To find the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = -3 \), \( b = 12 \), and \( c = 15 \).
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 12^2 = 144 \] \[ 4ac = 4 \cdot (-3) \cdot 15 = -180 \] \[ b^2 - 4ac = 144 - (-180) = 144 + 180 = 324 \]
Now we can plug the values into the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{324}}{2 \cdot -3} \]
Calculate \( \sqrt{324} \):
\[ \sqrt{324} = 18 \]
Then we have:
\[ x = \frac{-12 \pm 18}{-6} \]
Calculating the two potential solutions:
- For \( x = \frac{-12 + 18}{-6} \):
\[ x = \frac{6}{-6} = -1 \]
- For \( x = \frac{-12 - 18}{-6} \):
\[ x = \frac{-30}{-6} = 5 \]
Thus, the zeros of the equation are \( x = -1 \) and \( x = 5 \).
The zeros of the function are \( (-1, 0) \) and \( (5, 0) \).
So the correct response is: \[ (-1, 0)(5, 0) \]