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 find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula, which is:

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

For the equation \( y = -3x^2 + 12x + 15 \), the coefficients are:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)

Now we can substitute these values into the quadratic formula:

1. Calculate the discriminant (\( b^2 - 4ac \)):
\[
b^2 - 4ac = 12^2 - 4 \cdot (-3) \cdot 15 = 144 + 180 = 324
\]

2. Since the discriminant is positive, we know there will be two real zeros.

3. Now substitute into the quadratic formula:
\[
x = \frac{-12 \pm \sqrt{324}}{2 \cdot -3}
\]

4. Simplify \(\sqrt{324}\):
\[
\sqrt{324} = 18
\]

5. Substitute this back into the formula:
\[
x = \frac{-12 \pm 18}{-6}
\]

6. Calculate the two values for \(x\):

For the positive case:
\[
x_1 = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1
\]

For the negative case:
\[
x_2 = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5
\]

Thus, the zeros of the function \( y = -3x^2 + 12x + 15 \) are \( (-1, 0) \) and \( (5, 0) \).

The correct answer is:
**b (-1,0)(5,0)**.