Graphing Logarithm Question (1 Viewer)

starryblue · Aug 14, 2011

can anyone please tell me how to graph the curve y=log10(x+1)?

thank you~

SpiralFlex · Aug 14, 2011

First of all you have to see what a general logarithmic curve looks like. It will look something like this.

You will see that the graph is this shape if you graph some points.

The asymptote is at $x=0$ and it must go past through the point $(1, 0)$ . So why is the asymptote zero?

If we look at our graph $y=\log_{10}x$ and rewrite it as indice form,

$10^y=x$

If $x$ was zero, then,

$10^y=0$

This value of $y$ does not exist. Since you always get a number above zero.

Alternatively,

As $y \rightarrow -\infty$ [As the $y$ value tends to minus infinity, the graph approaches $x=-1$ . A bit right of it.]

$x \rightarrow -1^+$

Let's think back to parabolas.

Say we have the curve $y=(x+1)^2$ . Our original graph was $y=x^2$

We can see that the addition of 1 moves it to the left ONE unit.

Back to our curve. Applying this, we can see it will move one space to the left.

Normally that logarithmic curve will have an asmyptote at $x=0$ and must go through $(1,0)$ . In this case, the asymptote moves one place to the left, hence it is now at $x=-1$ and the $x$ intercept is now at $(0, 0)$

Your graph should look something like this.

http://imageshack.us/photo/my-images/193/forstarryblue.png/

Note: Why did I plot one point? This is to avoid confusion since graphs of logarithms with different bases look similar.

Example - Unless you have a good eye, when graphing separately $y=\log_{10}x$ or $y=\log_{2}x$ , it is quite difficult to tell which graph is which, but if a point was plotted, this gives the marker a general idea.

Alkanes · Aug 14, 2011

What program is that spi?

SpiralFlex · Aug 14, 2011

Alkanes said:
What program is that spi?

Google images.

starryblue · Aug 14, 2011

SpiralFlex said:
First of all you have to see what a general logarithmic curve looks like. It will look something like this.

You will see that the graph is this shape if you graph some points.

The asymptote is at $x=0$ and it must go past through the point $(1, 0)$ . So why is the asymptote zero?

If we look at our graph $y=\log_{10}x$ and rewrite it as indice form,

$10^y=x$

If $x$ was zero, then,

$10^y=0$

This value of $y$ does not exist. Since you always get a number above zero.

Let's think back to parabolas.

Say we have the curve $y=(x+1)^2$ . Our original graph was $y=x^2$

We can see that the addition of 1 moves it to the left ONE unit.

Back to our curve. Applying this, we can see it will move one space to the left.

Normally that logarithmic curve will have an asmyptote at $x=0$ and must go through $(1,0)$ . In this case, the asymptote moves one place, hence it is now at $x=-1$ and the $x$ intercept is now at $(0, 0)$

Your graph should look something like this.

http://imageshack.us/photo/my-images/193/forstarryblue.png/

Note: Why did I plot one point? This is to avoid confusion since graphs of logarithms with different bases look similar.

Example - Unless you have a good eye, when graphing separately $y=\log_{10}x$ or $y=\log_{2}x$ , it is quite difficult to tell which graph is which, but if a point was plotted, this gives the marker a general idea.

oh! i get it! *sigh* i got saved by you again!

SpiralFlex · Aug 14, 2011

starryblue said:
oh! i get it! *sigh* i got saved by you again!

More general, if we have the curve $y=f(x)$

$f(x+c)$ moves to the left $c$ units if $c>0$

$f(x+c)$ moves to the right $c$ units if $c<0$

$f(x)+c$ moves it up $c$ units if $c>0$

$f(x)+c$ moves it down $c$ units if $c<0$

starryblue · Aug 14, 2011

SpiralFlex said:
More general, if we have the curve $y=f(x)$

$f(x+c)$ moves to the left $c$ units if $c>0$

$f(x+c)$ moves to the right $c$ units if $c<0$

$f(x)+c$ moves it up $c$ units if $c>0$

$f(x)+c$ moves it down $c$ units if $c<0$

ahh == mayb i should of mentioned earlier but i knew that, it's just that i didn't know what to plot but you answered that...how about for hyperbolas? if it was eg. y=2/x, then what would u plot?

SpiralFlex · Aug 14, 2011

starryblue said:
ahh == mayb i should of mentioned earlier but i knew that, it's just that i didn't know what to plot but you answered that...how about for hyperbolas? if it was eg. y=2/x, then what would u plot?

Does your coaching college expect you to use the formal way to graph this? In year 10 you only do a generalised way. I will teach you the formal way in a second. When my hand does not feel sore.

starryblue · Aug 14, 2011

SpiralFlex said:
Does your coaching college expect you to use the formal way to graph this? In year 10 you only do a generalised way. I will teach you the formal way in a second. When my hand does not feel sore.

