty } \correct)\)
Range: \(\left( {-\infty ,\infty } \right)\)

Cease Behavior:
\(\begin{array}{l}x\to {{0}^{+}}\text{, }\,y\to -\infty \\10\to \infty \text{, }\,y\to \infty \end{array}\)

Disquisitional points:

\(\displaystyle \left( {\frac{1}{b},-1} \right),\,\left( {1,0} \right),\,\left( {b,i} \right)\)

Asymptote: \(10=0\)

\(\displaystyle y=\frac{1}{x}\)

Rational (Changed), Odd

Domain: \(\left( {-\infty ,0} \correct)\cup \left( {0,\infty } \right)\)
Range: \(\left( {-\infty ,0} \correct)\loving cup \left( {0,\infty } \right)\)

Terminate Beliefs:
\(\begin{array}{l}ten\to -\infty \text{, }\,y\to 0\\10\to \infty \text{, }\,\,\,y\to 0\end{array}\)

Critical points:

\(\displaystyle \left( {-1,-1} \right),\,\left( {ane,ane} \right)\)

Asymptotes: \(y=0,\,\,x=0\)

**Note that this function is the
inverse
of itself!

\(\displaystyle y=\frac{one}{{{{x}^{two}}}}\)

Rational (Inverse Squared), Fifty-fifty

Domain: \(\left( {-\infty ,0} \right)\loving cup \left( {0,\infty } \right)\)
Range: \(\left( {0,\infty } \right)\)

End Behavior:
\(\begin{array}{l}x\to -\infty \text{, }\,y\to 0\\x\to \infty \text{, }\,\,\,y\to 0\end{array}\)

Critical points:

\(\displaystyle \left( {-1,\,1} \right),\left( {i,1} \right)\)

Asymptotes: \(x=0,\,\,y=0\)

\(y=\text{int}\left( x \right)=\left\lfloor x \correct\rfloor \)

Greatest Integer*
, Neither

Domain: \(\left( {-\infty ,\infty } \correct)\)
Range: \(\{y:y\in \mathbb{Z}\}\text{ (integers)}\)

End Beliefs:
\(\brainstorm{assortment}{l}10\to -\infty \text{, }\,y\to -\infty \\ten\to \infty \text{, }\,\,\,y\to \infty \finish{array}\)

Disquisitional points:

\(\displaystyle \begin{array}{50}x:\left[ {-1,0} \right)\,\,\,y:-ane\\x:\left[ {0,1} \right)\,\,\,y:0\\x:\left[ {1,2} \right)\,\,\,y:1\cease{array}\)

\(y=C\)   (\(y=2\))

Abiding, Even

Domain: \(\left( {-\infty ,\infty } \right)\)
Range: \(\{y:y=C\}\)

End Behavior:
\(\begin{array}{l}ten\to -\infty \text{, }\,y\to C\\10\to \infty \text{, }\,\,\,y\to C\finish{array}\)

Critical points:

\(\displaystyle \left( {-ane,C} \right),\,\left( {0,C} \correct),\,\left( {one,C} \correct)\)

*The
Greatest Integer
Function, sometimes called the
Footstep Function, returns the greatest integer less than or equal to a number (think of rounding downwards to an integer). There’due south as well a
Least Integer
Role, indicated past \(y=\left\lceil x \right\rceil \), which returns the least integer greater than or equal to a number (recall of rounding upwards to an integer).




Sponsored Links

**Notes on End Behavior: To go theend behavior of a office, nosotros just expect at thesmallest andlargest
values of \(ten\), and encounter which way the \(y\) is going. Non all functions take end behavior defined; for example, those that get dorsum and forth with the \(y\) values (called “periodic functions”) don’t have end behaviors.
Nearly of the time, our end behavior looks something similar this: \(\displaystyle \begin{array}{fifty}x\to -\infty \text{, }\,y\to \,\,?\\x\to \infty \text{, }\,\,\,y\to \,\,?\cease{array}\) and nosotros have to fill in the \(y\) office. For example, the end behavior for a line with a positive gradient is: \(\begin{assortment}{l}10\to -\infty \text{, }\,y\to -\infty \\10\to \infty \text{, }\,\,\,y\to \infty \end{array}\), and the end behavior for a line with a negative slope is: \(\begin{assortment}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to -\infty \end{assortment}\). One way to think of terminate behavior is that for \(\displaystyle x\to -\infty \), we look at what’south going on with the \(y\) on the left-hand side of the graph, and for \(\displaystyle x\to \infty \), we expect at what’s happening with \(y\) on the right-hand side of the graph.

There are a couple of exceptions; for example, sometimes the \(10\) starts at



(such as in theradical function), we don’t have the negative portion of the \(10\) finish behavior. As well, when \(x\) starts very close to



(such as in in thelog function), we indicate that \(10\) is starting from the positive (right) side of



(and the \(y\) is going down); nosotros indicate this by \(\displaystyle ten\to {{0}^{+}}\text{, }\,y\to -\infty \).

Generic Transformations of Functions

Again, the “parent functions” presume that we have the simplest grade of the office; in other words, the function either goes through the origin \(\left( {0,0} \right)\), or if it doesn’t go through the origin, information technology isn’t shifted in whatever way. When a function is
shifted, stretched
(or
compressed), or flipped in any style from its “parent part“, information technology is said to be
transformed, and is a
transformation of a function.


