Question

Use any method to locate the zeros of y = -3x^2 + 12x + 15.

A: ( 0,15)

B: ( 1,0) ( -5,0)

C: ( -1,0) ( 5,0)

D: There are no zeros

Answer 1

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:

1. \( x - 5 = 0 \) ⇒ \( x = 5 \)
2. \( x + 1 = 0 \) ⇒ \( x = -1 \)

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)**