To locate the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \), we need to set the equation equal to zero and solve for \( x \):
\[ -3x^2 + 12x + 15 = 0 \]
To simplify, we can divide all terms by -3:
\[ x^2 - 4x - 5 = 0 \]
Now we can factor the quadratic:
\[ (x - 5)(x + 1) = 0 \]
Setting each factor equal to zero gives us the solutions:
- \( x - 5 = 0 \) ⇒ \( x = 5 \)
- \( x + 1 = 0 \) ⇒ \( x = -1 \)
So the zeros of the equation are \( x = 5 \) and \( x = -1 \).
Thus, the points at which the function intersects the x-axis (the zeros) are:
\[ (-1, 0) \quad \text{and} \quad (5, 0) \]
Looking at the answer choices provided:
A: (0, 15)
B: (1, 0) ( -5, 0)
C: ( -1, 0) ( 5, 0)
D: There are no zeros
The correct answer is:
C: ( -1, 0) ( 5, 0)
1 answer