T-charts

are extremely useful tools when dealing with transformations of functions. For example, if y’all know that the quadratic parent function \(y={{ten}^{two}}\) is being transformed

ii
units to the right
, and

ane
unit of measurement down

(merely a shift, non a stretch or a flip), we can create the original

t
-nautical chart, following by the transformation points on the outside of the original points. Then we can plot the “exterior” (new) points to become the newly transformed function:

Transformation
T-chart
Graph

Quadratic Office

\(y={{10}^{two}}\)

Transform function

ii
units to the correct
, and

ane
unit down
.

This turns into the function \(y={{\left( {x-2} \right)}^{two}}-1\), oddly enough!



10 +


2
x





y   y  –


1

      one    –one 1      0
      ii     –1
      iiione i      0

Transformed:

Domain:  \(\left( {-\infty ,\infty } \correct)\)

Range:   \(\left[ {-1,\,\,\infty } \right)\)

When looking at
the equation of the transformed part, notwithstanding, we have to be careful. When functions are transformed on the
exterior
of the \(f(x)\) part, you motility the function up and down and do the “regular” math, as we’ll see in the examples below. These are
vertical
transformations
or
translations, and touch the \(y\) part of the function. When transformations are made on the
within

of the \(f(ten)\) part, y’all move the function
back and forth
(but practice the “contrary” math – since if you were to isolate the \(x\), you lot’d move everything to the other side). These are
horizontal transformations
or
translations
, and affect the \(x\) office of the part.

There are several means to perform transformations of parent functions; I like to apply


t

-charts
, since they work consistently with ever function. And note that in most

t-charts
,

I’ve included more just the
critical points
to a higher place, just to evidence the graphs ameliorate.

Vertical Transformations

Here are the rules and examples of when functions are transformed on the “outside” (find that the \(y\)values are affected). The

t
-charts

include the points (ordered pairs) of the original parent functions, and also the transformed or shifted points. The first 2 transformations are
translations, the third is a
dilation, and the last are forms of
reflections
. Absolute value transformations will be discussed more expensively in the
Absolute Value Transformations section!

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Trans
formation
What Information technology Does Example Graph

\(f\left( x \right)+b\)

Translation

Move graph
upwardly
\(b\) units

Every signal on the graph is shifted up \(b\) units.

The \(x\)’s stay the aforementioned; add together \(b\) to the \(y\) values.

Parent:
\(y={{x}^{ii}}\)

Transformed:
\(y={{ten}^{2}}+ \,2\)



  x



  y  y

+2
–1 1    3
 0     2
 1 1    3

Domain:  \(\left( {-\infty ,\infty } \right)\)
Range:  \(\left[ {2,\infty } \right)\)

\(f\left( x \correct)-b\)

Translation

Move graph
downwardly
\(b\) units

Every point on the graph is shifted down \(b\) units.

The \(x\)’s stay the same; subtract \(b\) from the \(y\) values.

Parent: \(y=\sqrt{10}\)

Transformed: \(y=\sqrt{x}- \,iii\)


x

y   y

–3

–3
1 1   –2
iv 2   –1


Domain:
  \(\left[ {0,\infty } \right)\)
Range:
\(\left[ {-3,\infty } \right)\)

\(a\,\cdot f\left( x \right)\)

Dilation

Stretch
graph vertically by a calibration factor of \(a\) (sometimes called a
dilation). Note that if \(a<i\), the graph is
compressed
or shrunk.

Every betoken on the graph is stretched \(a\) units.

The \(x\)’s stay the aforementioned; multiply the \(y\) values past \(a\).

Parent: \(y={{10}^{three}}\)

Transformed: \(y={{4x}^{3}}\)


  x

  y

ivy
–ane –i  –iv
  0     0
  1 1    4


Domain:

\(\left( {-\infty ,\infty } \right)\)Range:
\(\left( {-\infty ,\,\infty } \right)\)

\(-f\left( x \right)\)

Reflection

Flip
graph around the \(ten\)-axis.

Every point on the graph is flipped vertically.

The \(x\)’s stay the aforementioned; multiply the \(y\) values by \(-i\).

Parent: \(y=\left| x \right|\)

Transformed: \(y=-\left| 10 \right|\)


  ten

y   –y
–1 ane   –1
  0      0
  i 1   –one


Domain:

\(\left( {-\infty ,\infty } \right)\)Range:
\(\left( {-\infty\,,0} \correct]\)

\(\left| {f\left( ten \right)} \correct|\)

Absolute Value
on the \(y\)

(More examples here in the
Absolute Value Transformation
department)

Reflect part of graph underneath the \(x\)-centrality (negative \(y\)’due south) across the \(x\)-axis. Leave positive \(y\)’south the same.

The \(x\)’s stay the same;
accept the absolute value of the
\(y\)’s.

Parent:  \(y=\sqrt[3]{x}\)

Transformed: \(y=\left| {\sqrt[3]{x}} \right|\)


ten

 y    |y|
–1 –1    ane
  0       0
  1 one      1

Domain:
  \(\left( {-\infty ,\infty } \right)\)
Range:\(\left[ {0,\infty } \correct)\)

Horizontal Transformations

Hither are the rules and examples of when functions are transformed on the “inside” (notice that the \(x\)-values are affected). Find that when the \(ten\)-values are affected, you
do the math in the “opposite” manner from what the function looks like: if you lot’re adding on the inside, you subtract from the \(x\); if you’re subtracting on the within, you lot add to the \(ten\); if you’re multiplying on the inside, you divide from the \(x\); if you’re dividing on the inside, y’all multiply to the \(10\). If yous accept a negative value on the inside, you
flip across the
\(\boldsymbol{y}\) axis
(notice that you nevertheless multiply the \(x\) by \(-one\) simply like you exercise for with the \(y\) for vertical flips). The first two transformations are
translations, the third is a
dilation, and the last are forms of
reflections
.

