# Finding Intercepts

• Sep 22nd 2008, 10:58 AM
Cam5
Finding Intercepts
Hey there, I've been asked to find the intercepts of:

$\displaystyle x^2 + y^2 = 100$

If I were to just set one value to zero, and then solve for the other one, would it still be correct, even in this form?

For example,
$\displaystyle 0^2 + y^2 = 100$
$\displaystyle y^2 = 100$
(take the square root of both $\displaystyle y^2$ and 100 - I don't know the code)

So then y = 10?

Is this the correct way to go about solving for the intercepts here?

- Cam

P.S. Maybe it's best done by graphing, but how would I determine the domain and range of this?
• Sep 22nd 2008, 11:05 AM
Moo
Hello !

Same problem as in your last thread (if my memory's fine) :

$\displaystyle y^2=a^2 \Longleftrightarrow y=\pm a \text{ that is to say } y=a \text{ or } y=-a$

This comes from : $\displaystyle y^2=a^2 \Longleftrightarrow y^2-a^2=0 \Longleftrightarrow (y-a)(y+a)=0 \Longleftrightarrow y-a=0 \text{ or } y+a=0$

Otherwise, it is the correct method.
(Wink)
• Sep 22nd 2008, 11:14 AM
Cam5
Thanks, once again, moo. :)

I've got another one, if you'll humor me. This is an extension of my last question before this...

If f(x) = $\displaystyle \frac{2x+1}{3}$ and g(x) = $\displaystyle \frac{x-1}{2}$ determine m so that f(m) = g(m).

Now, this much I know:

f(m) = (gm)

m and x are both images under the functions f, and g, but unlike before, I don't have a good idea about how to begin determining m.

Perhaps
$\displaystyle \frac{2x+1}{3}=\frac{x-1}{x}$ ?

Quote:

Originally Posted by Moo
Hello !

Same problem as in your last thread (if my memory's fine) :

$\displaystyle y^2=a^2 \Longleftrightarrow y=\pm a \text{ that is to say } y=a \text{ or } y=-a$

This comes from : $\displaystyle y^2=a^2 \Longleftrightarrow y^2-a^2=0 \Longleftrightarrow (y-a)(y+a)=0 \Longleftrightarrow y-a=0 \text{ or } y+a=0$

Otherwise, it is the correct method.
(Wink)

• Sep 22nd 2008, 11:20 AM
Moo
Quote:

Originally Posted by Cam5
Thanks, once again, moo. :)

I've got another one, if you'll humor me. This is an extension of my last question before this...

If f(x) = $\displaystyle \frac{2x+1}{3}$ and g(x) = $\displaystyle \frac{x-1}{2}$ determine m so that f(m) = g(m).

Now, this much I know:

f(m) = (gm)

m and x are both images under the functions f, and g, but unlike before, I don't have a good idea about how to begin determining m.

No, it means that the image of m under the function f is also the image of m under the function g.
Basically, what's in brackets f(...) or g(...) will be the element that will have the image. f(...) and g(...) represent the images.

Quote:

Perhaps
$\displaystyle \frac{2x+1}{3}=\frac{x-1}{x}$ ?
Almost !
It's not x, it's m, since it is said that $\displaystyle f(m)=g(m)$

$\displaystyle f(m)$ is rewritten : $\displaystyle \frac{2m+1}{3}$ and g(m)...

It's not quite the same as your question of several days ago, because here, you're dealing with two functions. And there is no real question to ask :)
• Sep 22nd 2008, 11:23 AM
Cam5
Alright. That gives me something to work with. You're a great help!

Quote:

Originally Posted by Moo
No, it means that the image of m under the function f is also the image of m under the function g.
Basically, what's in brackets f(...) or g(...) will be the element that will have the image. f(...) and g(...) represent the images.

Almost !
It's not x, it's m, since it is said that $\displaystyle f(m)=g(m)$

$\displaystyle f(m)$ is rewritten : $\displaystyle \frac{2m+1}{3}$ and g(m)...

It's not quite the same as your question of several days ago, because here, you're dealing with two functions. And there is no real question to ask :)