O.O" there's a formal way? is that more difficult to do? btw, how do you find the range of a graph?

SpiralFlex · Aug 14, 2011

$y=\frac{2}{x}$

First step is always find the $x$ and $y$ intercepts, if there are any.

$x$ and $y$ intercepts: When you let $y=0$ to find the $x$ intercept and $x=0$ to find the $y$ intercept, you can see that the value does not exist.

The next step is to find any asymptotes. It is convenient for me to find the vertical asymptote first.

VERTICAL ASYMPTOTE

Now, $y=\frac{2}{x}$

A vertical asymptote occurs when the denominator equals to zero.

$x=0$ [That is our asymptote.]

Your graph should look like this:

http://www.mediafire.com/?faa36hb8slbncv2

Now next stage is to find how the graph behaves near the vertical asymptote.

Behaviour: To do this we must see the tendency of the curve as it approaches the asymptote to the left and to the right.

BEHAVIOUR

We will test the behaviour near its immediate left first. So what is a number close to the left of zero? I know -0.1.

$y=\frac{2}{x}$

$y=\frac{2}{-0.1}$

$y=\frac{+}{-}$ [A positive number divide a negative number is a negative number.]

You get a negative number.

So,

As $x \rightarrow 0^-$ [As $x$ gets close to zero from the left hand side.]

$y \rightarrow - \infty$ [ $y$ tends to negative infinity.]

We will now test the behaviour to the immediate right. So what is a number close to the right of zero? I know 0.1.

$y=\frac{2}{x}$

$y=\frac{2}{0.1}$

$y=\frac{+}{+}$ [A positive number divide a positive number is a positive number.]

You get a positive number.

So,

As $x \rightarrow 0^+$ [As $x$ gets close to zero from the right hand side.]

$y \rightarrow \infty$ [ $y$ tends to positive infinity.]

Your graph should look something like this:

http://www.mediafire.com/?0o2nrnamc3qc3xe

Now horizontal asymptote.

HORIZONTAL ASYMPTOTE AND TENDENCY

We must find what happens when the graph tends to positive infinity. To help you understand this, substitute a large positive number into the equation of graph. Let's use 1000.

$y=\frac{2}{x}$

$y=\frac{2}{1000}$

$y= 0.0001111$ [Something like that, we can see it's a small figure close to zero above.]

So,

As $x \rightarrow \infty$ . [As $x$ approaches positive infinity.]

$y \rightarrow 0^+$ [ $y$ approaches zero a bit above it.]

We must find what happens when the graph tends to negative infinity. To help you understand this, substitute a large negative number into the equation of graph. Let's use -1000.

$y=\frac{2}{x}$

$y=\frac{2}{-1000}$

$y= -0.0001111$ [Something like that, we can see it's a small figure close to zero below it.]

So,

As $x \rightarrow -\infty$ . [As $x$ approaches negative infinity.]

$y \rightarrow 0^-$ [ $y$ approaches zero a bit below it.]

Now your graph should look like:

http://www.mediafire.com/?h5c4kcmekh3dgcl

Now all you have to do is connect the points. Make sure you indicate a point. Use something like $(1, 2)$

http://www.mediafire.com/?d8uuceefrz5t3nc

starryblue · Aug 14, 2011

SpiralFlex said:
$y=\frac{2}{x}$

First step is always find the $x$ and $y$ intercepts, if there are any.

$x$ and $y$ intercepts: When you let[ $y=0$ to find the $x$ intercept and $x=0$ to find the $y$ intercept, you can see that the value does not exist.

The next step is to find any asymptotes. It is convenient for me to find the vertical asymptote first.

Vertical asymptote:

Now, $y=\frac{2}{x}$

A vertical asymptote occurs when the denominator equals to zero.

$x=0$ [That is our asymptote.]

Your graph should look like this:

Now next stage is to find how the graph behaves near the vertical asymptote.

Behaviour: To do this we must see the tendency of the curve as it approaches the asymptote to the left and to the right.

We will test the behaviour near its immediate left first. So what is a number close to the left of zero? I know -0.1.

$y=\frac{2}{x}$

$y=\frac{2}{0.1}$

$y=\frac{+}{+}$ [A positive number divide a positive number is a positive number.]

You get a positive number.

So,

As $x \rightarrow 0^-$ [As $x$ gets close to zero from the left hand side.]

$y \rightarrow \infty$ [ $y$ tends to positive infinity.]

EDITING: Don't read until I am done and finished editing my mistakes.

oh ok, no wonder it doesn't make sense...do you need to dot x=0 since its the asymptote?

SpiralFlex · Aug 14, 2011

starryblue said:
oh ok, no wonder it doesn't make sense...do you need to dot x=0 since its the asymptote?

