# Finding Intercepts of an Equation

1. Sep 2, 2014

### Staff: Mentor

1. The problem statement, all variables and given/known data
Find the intercepts of the following equation: $y=\frac{-x^3}{x^2-9}$

2. Relevant equations

To find the X-Intercept, set Y equal to zero and solve.
To find the Y-Intercept, set X equal to zero and solve.

3. The attempt at a solution

Setting Y equal to zero gives me: $0=\frac{-x^3}{x^2-9}$
This simplifies to: $0=\frac{-x^3}{(x+3)(x-3)}$

And now I'm stuck. :uhh:

2. Sep 2, 2014

### Mentallic

The first thing you should remind yourself of is that we can't divide by 0. Clearly since the denominator is

$$x^2-9=(x-3)(x+3)$$

and this quadratic is equal to 0 when $x=\pm 3$ hence our original equation does not exist at those x values.

So now, we consider when x is any other value.

If we have some fraction a/b for any numbers a and b where b is anything except 0, then when is this fraction equal to 0?

3. Sep 2, 2014

Never?

4. Sep 2, 2014

### Mentallic

What does 0/5 equal to? 0/100? 0/k? for any k that isn't 0?

5. Sep 2, 2014

### Staff: Mentor

Either zero or undefined. I'm guessing zero?

6. Sep 3, 2014

### Staff: Mentor

Okay, I looked it up and it's definitely zero.

7. Sep 3, 2014

### Staff: Mentor

Yes, definitely. Zero divided by any nonzero number is zero.

8. Sep 3, 2014

### Staff: Mentor

Yeah, I feel bad that I had to go look that up. I guess it's been a while since I did math.

9. Sep 3, 2014

### Mentallic

It's ok, that's what we're here for!

So you can look at this problem in two ways, either notice that a fraction of the form a/b is equal to zero when the numerator is equal to 0, so in your equation

$$y=\frac{-x^3}{x^2-9}$$

To find when y=0, i.e. to solve

$$0=\frac{-x^3}{x^2-9}$$

we just need to find when the numerator -x3=0.

Or the other way, simply multiply both sides of the equation by the denominator of the fraction.

$$0\times (x^2-9) = \frac{-x^3}{x^2-9}\times (x^2-9)$$

Now on the RHS we can cancel the factor of x2-9 in both the numerator and denominator to end up with just -x^3, and on the LHS, 0 times anything is still 0, so of course again we end up solving 0=-x3.

10. Sep 3, 2014

### Staff: Mentor

Hmmm. The only thing I can think of that makes -x3 zero is x = 0.

So that would make the X intercept zero then, correct?

11. Sep 3, 2014

### Mentallic

Yes and yes.

12. Sep 3, 2014

### Staff: Mentor

Okay, then if we set X = 0 to find the Y intercept: Y=0/-9, which would make Y=0.

So the intercepts are (0,0)?

13. Sep 3, 2014

### Mentallic

Yes, but I wouldn't call it two intercepts

Just a little extra in case you're curious:
We didn't even need to check x=0 (the intercept of the y-axis) because we already have from the x-axis intercept that (0,0) is satisfied.

The alternative doesn't always suffice though. If we first checked x=0 (the y-axis intercept) and found y=0 satisfied this, we would still need to solve for y=0 because we could possibly have more solutions for x such as in the case of quadratics.

Now, the reason why we wouldn't need to check x=0 (as opposed needing to check y=0) is because y is a function of x. Functions have a very important property that for every x-value, there is only ever one y-value, so when we found that for x=0, y=0, we couldn't possibly have another value for y given x=0, so we didn't need to test x=0.

14. Sep 3, 2014

### Staff: Mentor

Interesting. I didn't even realize that having the X-intercept as 0 meant that we also had the Y intercept. Makes sense. Thanks, Mentallic!