To determine the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the given quadratic \( ax^2 + bx + c \):
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
Now we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
Next, we apply the quadratic formula:
\[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} = \frac{-15 \pm 9}{2} \]
Now we calculate the two possible values of \( x \):
-
For the positive root: \[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]
-
For the negative root: \[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]
Thus, the zeros of the function are \( x = -12 \) and \( x = -3 \).
Now, we identify the smaller and greater zeros:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
Therefore, the final answers are:
smaller zero: \( x = -12 \)
greater zero: \( x = -3 \)