Absolute value transformations will be discussed more than expensively in the
Absolute Value Transformations
department!

(Yous may observe it interesting is that a vertical stretch behaves the aforementioned mode as a horizontal compression, and vice versa, since when stretch something upward, we are making it skinnier.)

Trans
formation
What It Does Example Graph
 \(f\left( {10+b} \right)\)

Translation

Move graph
left \(b\)  units

(Exercise the “contrary” when change is within the parentheses or underneath radical sign.)

Every signal on the graph is shifted left  \(b\)  units.

The \(y\)’s stay the same; subtract  \(b\)


from the \(x\) values.

Parent: \(y={{ten}^{2}}\)

Transformed: \(y={{\left( {x+two} \right)}^{2}}\)


 x


– 2    x

y
   –three–i 1
   –two
   –iane 1


Domain:
 \(\left( {-\infty ,\infty } \right)\)Range:
\(\left[ {0,\infty } \correct)\)

 \(f\left( {x-b} \right)\)

Translation

Move graph
correct \(b\)
units

Every point on the graph is shifted correct \(b\) units.

The \(y\)’southward stay the aforementioned; add \(b\)


to the \(x\) values.

Parent: \(y=\sqrt{x}\)

Transformed: \(y=\sqrt{{x- \,3}}\)


x


+ three   x

y
   iii
   4
one
i
   7
4
2


Domain:
 \(\left[ {-three,\infty } \right)\)Range:
\(\left[ {0,\infty } \right)\)

 \(f\left( {a\cdot 10} \right)\)

Dialation

Compress
graph horizontally by a scale factor of \(a\) units (stretch or multiply past \(\displaystyle \frac{one}{a}\))

Every bespeak on the graph is compressed \(a\)


units horizontally.

The \(y\)’southward stay the same; multiply the \(x\)-values by \(\displaystyle \frac{1}{a}\).

Parent: \(y={{x}^{iii}}\)

T
ransformed: \(y={{\left( {4x} \right)}^{3}}\)

\(\frac{1}{four}x\)
ten

y
\(-\frac{1}{4}\)–1 –ane
    0
   \(\frac{1}{four}\)1 ane


Domain:
 \(\left( {-\infty ,\infty } \right)\)Range: \(\left( {-\infty ,\infty } \right)\)

 \(f\left( {-10} \right)\)

Reflection

Flip
graph around the \(y\)-axis

Every betoken on the graph is flipped around the \(y\) axis.

The \(y\)’south stay the same; multiply the \(10\)-values by \(-1\).

Parent: \(y=\sqrt{10}\)

Transformed: \(y=\sqrt{{-x}}\)


 –x


    x

y
   0
 –ane1 ane
 –4iv 2


Domain:
  \(\left( {-\infty ,0} \right]\)Range:
\(\left[ {0,\infty } \right)\)

\(f\left( {\left| x \right|} \right)\)

Absolute Value
on the \(x\)

(More examples here in the
Absolute Value Transformation
department)

“Throw abroad” the negative \(ten\)’due south; reflect the positive \(x\)’s beyond the \(y\)-axis.

The positive \(x\)’southward stay the aforementioned; the
negative
\(x\)’s take on the
\(y\)’s of the positive
\(10\)’s.

Parent: \(y=\sqrt{x}\)

Transformed: \(y=\sqrt{{\left| 10 \right|}}\)


10

y

New
y

–4 NA
2
 –ane NA
1
  0
 1
1
 4
ii

Domain:  \(\left( {-\infty ,\infty } \right)\)Range:\(\left[ {0,\infty } \correct)\)

Mixed Transformations

Most of the bug you’ll get volition involve
mixed transformations, or multiple transformations, and we practise need to worry about
the order
in which we perform the transformations. It commonly doesn’t matter if we make the \(x\) changes or the \(y\) changes beginning, merely within the \(x\)’s and \(y\)’due south, we need to perform the transformations in the order below. Note that this is sort of similar to the social club with
PEMDAS(parentheses, exponents, multiplication/partitioning, and improver/subtraction). When performing these rules, the coefficients of the within \(x\) must exist

1
; for instance, nosotros would need to have \(y={{\left( {4\left( {x+2} \correct)} \right)}^{2}}\) instead of \(y={{\left( {4x+eight} \right)}^{2}}\) (by factoring). If yous didn’t learn it this way, meet


IMPORTANT NOTE


below.

Here is the order. We can do
steps i and 2 together
(society doesn’t really matter), since nosotros can recall of the commencement two steps as a “negative stretch/compression.”

  1. Perform
    Flipping across the axes first (negative signs).
  2. Perform
    Stretching and Shrinking next
    (multiplying and dividing).
  3. Perform
    Horizontal and Vertical shifts last
    (adding and subtracting).

I similar to take the
disquisitional points
and maybe a few more points of the parent functions, and perform all thetransformations at the same time
with a

t-chart
! We merely exercise the multiplication/division first on the \(x\) or \(y\) points, followed by addition/subtraction. It makes information technology much easier!Note again that since we don’t accept an
\(\boldsymbol {ten}\)
“by itself” (coefficient of
one) on the inside, we have to get it that mode by factoring!

