Can you help me with this math problem?

Question

Find all x- and y- intercepts
[(x-3^2)/4]-[y^2/9]=1
or it might be
[((x-3)^2)/4]-[(y^2)/9]=1
does one make more sense than the other?
I don't know, sorry, the problem was told to me over the phone.

Answer 1

The second one makes a lot more sense, yes.

Rule of thumb: To find the x-intercepts, make y equal to zero and solve for y. To find the y-intercepts, make x equal to zero and solve for x.

So for the equation

[(x - 3)^2]/4 - (y^2)/9 = 1

Let's find the y-intercepts. Make x = 0.

[(0 - 3)^2]/4 - (y^2)/9 = 1
[(-3)^2]/4 - (y^2)/9 = 1
[9/4] - (y^2)/9 = 1
-(y^2)/9 = 1 - 9/4
-(y^2)/9 = -5/4
(y^2)/9 = 5/4
y^2 = 45/4

Remember that whenever we take the square root of both sides, we have to add a "plus or minus" sign.
y = +/- sqrt(45)/2

To solve for the x-intercepts, make y=0

[(x - 3)^2]/4 - (0^2)/9 = 1
[(x - 3)^2]/4 - 0 = 1
[(x - 3)^2]/4 = 1
(x - 3)^2 = 4
x - 3 = +/- 2
x = 3 +/- 2
x = 5, 1
So the y intercepts are sqrt(45)/2 and -sqrt(45)/2.
The x-intercepts are 5 and 1.

Answer 2

The second equation makes much more sense.

[((x-3)^2)/4]-[(y^2)/9]=1
Set x = 0 to find y-intercepts, so
((0-3)^2)/4 -(y^2)/9 = 1
9/4 -1 = (y^2)/9
y^2 = (5/4)*9
y = + or - (3/2)sqrt(5)

Set y = 0 to find x-intercepts, so
((x-3)^2)/4 -(0^2)/9 = 1
((x-3)^2)/4 = 1
(x-3)^2 = 4
x-3 = + or - 2
x = 3 + or - 2
x= 5 or 1

Answer 3

These both problems are same.

Intercept means the point at which the function becomes zero.

So, x intercept is x=3 and y intercept is y=0.

Answer 4

I expect it to be:

(x-3)² / 4 - y² / 9 = 1

This is the formula of a hyperbola with its center at the point (3,0).

See: http://en.wikipedia.org/wiki/Hyperbola#Cartesian