# Math Help - Finding Intercepts

1. ## Finding Intercepts

Hey there, I've been asked to find the intercepts of:

$x^2 + y^2 = 100$

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,
$0^2 + y^2 = 100$
$y^2 = 100$
(take the square root of both $y^2$ 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?

2. Hello !

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

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

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

Otherwise, it is the correct method.

3. 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) = $\frac{2x+1}{3}$ and g(x) = $\frac{x-1}{2}$ 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
$\frac{2x+1}{3}=\frac{x-1}{x}$ ?

Originally Posted by Moo
Hello !

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

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

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

Otherwise, it is the correct method.

4. 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) = $\frac{2x+1}{3}$ and g(x) = $\frac{x-1}{2}$ 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.

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

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

5. Alright. That gives me something to work with. You're a great help!

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 $f(m)=g(m)$

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