For instance,we’d have to change \(y={{\left( {4x+8} \right)}^{2}}\text{ to }y={{\left( {iv\left( {x+ii} \right)} \right)}^{two}}\).


Let’south try to graph this “complicated” equation and I’ll evidence y’all how piece of cake it is to do with a
t-chart
:

\(\displaystyle f(ten)=-three{{\left( {2x+8} \right)}^{2}}+10\).
(Note that for this example, we could motion the \({{two}^{two}}\) to the outside to get a vertical stretch of \(iii\left( {{{2}^{ii}}} \correct)=12\), but we can’t exercise that for many functions.) We first need to go the \(x\) by itself
on the inside past
factoring, so we tin perform the horizontal translations. This is what nosotros end upward with:
\(\displaystyle f(x)=-3{{\left( {2\left( {x+4} \right)} \right)}^{2}}+10\).
Wait at what’south done on the “outside” (for the \(y\)’due south) and make all the moves at once, by following the
exact math. Then look at what we do on the “inside” (for the \(ten\)’due south) and make all the moves at once,
merely exercise the

opposite math. We do this with a

t
-chart.

Beginning with the parent function \(f(x)={{x}^{ii}}\). If we await at what we’re doing on the
outside
of what is being squared, which is the \(\displaystyle \left( {two\left( {ten+4} \correct)} \right)\), we’re
flipping
it across the \(x\)-axis (the minus sign),
stretching
information technology by a
factor of
three
, and
adding
10

(shifting upward

10
). These are the things that we are doing
vertically, or to the \(y\). If nosotros look at what we are doing on the inside of what we’re squaring, we’re multiplying it by

ii
, which means we have to
divide by
2
 (horizontal compression by a factor of \(\displaystyle \frac{i}{2}\)), and we’re adding

4
, which means nosotros have to
decrease
4

(a left shift of

4
). Remember that we do the
contrary
when we’re dealing with the \(ten\). Also recollect that nosotros always have to do the
multiplication or segmentation first
with our points, and so the
adding and subtracting
(sort of like
PEMDAS
).

Hither is the

t-chart

with the original function, and then the transformations on the outsides. Now we tin graph the outside points (points that aren’t crossed out) to get the graph of the transformation. I’ve also included an explanation of how to transform this parabola
without a
t-nautical chart
, as we did in the here in the
Introduction to Quadratics
section.


t-chart
Transformed Graph

Parent:  \(y={{x}^{ii}}\)  (Quadratic)

 Transformed:\(\displaystyle f(x)=-iii{{\left( {ii\left( {10+4} \right)} \right)}^{2}}+10\)


y


changes:\(\displaystyle f(x)=\color{bluish}{{-3}}{{\left( {2\left( {ten+four} \right)} \right)}^{2}}\color{blue}{+10}\)


x


changes:
\(\displaystyle f(x)=-3{{\left( {\color{blue}{2}\left( {x\text{ }\color{blueish}{{+\text{ }four}}} \right)} \right)}^{2}}+10\)

Opposite for \(x\), “regular” for \(y\), multiplying/dividing outset:

Coordinate Rule: \(\left( {10,\,y} \right)\to \left( {.5x-four,-3y+ten} \right)\)





.vx






4
    x

 y


–iiiy
+ 10
      –5        –2  4        –2
    –4.5
–one
one          7
      –4         10
    –3.5one  1          7
      –three2  4        –2

Domain:
\(\left( {-\infty ,\infty } \correct)\)
Range: \(\left( {-\infty ,ten} \right]\)

How to graphwithout a
t-chart:

\(\displaystyle f(ten)=-3{{\left( {2\left( {x+4} \right)} \right)}^{two}}+x\)

Since this is a
parabola
and it’due south in
vertex

form (\(y=a{{\left( {x-h} \right)}^{2}}+chiliad,\,\,\left( {h,thou} \correct)\,\text{vertex}\)), the
vertex
of the transformation is \(\left( {-4,ten} \right)\).

Notice that the coefficient of  is

–12

(past moving the \({{2}^{ii}}\) outside and multiplying it by the

–three
). Then the
vertical stretch
is

12
, and the
parabola faces down
because of the negative sign.

The parent graph quadratic goes up

1

and over (and back)

1

to get two more points, merely with a vertical stretch of

12
, we go over (and dorsum)

1

and downward

12

from the vertex. Now we have two points from which you lot can draw the parabola from the vertex.

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IMPORTANT NOTE:
In some books, for
\(\displaystyle f\left( ten \right)=-3{{\left( {2x+8} \right)}^{ii}}+x\)
, they may Non have you factor out thetwo on the inside, but just switch the guild of the transformation on the
\(\boldsymbol{x}\).




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In this case, the order of transformations would be horizontal shifts, horizontal reflections/stretches, vertical reflections/stretches, and and so vertical shifts. For example, for this problem, y’all would move to the left
eight
offset for the
\(\boldsymbol{x}\), and then compress with a factor of
\(\displaystyle \frac {1}{two}\)
for the
\(\boldsymbol{10}\) (which is contrary of PEMDAS). And so you would perform the
\(\boldsymbol{y}\)
(vertical) changes the regular way: reflect and stretch by
3
showtime, and so shift up
10. So, you would accept
\(\displaystyle {\left( {10,\,y} \right)\to \left( {\frac{one}{2}\left( {ten-8} \correct),-3y+ten} \right)}\). Try a
t-chart; you’ll get the same
t-nautical chart as above!

