Question

use any me thod to locate the zeros of y = -3x^2 + 12x + 15

Answer 1

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

The general form of a quadratic equation is:

\[
ax^2 + bx + c = 0
\]

In this case, we have:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)

The quadratic formula is given by:

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

First, we need to calculate the discriminant \( b^2 - 4ac \):

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

Now, we substitute the values into the quadratic formula:

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

Calculating \( \sqrt{324} \):

\[
\sqrt{324} = 18
\]

Now substituting back into the formula:

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

This gives us two potential zeros:

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

2. **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:

\[
x = -1 \quad \text{and} \quad x = 5
\]

In summary, the roots (or zeros) of the equation are \( x = -1 \) and \( x = 5 \).