Question

use any method to locate the zeros of y = -3x^2 + 12x + 15
A. There are no zeros
B. (1. 0) (-5, 0)
C. (-1, 0) (5, 0)
D. (0, 15)

Answer 1

To find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we set the equation equal to zero:

\[
-3x^2 + 12x + 15 = 0
\]

Next, we can either factor the quadratic, use the quadratic formula, or complete the square. Here, we will use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation, the coefficients are:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)

We first calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = 12^2 = 144
\]
\[
4ac = 4 \cdot (-3) \cdot 15 = -180
\]
\[
b^2 - 4ac = 144 - (-180) = 144 + 180 = 324
\]

Since the discriminant is positive (324), there are two distinct real roots.

Now we use the quadratic formula:

\[
x = \frac{-12 \pm \sqrt{324}}{2 \cdot -3}
\]

Calculate \( \sqrt{324} = 18 \):

\[
x = \frac{-12 \pm 18}{-6}
\]

Calculating the two possible solutions:

1. \( x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1 \)
2. \( x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5 \)

Thus, the zeros of the function are at:

\[
(-1, 0) \quad \text{and} \quad (5, 0)
\]

Now, we can match our result to the provided options:

- A. There are no zeros
- B. (1, 0) and (-5, 0)
- C. (-1, 0) and (5, 0)
- D. (0, 15)

The correct answer is **C. (-1, 0) and (5, 0)**.