# 6.2 Graphs of exponential functions  (Page 4/6)

 Page 4 / 6

## Stretches and compressions of the parent function f ( x ) = b x

For any factor $\text{\hspace{0.17em}}a>0,$ the function $\text{\hspace{0.17em}}f\left(x\right)=a{\left(b\right)}^{x}$

• is stretched vertically by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}|a|>1.$
• is compressed vertically by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}|a|<1.$
• has a y -intercept of $\text{\hspace{0.17em}}\left(0,a\right).$
• has a horizontal asymptote at $\text{\hspace{0.17em}}y=0,$ a range of $\text{\hspace{0.17em}}\left(0,\infty \right),$ and a domain of $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ which are unchanged from the parent function.

## Graphing the stretch of an exponential function

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=4{\left(\frac{1}{2}\right)}^{x}.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

Before graphing, identify the behavior and key points on the graph.

• Since $\text{\hspace{0.17em}}b=\frac{1}{2}\text{\hspace{0.17em}}$ is between zero and one, the left tail of the graph will increase without bound as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ decreases, and the right tail will approach the x -axis as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases.
• Since $\text{\hspace{0.17em}}a=4,$ the graph of $\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{2}\right)}^{x}\text{\hspace{0.17em}}$ will be stretched by a factor of $\text{\hspace{0.17em}}4.$
• Create a table of points as shown in [link] .  $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $f\left(x\right)=4\left(\frac{1}{2}{\right)}^{x}$ $32$ $16$ $8$ $4$ $2$ $1$ $0.5$
• Plot the y- intercept, $\text{\hspace{0.17em}}\left(0,4\right),$ along with two other points. We can use $\text{\hspace{0.17em}}\left(-1,8\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,2\right).$

Draw a smooth curve connecting the points, as shown in [link] .

The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$

Sketch the graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}{\left(4\right)}^{x}.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.\text{\hspace{0.17em}}$

## Graphing reflections

In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x -axis or the y -axis. When we multiply the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}-1,$ we get a reflection about the x -axis. When we multiply the input by $\text{\hspace{0.17em}}-1,$ we get a reflection about the y -axis. For example, if we begin by graphing the parent function $\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ we can then graph the two reflections alongside it. The reflection about the x -axis, $\text{\hspace{0.17em}}g\left(x\right)={-2}^{x},$ is shown on the left side of [link] , and the reflection about the y -axis $\text{\hspace{0.17em}}h\left(x\right)={2}^{-x},$ is shown on the right side of [link] .

## Reflections of the parent function f ( x ) = b x

The function $\text{\hspace{0.17em}}f\left(x\right)=-{b}^{x}$

• reflects the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ about the x -axis.
• has a y -intercept of $\text{\hspace{0.17em}}\left(0,-1\right).$
• has a range of $\text{\hspace{0.17em}}\left(-\infty ,0\right)$
• has a horizontal asymptote at $\text{\hspace{0.17em}}y=0\text{\hspace{0.17em}}$ and domain of $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ which are unchanged from the parent function.

The function $\text{\hspace{0.17em}}f\left(x\right)={b}^{-x}$

• reflects the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ about the y -axis.
• has a y -intercept of $\text{\hspace{0.17em}}\left(0,1\right),$ a horizontal asymptote at $\text{\hspace{0.17em}}y=0,$ a range of $\text{\hspace{0.17em}}\left(0,\infty \right),$ and a domain of $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ which are unchanged from the parent function.

## Writing and graphing the reflection of an exponential function

Find and graph the equation for a function, $\text{\hspace{0.17em}}g\left(x\right),$ that reflects $\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{4}\right)}^{x}\text{\hspace{0.17em}}$ about the x -axis. State its domain, range, and asymptote.

Since we want to reflect the parent function $\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{4}\right)}^{x}\text{\hspace{0.17em}}$ about the x- axis, we multiply $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ to get, $\text{\hspace{0.17em}}g\left(x\right)=-{\left(\frac{1}{4}\right)}^{x}.\text{\hspace{0.17em}}$ Next we create a table of points as in [link] .

 $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $g\left(x\right)=-\left(\frac{1}{4}{\right)}^{x}$ $-64$ $-16$ $-4$ $-1$ $-0.25$ $-0.0625$ $-0.0156$

Plot the y- intercept, $\text{\hspace{0.17em}}\left(0,-1\right),$ along with two other points. We can use $\text{\hspace{0.17em}}\left(-1,-4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,-0.25\right).$

Draw a smooth curve connecting the points:

The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(-\infty ,0\right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
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?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
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
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar