To determine the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = 15 \), and \( c = 36 \).
First, we calculate the discriminant:
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
Next, we apply the quadratic formula:
\[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} \]
Since \( \sqrt{81} = 9 \), we have:
\[ x = \frac{-15 \pm 9}{2} \]
Now, we calculate the two possible values for \( x \):
- For the positive case:
\[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]
- For the negative case:
\[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]
Thus, the zeros of the function are \( x = -12 \) and \( x = -3 \).
Finally, we list the smaller and greater zeros:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
So, the final answer is:
smaller zero: \( x = -12 \); greater zero: \( x = -3 \)