More Examples of Mixed Transformations:

Here are a couple more examples (using

t
-charts), with different parent functions. Don’t worry if you are totally lost with the exponential and log functions; they will be discussed in the
Exponential Functions and
Logarithmic Functions
sections. Also, the concluding type of function is a rational part that will be discussed in the
Rational Functions
section.

Transformation
T-chart/Domain and Range
Graph

\(\displaystyle y=\frac{three}{two}{{\left( {-x} \right)}^{3}}+ii\)

Parent function:

\(y={{x}^{3}}\)

For this part, note that could accept also put the negative sign on the
outside
(thus affecting the \(y\)), and nosotros would have gotten the aforementioned graph.

x    x

y
\(\frac{3}{2}y+2\)
1
–1
–ane       .5

2
–ane
one
1       iii.five

Domain:  \(\left( {-\infty ,\infty } \right)\)

Range:\(\left( {-\infty ,\infty } \right)\)



\(\displaystyle y=\frac{1}{2}\sqrt{{-ten}}\)

Parent function:

\(y=\sqrt{x}\)

x
 ten



y

\(\frac{ane}{two}y\)
    0

     0
   –1i

1
.5
   –4iv

2     ane

Domain:  \(\left( {-\infty ,0} \right]\)

Range:\(\left[ {0,\infty } \correct)\)

\(y={{2}^{{x-4}}}+iii\)

Parent function:

\(y={{two}^{x}}\)

For
exponential
functions, use

–one
,


, and

ane

for the \(10\)-values for the parent office. (Piece of cake way to remember:
exponent is similar \(10\)).



x


+ iv
 ten



y    y


+ three

    3
–1
.5    3.five
    iv
1
4
    5
1
2      5

Domain:  \(\left( {-\infty ,\infty } \correct)\)

Range:\(\left( {3,\infty } \right)\)

Asymptote:  \(y=3\)

\(\begin{array}{l}y=\log \left( {2x-two} \correct)-1\\y=\log \left( {2\left( {x-i} \correct)} \correct)-1\end{array}\)

Parent part:

\(y=\log \left( ten \correct)={{\log }_{{10}}}\left( x \correct)\)

For
log
and
ln
functions, use –
1
,


, and

1

for the \(y\)-values for the parent office For instance, for
\(y={{\log }_{3}}\left( {ii\left( {x-ane} \right)} \right)-1\), the \(ten\) values for the parent part would be \(\displaystyle \frac{1}{iii},\,one,\,\text{and}\,3\).)


.5x



+


1ten



 y    y


– 1

 ane.05.ane  1      –two
  1.vane       –ane
   6ten one         0

Domain:  \(\left( {1,\infty } \correct)\)

Range:\(\left( {-\infty ,\infty } \right)\)

Asymptote:  \(x=1\)

\(\displaystyle y=\frac{3}{{2-x}}\,\,\,\,\,\,\,\,\,\,\,y=\frac{3}{{-\left( {x-ii} \right)}}\)

Parent role:

\(\displaystyle y=\frac{ane}{x}\)

For this function, notation that could take as well put the negative sign on the
exterior
(thus, used \(x+2\) and \(-3y\)).


–x


+ 2x

y

     3y
3–one –1
–3
11 i
3

Domain:\(\left( {-\infty ,ii} \right)\loving cup \left( {2,\infty } \right)\)

Range:\(\left( {-\infty ,0} \right)\loving cup \left( {0,\infty } \correct)\)

Asymptotes: \(y=0\) and \(x=2\)

Here’due south a mixed transformation with the
Greatest
Integer Function
(sometimes called the
Flooring Role
). Note how we can employ intervals every bit the \(ten\) values to make the transformed function easier to draw:

Transformation
T-chart/Domain and Range
Graph

\(\displaystyle y=\left[ {\frac{i}{2}x-ii} \right]+3\)

\(\displaystyle y=\left[ {\frac{i}{ii}\left( {x-four} \right)} \right]+3\)

Parent function:

\(y=\left[ x \right]\)

Annotation how we had to have out the \(\displaystyle \frac{i}{two}\) to make information technology in the correct course.

2ten
+ 4        ten

  y


y


+ three
  [0, ii)[–two, –ane)


ii    ane
  [two, 4)[–i, 0)


1    2
 [4, 6)[0, 1)      3
 [6, 8)[1, 2) i     4
 [8, 10)[two, 3) 2     v

Domain:\(\left( {-\infty ,\infty } \right)\)

Range:\(\{y:y\in \mathbb{Z}\}\text{ (integers)}\)



Transformations Using Functional Note

Yous might run into mixed transformations in the class \(\displaystyle g\left( x \right)=a\cdot f\left( {\left( {\frac{ane}{b}} \correct)\left( {x-h} \right)} \correct)+g\), where \(a\) is the vertical stretch, \(b\) is the horizontal stretch, \(h\) is the horizontal shift to the right, and \(k\) is the vertical shift upwards. In this case, nosotros have the coordinate dominion \(\displaystyle \left( {ten,y} \right)\to \left( {bx+h,\,ay+k} \right)\). For example, for the transformation
\(\displaystyle f(ten)=-3{{\left( {two\left( {x+iv} \right)} \right)}^{2}}+10\), we take \(a=-3\), \(\displaystyle b=\frac{ane}{2}\,\,\text{or}\,\,.five\), \(h=-4\), and \(1000=10\). Our transformation \(\displaystyle g\left( ten \correct)=-3f\left( {two\left( {x+four} \correct)} \right)+10=thousand\left( x \correct)=-3f\left( {\left( {\frac{ane}{{\frac{1}{2}}}} \right)\left( {x-\left( {-iv} \correct)} \right)} \right)+x\) would result in a coordinate rule of \({\left( {x,\,y} \correct)\to \left( {.5x-4,-3y+10} \right)}\).
(Y’all may besides see this every bit \(m\left( x \right)=a\cdot f\left( {b\left( {x-h} \right)} \right)+thou\), with coordinate rule \(\displaystyle \left( {x,\,y} \right)\to \left( {\frac{i}{b}x+h,\,ay+1000} \right)\); the end effect will be the same.)

