To determine the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula, which is given by:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
In this case, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
Now, we'll calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
Since the discriminant is positive, we will have two real and distinct roots.
Now, we can substitute these values into the quadratic formula:
\[ x = \frac{{-15 \pm \sqrt{81}}}{2 \cdot 1} \] \[ x = \frac{{-15 \pm 9}}{2} \]
This gives us two possible solutions for \( x \):
-
Using the positive root: \[ x = \frac{{-15 + 9}}{2} = \frac{{-6}}{2} = -3 \]
-
Using the negative root: \[ x = \frac{{-15 - 9}}{2} = \frac{{-24}}{2} = -12 \]
Thus, the zeros of the function are:
\[ x = -3 \quad \text{and} \quad x = -12 \]