# 4.4 Graphs of logarithmic functions  (Page 5/8)

 Page 5 / 8

Given a logarithmic function with the form $\text{\hspace{0.17em}}f\left(x\right)=a{\mathrm{log}}_{b}\left(x\right),$ $a>0,$ graph the translation.

1. Identify the vertical stretch or compressions:
• If $\text{\hspace{0.17em}}|a|>1,$ the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ is stretched by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units.
• If $\text{\hspace{0.17em}}|a|<1,$ the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ is compressed by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units.
2. Draw the vertical asymptote $\text{\hspace{0.17em}}x=0.$
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ coordinates by $\text{\hspace{0.17em}}a.$
4. Label the three points.
5. 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 a stretch or compression of the parent function y = log b ( x )

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=2{\mathrm{log}}_{4}\left(x\right)\text{\hspace{0.17em}}$ alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Since the function is $\text{\hspace{0.17em}}f\left(x\right)=2{\mathrm{log}}_{4}\left(x\right),$ we will notice $\text{\hspace{0.17em}}a=2.$

This means we will stretch the function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{4}\left(x\right)\text{\hspace{0.17em}}$ by a factor of 2.

The vertical asymptote is $\text{\hspace{0.17em}}x=0.$

Consider the three key points from the parent function, $\text{\hspace{0.17em}}\left(\frac{1}{4},-1\right),$ $\left(1,0\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(4,1\right).$

The new coordinates are found by multiplying the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ coordinates by 2.

Label the points $\text{\hspace{0.17em}}\left(\frac{1}{4},-2\right),$ $\left(1,0\right)\text{\hspace{0.17em}},$ and $\text{\hspace{0.17em}}\left(4,\text{2}\right).$

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

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}\text{\hspace{0.17em}}{\mathrm{log}}_{4}\left(x\right)\text{\hspace{0.17em}}$ alongside its parent function. Include the key points and asymptote on the graph. 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 $\text{\hspace{0.17em}}x=0.$

## Combining a shift and a stretch

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=5\mathrm{log}\left(x+2\right).\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5, as in [link] . The vertical asymptote will be shifted to $\text{\hspace{0.17em}}x=-2.\text{\hspace{0.17em}}$ The x -intercept will be $\text{\hspace{0.17em}}\left(-1,0\right).\text{\hspace{0.17em}}$ The domain will be $\text{\hspace{0.17em}}\left(-2,\infty \right).\text{\hspace{0.17em}}$ Two points will help give the shape of the graph: $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(8,5\right).\text{\hspace{0.17em}}$ We chose $\text{\hspace{0.17em}}x=8\text{\hspace{0.17em}}$ as the x -coordinate of one point to graph because when $\text{\hspace{0.17em}}x=8,\text{\hspace{0.17em}}$ $\text{\hspace{0.17em}}x+2=10,\text{\hspace{0.17em}}$ the base of the common logarithm.

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 $\text{\hspace{0.17em}}x=-2.$

Sketch a graph of the function $\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{log}\left(x-2\right)+1.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

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 $\text{\hspace{0.17em}}x=2.$

## Graphing reflections of f ( x ) = log b ( x )

When the parent function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ is multiplied by $\text{\hspace{0.17em}}-1,$ the result is a reflection about the x -axis. When the input is multiplied by $\text{\hspace{0.17em}}-1,$ the result is a reflection about the y -axis. To visualize reflections, we restrict $\text{\hspace{0.17em}}b>1,\text{\hspace{0.17em}}$ and observe the general graph of the parent function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ alongside the reflection about the x -axis, $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{-log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ and the reflection about the y -axis, $\text{\hspace{0.17em}}h\left(x\right)={\mathrm{log}}_{b}\left(-x\right).$

## Reflections of the parent function y = log b ( x )

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

• reflects the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ about the x -axis.
• has domain, $\text{\hspace{0.17em}}\left(0,\infty \right),$ range, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and vertical asymptote, $\text{\hspace{0.17em}}x=0,$ which are unchanged from the parent function.

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

• reflects the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ about the y -axis.
• has domain $\text{\hspace{0.17em}}\left(-\infty ,0\right).$
• has range, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and vertical asymptote, $\text{\hspace{0.17em}}x=0,$ which are unchanged from the parent function.

#### Questions & Answers

how can are find the domain and range of a relations
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
i want to sure my answer of the exercise
what is the diameter of(x-2)²+(y-3)²=25
how to solve the Identity ?
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim