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, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
First, we calculate the discriminant (\( b^2 - 4ac \)):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
Next, we plug the values into the quadratic formula:
\[ x = \frac{{-15 \pm \sqrt{81}}}{2 \cdot 1} = \frac{{-15 \pm 9}}{2} \]
Now we can calculate the two zeros:
-
First zero (using the \( + \)): \[ x_1 = \frac{{-15 + 9}}{2} = \frac{{-6}}{2} = -3 \]
-
Second zero (using the \( - \)): \[ x_2 = \frac{{-15 - 9}}{2} = \frac{{-24}}{2} = -12 \]
Thus, the smaller and greater zeros of the quadratic function are:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
So the final answer is:
Smaller zero: \( x = -12 \); Greater zero: \( x = -3 \)