The function for the Maclaurin Series is the following sum:

SUM{n=0:infinity} (f^(n)(0)*x^n)/n! ... where f^(n) is the "n'th derivative of f," and n! is "n factorial" = n(n-1)(n-2)*...*(2)(1)

I suspect that sum doesn't make sense to you, so here's the easy way to remember it. We are basically adding all the derivatives of the function in the following way:

f(0) + f'(0)/(1!)*x + f''(0)/(2!)*x^2 + f'''(0)/(3!)*x^3 + ... + f^(n)(0)/(n!)*x^n + ...

All we need in this case is the first 4 terms of the series:

f(0) + f'(0)*x + f''(0)/2*x^2 + f'''(0)/6*x^3

For the 4 functions you've listed, find up to the third derivative and plug those into the above sum. You should see which match with which.

I'll do the first for you:

f(x) = e^(-x) - 1

f'(x) = -e^(-x)

f''(x) = e^(-x)

f'''(x) = -e^(-x)

f(0) = 0

f'(0) = -1

f''(0) = 1

f'''(0) = -1

Let's plug these into the sum to get:

0 + (-1)x + (1)/2*x^2 + (-1)/6*x^3 = -x +1/2*x^2 - 1/6*x^3

This looks like answer 3, except mine is divided by 6. Check to see that number 3 is typed correctly.