To determine the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
In this case, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 15^2 = 225 \] \[ 4ac = 4 \cdot 1 \cdot 36 = 144 \] \[ b^2 - 4ac = 225 - 144 = 81 \]
Since the discriminant is positive, there are two distinct real roots. Now we can find the roots using the quadratic formula:
\[ x = \frac{{-15 \pm \sqrt{81}}}{2 \cdot 1} \] \[ x = \frac{{-15 \pm 9}}{2} \]
Now we calculate the two possible values for \( x \):
- For the plus sign \( ( -15 + 9 ) \):
\[ x = \frac{{-6}}{2} = -3 \]
- For the minus sign \( ( -15 - 9 ) \):
\[ x = \frac{{-24}}{2} = -12 \]
Thus, the zeros of the function are \( x = -12 \) and \( x = -3 \).
Now we can list them as follows:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
Therefore, the final answer is:
smaller zero: \( x = -12 \) ; greater zero: \( x = -3 \)