How to find horizontal asymptote

f(x) = x^3-1/x^2+3x-10

I am studying for my test, and my study guide has the answer as y=x-3, but doesn't the horizontal asymptote = 0 if the numerator power is bigger?

Thanks!

No, it isn't. It's 0 if the denominator power is bigger. You need to synthetically divide the numerator by the denominator

x^3 / x^2 = x
x * (x^2 + 3x - 10) = x^3 + 3x^2 - 10x
x^3 + 0x^2 + 0x - x^3 - 3x^2 + 10x = -3x^2 + 10x

-3x^2 / x^2 = -3
-3 * (x^2 + 3x - 10) = -3x^2 - 9x + 30
-3x^2 + 10x - 1 + 3x^2 + 9x - 30 = 19x - 31

So we have:

(x - 3) + (19x - 31) / (x^2 + 3x - 10)

As x goes to infinity, (19x - 31) / (x^2 + 3x - 10) goes to 0, so we're left with:

y = x - 3

That's not a horizontal asymptote either. It's a slant asymptote

Could you clarify f(x) with brackets?

As you've written it ... f(x) = x^3 - (1/x^2) + 3x - 10 ... there's no horizontal asymptote. The function just approaches f(x) = x^3 for large x.

Do you mean f(x) = (x^3 - 1)/(x^2 + 3x - 10) ? If so, long division gives you:

f(x) = x - 3 + [(19x -31)/(x^2 + 3x - 10)]

As x becomes large, the bit in square brackets tends to zero, giving y = x - 3 as the asymptote.