To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we need to use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
Now we can substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} \]
Calculating \( \sqrt{81} \):
\[ \sqrt{81} = 9 \]
Now substitute this back into the formula:
\[ x = \frac{-15 \pm 9}{2} \]
This gives us two solutions:
- For the positive case:
\[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]
- For the negative case:
\[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]
So the zeros of the quadratic function are:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
Thus, we can summarize the results:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)