You may be given a
random point
and give the transformed coordinates for the point of the graph. For example, if the point \(\left( {8,-2} \right)\) is on the graph \(y=g\left( ten \right)\), requite the transformed coordinates for the betoken on the graph \(y=-6g\left( {-2x} \correct)-ii\). To do this, to get the transformed \(y\), multiply the \(y\) part of the point by

–6

and and then subtract

2
. To get the transformed \(ten\), multiply the \(ten\) part of the betoken by \(\displaystyle -\frac{1}{2}\) (contrary math). The new bespeak is \(\left( {-4,10} \correct)\). Let’s do another instance: If the point \(\left( {-iv,1} \right)\) is on the graph \(y=thousand\left( x \right)\), the transformed coordinates for the signal on the graph of \(\displaystyle y=2g\left( {-3x-two} \right)+three=2g\left( {-3\left( {x+\frac{2}{three}} \right)} \right)+3\) is \(\displaystyle \left( {-iv,one} \right)\to \left( {-iv\left( {-\frac{1}{three}} \right)-\frac{2}{three},2\left( 1 \right)+3} \right)=\left( {\frac{two}{iii},5} \correct)\) (using coordinate rules \(\displaystyle \left( {ten,\,y} \correct)\to \left( {\frac{1}{b}ten+h,\,\,ay+k} \correct)=\left( {-\frac{1}{3}x-\frac{2}{iii},\,\,2y+3} \right)\)).

You may likewise be asked to transform a parent or non-parent equation
to get a new equation. Nosotros can exercise this
without using a
t-chart
, only by using
substitution
and
algebra. For case, if we desire to transform \(f\left( x \right)={{ten}^{2}}+4\) using the transformation \(\displaystyle -2f\left( {x-i} \right)+3\), nosotros can only substitute “\(x-1\)” for “\(x\)” in the original equation, multiply by

–ii
, and then add

3
. For example: \(\displaystyle -2f\left( {x-ane} \correct)+iii=-ii\left[ {{{{\left( {x-ane} \right)}}^{2}}+4} \right]+3=-ii\left( {{{x}^{two}}-2x+ane+4} \right)+3=-2{{ten}^{2}}+4x-7\). We used this method to assist transform a
piecewise role
here.

Transformations in Role Notation (based on Graph and/or Points).

You may also be asked to perform a transformation of a function
using a graph and individual points; in this instance, you’ll probably exist given the transformation in
function notation. Note that we may need to use several points from the graph and “transform” them, to make sure that the transformed function has the right “shape”.

Here are some examples; the 2d instance is the transformation with an absolute value on the \(x\); see the
Absolute Value Transformations
section for more item.

Original Graph and Points of Function

Transformation Example

Transformation Example

Original Part:

Domain:
\(\left[ {-4,v} \right]\)
Range: \(\left[ {-7,5} \correct]\)

Key Points:


  x

  y
–four   5
 0   1
 2   2
 5 –7

Remember to draw the points in the same order every bit the original to make information technology easier! If you lot’re having problem drawing the graph from the transformed ordered pairs, just take more points from the original graph to map to the new one!

Transformation:\(\displaystyle f\left( {-\frac{one}{2}\left( {ten-ane} \correct)} \right)-three\)

\(y\) changes:\(\displaystyle f\left( {-\frac{ane}{2}\left( {ten-1} \right)} \right)\color{blue}{{-\text{ }3}}\)

\(10\) changes:\(\displaystyle f\left( {\color{bluish}{{-\frac{1}{2}}}\left( {x\text{ }\color{blue}{{-\text{ }ane}}} \right)} \right)-iii\)

Note that this transformation
flips around the
\(\boldsymbol{y}\)
axis, has a
horizontal stretch of
2
, moves
correct past
ane
, anddown by
3
.

Key Points Transformed:

(we practise the “contrary” math with the “\(x\)”)


2ten
+ 1

  x

  y       y


– iii
      9
–four
5          2
      1 1        –2
    –3ii ii        –one
    –ix5 –7       –10

Transformed Function:



Domain:  \(\left[ {-9,9} \right]\)
Range: \(\left[ {-x,ii} \correct]\)

Transformation:
\(\displaystyle f\left( {\left| x \right|+1} \correct)-two\)

\(y\) changes:
\(\displaystyle f\left( {\left| x \right|+1} \right)\color{blueish}{{\underline{{-\text{ }2}}}}\)

\(x\) changes:\(\displaystyle f\left( {\colour{blue}{{\underline{{\left| x \correct|+i}}}}} \right)-2\):

Note that this transformation
moves down past
2
, and
left
1
. Then, for the within accented value, we will “get rid of” whatsoever values to the left of the \(y\)-axis and replace with values to the right of the \(y\)-axis, to brand the graph symmetrical with the \(y\)-axis. We do the absolute value part last, since it’s but around the \(x\) on the inside.

