To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
In this case, we have:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
Now, we can calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
Now, we can substitute the values into the quadratic formula:
\[ x = \frac{{-15 \pm \sqrt{81}}}{2 \cdot 1} = \frac{{-15 \pm 9}}{2} \]
This gives us two solutions:
- \( x = \frac{{-15 + 9}}{2} = \frac{{-6}}{2} = -3 \)
- \( x = \frac{{-15 - 9}}{2} = \frac{{-24}}{2} = -12 \)
The zeros of the quadratic function are \( -12 \) and \( -3 \).
When listing the smaller zero first, we would write:
\[ -12, -3 \]