6.4 Graphs of logarithmic functions  (Page 3/8)

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Given a logarithmic function with the form $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right),$ graph the function.

1. Draw and label the vertical asymptote, $\text{\hspace{0.17em}}x=0.$
2. Plot the x- intercept, $\text{\hspace{0.17em}}\left(1,0\right).$
3. Plot the key point $\text{\hspace{0.17em}}\left(b,1\right).$
4. Draw a smooth curve through the points.
5. State the domain, $\text{\hspace{0.17em}}\left(0,\infty \right),$ the range, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote, $\text{\hspace{0.17em}}x=0.$

Graphing a logarithmic function with the form f ( x ) = log b ( x ).

Graph $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{5}\left(x\right).\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

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

• Since $\text{\hspace{0.17em}}b=5\text{\hspace{0.17em}}$ is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote $\text{\hspace{0.17em}}x=0,$ and the right tail will increase slowly without bound.
• The x -intercept is $\text{\hspace{0.17em}}\left(1,0\right).$
• The key point $\text{\hspace{0.17em}}\left(5,1\right)\text{\hspace{0.17em}}$ is on the graph.
• We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see [link] ).

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

Graph $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{\frac{1}{5}}\left(x\right).\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

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

Graphing transformations of logarithmic functions

As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ without loss of shape.

Graphing a horizontal shift of f ( x ) = log b ( x )

When a constant $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is added to the input of the parent function $\text{\hspace{0.17em}}f\left(x\right)=lo{g}_{b}\left(x\right),$ the result is a horizontal shift     $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units in the opposite direction of the sign on $\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$ To visualize horizontal shifts, we can observe the general graph of the parent function $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ and for $\text{\hspace{0.17em}}c>0\text{\hspace{0.17em}}$ alongside the shift left, $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{log}}_{b}\left(x+c\right),$ and the shift right, $\text{\hspace{0.17em}}h\left(x\right)={\mathrm{log}}_{b}\left(x-c\right).$ See [link] .

Horizontal shifts of the parent function y = log b ( x )

For any constant $\text{\hspace{0.17em}}c,$ the function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x+c\right)$

• shifts the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ left $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}c>0.$
• shifts the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ right $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}c<0.$
• has the vertical asymptote $\text{\hspace{0.17em}}x=-c.$
• has domain $\text{\hspace{0.17em}}\left(-c,\infty \right).$
• has range $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).$

Given a logarithmic function with the form $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x+c\right),$ graph the translation.

1. Identify the horizontal shift:
1. If $\text{\hspace{0.17em}}c>0,$ shift the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ left $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units.
2. If $\text{\hspace{0.17em}}c<0,$ shift the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ right $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units.
2. Draw the vertical asymptote $\text{\hspace{0.17em}}x=-c.$
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ from the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ coordinate.
4. Label the three points.
5. The Domain is $\text{\hspace{0.17em}}\left(-c,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=-c.$

Graphing a horizontal shift of the parent function y = log b ( x )

Sketch the horizontal shift $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{3}\left(x-2\right)\text{\hspace{0.17em}}$ alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.

Since the function is $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{3}\left(x-2\right),$ we notice $\text{\hspace{0.17em}}x+\left(-2\right)=x–2.$

Thus $\text{\hspace{0.17em}}c=-2,$ so $\text{\hspace{0.17em}}c<0.\text{\hspace{0.17em}}$ This means we will shift the function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{3}\left(x\right)\text{\hspace{0.17em}}$ right 2 units.

The vertical asymptote is $\text{\hspace{0.17em}}x=-\left(-2\right)\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}x=2.$

Consider the three key points from the parent function, $\text{\hspace{0.17em}}\left(\frac{1}{3},-1\right),$ $\left(1,0\right),$ and $\text{\hspace{0.17em}}\left(3,1\right).$

The new coordinates are found by adding 2 to the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ coordinates.

Label the points $\text{\hspace{0.17em}}\left(\frac{7}{3},-1\right),$ $\left(3,0\right),$ and $\text{\hspace{0.17em}}\left(5,1\right).$

The domain is $\text{\hspace{0.17em}}\left(2,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=2.$

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