Which Parent Function is Represented by the 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:

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

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!

Transformed:

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

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

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

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

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

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

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

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

Absolute Value
on the \(y\)

(More examples here in the
Absolute Value Transformation
department)

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

Parent:  \(y=\sqrt[3]{x}\)

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

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

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

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

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

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

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

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

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)

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

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

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

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

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.

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

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.

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

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

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

Parent function:

\(y=\sqrt{x}\)

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 \(10\)).

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 \(\displaystyle \frac{1}{iii},\,one,\,\text{and}\,3\).)

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 \(-3y\)).

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 \(x=2\)

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

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

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.

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

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

Original Part:

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

Key Points:

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 “\(x\)”)

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:\(\left[ {-9,0} \right]\)

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 \(\displaystyle \frac{1}{{27}}\)!)

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! √

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 \(2\).

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
.

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.

Parent:  \(y={{ten}^{2}}\)

Replace \((x,y)\) with \((y,–10)\)

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

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 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?