To find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we can set the equation equal to zero and solve for \( x \):
\[ 0 = -3x^2 + 12x + 15 \]
To make it easier to solve, let's divide the entire equation by -3:
\[ 0 = x^2 - 4x - 5 \]
Next, we can factor the quadratic:
\[ 0 = (x - 5)(x + 1) \]
Now, we set each factor equal to zero:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
Thus, the zeros of the function are \( x = 5 \) and \( x = -1 \).
Therefore, the points (or zeros) are:
\[ (-1, 0) \quad \text{and} \quad (5, 0) \]
The correct response is:
(−1,0)(5,0)