# 6.2 Graphs of exponential functions  (Page 3/6)

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

For any constants $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}d,$ the function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d\text{\hspace{0.17em}}$ shifts the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}$

• vertically $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units, in the same direction of the sign of $\text{\hspace{0.17em}}d.$
• horizontally $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units, in the opposite direction of the sign of $\text{\hspace{0.17em}}c.$
• The y -intercept becomes $\text{\hspace{0.17em}}\left(0,{b}^{c}+d\right).$
• The horizontal asymptote becomes $\text{\hspace{0.17em}}y=d.$
• The range becomes $\text{\hspace{0.17em}}\left(d,\infty \right).$
• The domain, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ remains unchanged.

Given an exponential function with the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d,$ graph the translation.

1. Draw the horizontal asymptote $\text{\hspace{0.17em}}y=d.$
2. Identify the shift as $\text{\hspace{0.17em}}\left(-c,d\right).\text{\hspace{0.17em}}$ Shift the graph of $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ left $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is positive, and right $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $c\text{\hspace{0.17em}}$ is negative.
3. Shift the graph of $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ up $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ is positive, and down $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ is negative.
4. State the domain, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ the range, $\text{\hspace{0.17em}}\left(d,\infty \right),$ and the horizontal asymptote $\text{\hspace{0.17em}}y=d.$

## Graphing a shift of an exponential function

Graph $\text{\hspace{0.17em}}f\left(x\right)={2}^{x+1}-3.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

We have an exponential equation of the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d,$ with $\text{\hspace{0.17em}}b=2,$ $\text{\hspace{0.17em}}c=1,$ and $\text{\hspace{0.17em}}d=-3.$

Draw the horizontal asymptote $\text{\hspace{0.17em}}y=d$ , so draw $\text{\hspace{0.17em}}y=-3.$

Identify the shift as $\text{\hspace{0.17em}}\left(-c,d\right),$ so the shift is $\text{\hspace{0.17em}}\left(-1,-3\right).$

Shift the graph of $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ left 1 units and down 3 units.

The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(-3,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=-3.$

Graph $\text{\hspace{0.17em}}f\left(x\right)={2}^{x-1}+3.\text{\hspace{0.17em}}$ State domain, range, and asymptote.

The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(3,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=3.$

Given an equation of the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x,$ use a graphing calculator to approximate the solution.

• Press [Y=] . Enter the given exponential equation in the line headed “ Y 1 = ”.
• Enter the given value for $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ in the line headed “ Y 2 = ”.
• Press [WINDOW] . Adjust the y -axis so that it includes the value entered for “ Y 2 = ”.
• Press [GRAPH] to observe the graph of the exponential function along with the line for the specified value of $\text{\hspace{0.17em}}f\left(x\right).$
• To find the value of $\text{\hspace{0.17em}}x,$ we compute the point of intersection. Press [2ND] then [CALC] . Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x for the indicated value of the function.

## Approximating the solution of an exponential equation

Solve $\text{\hspace{0.17em}}42=1.2{\left(5\right)}^{x}+2.8\text{\hspace{0.17em}}$ graphically. Round to the nearest thousandth.

Press [Y=] and enter $\text{\hspace{0.17em}}1.2{\left(5\right)}^{x}+2.8\text{\hspace{0.17em}}$ next to Y 1 =. Then enter 42 next to Y2= . For a window, use the values –3 to 3 for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and –5 to 55 for $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Press [GRAPH] . The graphs should intersect somewhere near $\text{\hspace{0.17em}}x=2.$

For a better approximation, press [2ND] then [CALC] . Select [5: intersect] and press [ENTER] three times. The x -coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess? ) To the nearest thousandth, $\text{\hspace{0.17em}}x\approx 2.166.$

Solve $\text{\hspace{0.17em}}4=7.85{\left(1.15\right)}^{x}-2.27\text{\hspace{0.17em}}$ graphically. Round to the nearest thousandth.

$x\approx -1.608$

## Graphing a stretch or compression

While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ by a constant $\text{\hspace{0.17em}}|a|>0.\text{\hspace{0.17em}}$ 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 stretch, using $\text{\hspace{0.17em}}a=3,$ to get $\text{\hspace{0.17em}}g\left(x\right)=3{\left(2\right)}^{x}\text{\hspace{0.17em}}$ as shown on the left in [link] , and the compression, using $\text{\hspace{0.17em}}a=\frac{1}{3},$ to get $\text{\hspace{0.17em}}h\left(x\right)=\frac{1}{3}{\left(2\right)}^{x}\text{\hspace{0.17em}}$ as shown on the right in [link] .

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
hello
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