Allow’s just exercise this one via graphs. First, motility down

2
, and left

ane
:

And then reflect the right-manus side across the \(y\)-axis to make symmetrical.

Transformed Function:

Domain:  \(\left[ {-4,iv} \right]\)
Range:\(\left[ {-9,0} \right]\)

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Writing Transformed Equations from Graphs

You might be asked to
write a transformed equation, give a graph. A lot of times, you lot can merely tell past looking at it, but sometimes yous take to use a bespeak or two. And you do have to be conscientious and check your work, since the order of the transformations tin matter.

Note that when figuring out the transformations from a graph, information technology’s difficult to know whether you have an “\(a\)” (vertical stretch) or a “\(b\)” (horizontal stretch) in the equation \(\displaystyle g\left( x \correct)=a\cdot f\left( {\left( {\frac{1}{b}} \correct)\left( {x-h} \right)} \right)+k\). Sometimes the problem volition indicate what parameters (\(a\), \(b\), and then on) to await for. For others, similar polynomials (such as quadratics and cubics), a
vertical stretch
mimics a
horizontal compression, and then it’due south possible to factor out a coefficient to turn a horizontal stretch/compression to a vertical compression/stretch. (For more complicated graphs, yous may want to take several points and perform a
regression in your estimator
to become the office, if you’re allowed to do that).

Hither are some problems. Note that atransformed equation from an absolute value graph
is in the Absolute Value Transformationssection.

Transformed Graph Getting Equation
Write the full general equation for the
cubic equation
in the grade:

\(\displaystyle y={{\left( {\frac{ane}{b}\left( {10-h} \right)} \right)}^{3}}+k\).

We see that that the heart point, or
critical betoken
is at \(\left( {-four,-v} \right)\), so the
cubic
is in the course: \(\displaystyle y={{\left( {\frac{i}{b}\left( {x+four} \right)} \right)}^{3}}-5\).

Notice that to go dorsum and over to the next points, we get dorsum/over \(3\) and down/up \(1\), so we encounter in that location’s a horizontal stretch of \(three\), so \(b=3\). (We could have also used some other indicate on the graph to solve for \(b\)). We have \(\displaystyle y={{\left( {\frac{one}{three}\left( {10+4} \correct)} \correct)}^{3}}-5\). Try information technology – it works!

Annotation that if we wanted this function in the class \(\displaystyle y=a{{\left( {\left( {x-h} \right)} \right)}^{three}}+k\), we could use the point \(\left( {-seven,-half-dozen} \correct)\) to become \(\displaystyle y=a{{\left( {\left( {x+four} \right)} \right)}^{3}}-v;\,\,\,\,-6=a{{\left( {\left( {-vii+iv} \correct)} \correct)}^{3}}-v\), or \(\displaystyle a=\frac{1}{{27}}\). This makes sense, since if we brought the \(\displaystyle {{\left( {\frac{1}{three}} \correct)}^{3}}\) out from above, it would exist \(\displaystyle \frac{1}{{27}}\)!)

Find the equation of this graph in whatsoever course:

Here’south a
generic
method you can typically use:

We see that this is a
cubicpolynomial
graph (parent graph \(y={{x}^{3}}\)), merely flipped effectually either the \(x\) the \(y\)-axis, since it’southward an odd function; let’s utilise the \(10\)-centrality for simplicity’s sake. The equation will be in the form \(y=a{{\left( {x+b} \correct)}^{3}}+c\), where \(a\) is negative, and it is shifted upwards \(2\), and to the left \(1\). At present we have \(y=a{{\left( {x+1} \right)}^{iii}}+2\).

We demand to find \(a\); employ the bespeak \(\left( {i,-10} \right)\):
\(\begin{align}-x&=a{{\left( {1+1} \right)}^{iii}}+ii\\-10&=8a+two\\8a&=-12;\,\,a=-\frac{{12}}{8}=-\frac{3}{2}\end{align}\).
The equation of the graph is: \(\displaystyle y=-\frac{3}{ii}{{\left( {x+ane} \right)}^{3}}+2\). Be sure to check your answer by graphing or plugging in more points! √

Find the equation of this graph in any grade:

The graph looks similar a
quadratic
with
vertex
\(\left( {-ane,-eight} \correct)\), which is a shift of \(8\) down and \(1\) to the left. This volition give united states an equation of the form \(y=a{{\left( {x+ane} \correct)}^{2}}-eight\), which is (not and so coincidentally!) the
vertex form
for quadratics.

We demand to detect \(a\); utilise the betoken \(\left( {1,0} \right)\):\(\begin{marshal}y&=a{{\left( {x+1} \correct)}^{2}}-eight\\\,0&=a{{\left( {1+i} \right)}^{2}}-8\\8&=4a;\,\,a=two\finish{align}\).
The equation of the graph then is: \(y=2{{\left( {x+ane} \correct)}^{2}}-8\).

Note: we could have also noticed that the graph goes over \(1\) and up \(2\) from the vertex, instead of over \(1\) and upwards \(one\) usually with \(y={{x}^{two}}\). This would hateful that our vertical stretch is \(2\).

Find the equation of this graph in any course:

The graph looks like a
rational
with the
“center”
of asymptotes at \(\left( {-2,iii} \right)\), which is a shift of

2

to the left and

iii

upwards. This will give usa an equation of the form \(\displaystyle y=a\left( {\frac{1}{{10+2}}} \right)+iii\), with asymptotes at \(ten=-2\) and \(y=3\).

