To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we need to solve for \( x \) when \( y = 0 \). This involves solving the equation:
\[
x^2 + 15x + 36 = 0
\]
We can use the factoring method. We need two numbers that multiply to \( 36 \) and add up to \( 15 \). The numbers \( 3 \) and \( 12 \) meet this criterion:
\[
(x + 3)(x + 12) = 0
\]
Setting each factor to zero gives us:
1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x + 12 = 0 \) → \( x = -12 \)
Now we can list the zeros. To follow the instruction of smaller and greater:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
So:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)