Don't read just yet, I need to make a few adjustments.

starryblue · Aug 14, 2011

SpiralFlex said:
Don't read just yet, I need to make a few adjustments.

oks =) take your time~

SpiralFlex · Aug 14, 2011

starryblue said:
oks =) take your time~

Should be okay to read now, I will see if I made any silly errors.

starryblue · Aug 14, 2011

SpiralFlex said:
Should be okay to read now, I will see if I made any silly errors.

you rote: We will test the behaviour near its immediate left first. So what is a number close to the left of zero? I know -0.1.

[A positive number divide a positive number is a positive number.]

You get a positive number.

umm....wouldn't it be a positive number divided by a negative number since you said the left of 0, ie. -0.1?

SpiralFlex · Aug 14, 2011

starryblue said:
you rote: We will test the behaviour near its immediate left first. So what is a number close to the left of zero? I know -0.1.

[A positive number divide a positive number is a positive number.]

You get a positive number.

umm....wouldn't it be a positive number divided by a negative number since you said the left of 0, ie. -0.1?

Where did I write that? Quote and bold it.

starryblue · Aug 14, 2011

SpiralFlex said:
$y=\frac{2}{x}$

First step is always find the $x$ and $y$ intercepts, if there are any.

$x$ and $y$ intercepts: When you let $y=0$ to find the $x$ intercept and $x=0$ to find the $y$ intercept, you can see that the value does not exist.

The next step is to find any asymptotes. It is convenient for me to find the vertical asymptote first.

VERTICAL ASYMPTOTE

Now, $y=\frac{2}{x}$

A vertical asymptote occurs when the denominator equals to zero.

$x=0$ [That is our asymptote.]

Your graph should look like this:

http://www.mediafire.com/?faa36hb8slbncv2

Now next stage is to find how the graph behaves near the vertical asymptote.

Behaviour: To do this we must see the tendency of the curve as it approaches the asymptote to the left and to the right.

BEHAVIOUR

We will test the behaviour near its immediate left first. So what is a number close to the left of zero? I know -0.1.

$y=\frac{2}{x}$

$y=\frac{2}{-0.1}$

$y=\frac{+}{-}$ [A positive number divide a negative number is a negative number.]

You get a negative number.

So,

As $x \rightarrow 0^-$ [As $x$ gets close to zero from the left hand side.]

$y \rightarrow - \infty$ [ $y$ tends to negative infinity.]

We will now test the behaviour to the immediate right. So what is a number close to the right of zero? I know 0.1.

$y=\frac{2}{x}$

$y=\frac{2}{0.1}$

$y=\frac{+}{+}$ [A positive number divide a positive number is a positive number.]

You get a positive number.

So,

As $x \rightarrow 0^+$ [As $x$ gets close to zero from the right hand side.]

$y \rightarrow \infty$ [ $y$ tends to positive infinity.]

Your graph should look something like this:

http://www.mediafire.com/?0o2nrnamc3qc3xe

Now horizontal asymptote.

HORIZONTAL ASYMPTOTE AND TENDENCY

We must find what happens when the graph tends to positive infinity. To help you understand this, substitute a large positive number into the equation of graph. Let's use 1000.

$y=\frac{2}{x}$

$y=\frac{2}{1000}$

$y= 0.0001111$ [Something like that, we can see it's a small figure close to zero above.]

So,

As $x \rightarrow \infty$ . [As $x$ approaches positive infinity.]

$y \rightarrow 0^+$ [ $y$ approaches zero a bit above it.]

We must find what happens when the graph tends to negative infinity. To help you understand this, substitute a large negative number into the equation of graph. Let's use -1000.

$y=\frac{2}{x}$

$y=\frac{2}{-1000}$

$y= -0.0001111$ [Something like that, we can see it's a small figure close to zero below it.]

So,

As $x \rightarrow -\infty$ . [As $x$ approaches negative infinity.]

$y \rightarrow 0^-$ [ $y$ approaches zero a bit below it.]

Now your graph should look like:

http://www.mediafire.com/?h5c4kcmekh3dgcl

Now all you have to do is connect the points. Make sure you indicate a point. Use something like $(1, 2)$

http://www.mediafire.com/?d8uuceefrz5t3nc

==" i swear it was something else before~ nvm i got it! tys~

SpiralFlex · Aug 14, 2011

starryblue said:
==" i swear it was something else before~ nvm i got it! tys~

Are you sure? Feel free to ask any more questions. I said I was in the process of editing it.

Graphing Logarithm Question (1 Viewer)

starryblue

Member

SpiralFlex

Well-Known Member

Alkanes

Active Member

SpiralFlex

Well-Known Member

starryblue

Member

SpiralFlex

Well-Known Member

starryblue

Member

SpiralFlex

Well-Known Member

starryblue

Member

SpiralFlex

Well-Known Member

starryblue

Member

SpiralFlex

Well-Known Member

starryblue

Member

SpiralFlex

Well-Known Member

starryblue

Member

SpiralFlex

Well-Known Member

starryblue

Member

SpiralFlex

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)