We need to find \(a\); use the given bespeak \((0,4)\):  \(\begin{align}y&=a\left( {\frac{1}{{x+2}}} \right)+3\\four&=a\left( {\frac{ane}{{0+2}}} \right)+3\\1&=\frac{a}{2};\,a=2\terminate{align}\). The equation of the graph is: \(\displaystyle y=2\left( {\frac{1}{{x+two}}} \right)+three,\,\text{or }y=\frac{two}{{x+2}}+iii\).

Note: we could have also noticed that the graph goes over

1

and up

2

from the eye of asymptotes, instead of over

1

and up

1

normally with \(\displaystyle y=\frac{1}{x}\). This would hateful that our vertical stretch is

2
.

Find the equation of this graph with a base of \(.5\) and horizontal shift of \(-1\):

We see that thisexponential graph has a horizontal asymptote at \(y=-3\), and with the horizontal shift, we have \(y=a{{\left( {.5} \right)}^{{x+i}}}-3\) so far.

When you have a problem similar this,
beginning utilise whatever point that has a “



” in it
if you can; it will be easiest to solve the organization. Solve for \(a\) beginning using point \(\left( {0,-ane} \right)\):

\(\begin{array}{c}y=a{{\left( {.5} \correct)}^{{x+i}}}-3;\,\,-one=a{{\left( {.5} \right)}^{{0+one}}}-three;\,\,\,\,2=.5a;\,\,a=4\\y=four{{\left( {.five} \correct)}^{{ten+1}}}-3\terminate{assortment}\)

Note that there are more examples of exponential transformations hither in the
Exponential Functions
section, and logarithmic transformations
here
in the
Logarithmic Functions
section.

Rotational Transformations

Y’all may be asked to perform a
rotation transformation on a function (you commonly run into these in
Geometry
grade). A rotation of

90°

counterclockwise involves replacing \(\left( {x,y} \right)\) with \(\left( {-y,10} \correct)\), a rotation of

180°


counterclockwise involves replacing \(\left( {x,y} \right)\) with \(\left( {-ten,-y} \right)\), and a rotation of

270°

counterclockwise involves replacing \(\left( {x,y} \right)\) with \(\left( {y,-x} \right)\). Here is an instance:

Transformation Example Graph
Rotate graph

270°
counterclockwise

Parent:  \(y={{ten}^{2}}\)

Replace \((x,y)\) with \((y,–10)\)


y       10

 y


x
4
–two
4
     two
one
–1
1      1

      0
1
ane
1    –1
4ii 4    –2

Rotated Part Domain:
\(\left[ {0,\infty } \right)\)
Range:\(\left( {-\infty ,\infty } \right)\)

Transformations of Inverse Functions

We learned about
Changed Functions
here, and you might be asked to compare original functions and inverse functions, as far as their transformations are concerned. Call back that an changed function is ane where the \(x\) is switched by the \(y\), so the all the transformations originally performed on the \(x\) will be performed on the \(y\):
Problem:

If a cubic function is
vertically stretched by a factor of
3
,
reflected over the
\(\boldsymbol {y}\)-axis, and
shifted downwards
2
units
, what transformations are washed to its
changed role?


Solution:

We demand to do transformations on the
contrary variable. Thus, the inverse of this office will be
horizontally stretched by a cistron of
iii
,
reflected over the
\(\boldsymbol {ten}\)-axis, and
shifted to the left
ii
units
. Here is a graph of the two functions:

Note that examples of
Finding Inverses with Restricted Domains
can be constitute
here.

Applications of Parent Function Transformations

You may run across a “word problem” that used Parent Function Transformations, and you tin use what you know nigh how to shift a function. Here is an example:

Transformation Application Trouble Solution
The following polynomial graph shows the
profit
that results from selling math books afterwards September 1. The polynomial is \(p\left( x \right)=5{{10}^{iii}}-xx{{x}^{2}}+40x-1\), where \(x\) is the number of weeks after September 1.

The publisher of the math books were
one week behind
yet; describe how this new graph would look and what would be the new (transformed) part?

Since we’re moving the fourth dimension in weeks past

i

week, we are shifting the graph horizontally, or shifting the
inside, or \(ten\) values.

Since our showtime profits volition start a little after calendar week

1
, we tin see that we need to move the graph
to the correct. When we move the \(10\) role to the correct, we take the \(ten\) values and
decrease
from them, so the new polynomial will be \(d\left( x \right)=5{{\left( {x-one} \right)}^{3}}-20{{\left( {ten-1} \correct)}^{ii}}+forty\left( {x-1} \right)-one\). (I won’t multiply and simplify.) Run across how this was much easier, knowing what nosotros know about transforming parent functions?

Acquire these rules, and practise, practice, exercise!


For Practice: Use the
Mathway widget beneath to try aTransformation
trouble. Click on
Submit
(the blue arrow to the right of the problem) and click on
Describe the Transformation to run into the answer.

You tin can likewise type in your own problem, or click on the iii dots in the upper correct hand corner and click on “Examples” to drill downwardly past topic.

If y’all click on
Tap to view steps, or
Click Hither, yous can register at
Mathway
for a
free trial, and then upgrade to a paid subscription at any time (to get any type of math trouble solved!).

On to
Absolute Value Transformations
– you are gear up!

Which Parent Function is Represented by the Graph

Source: https://mathhints.com/parent-graphs-and-transformations/