Question

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

Answer 1

To locate the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula, factoring, or completing the square. Let's first use the quadratic formula:

The quadratic formula states that for any quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) can be found using:

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

In this case, the coefficients are:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)

### Step 1: Calculate the Discriminant

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

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

### Step 2: Apply the Quadratic Formula

Now, substituting the values into the quadratic formula:

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

Calculating \( \sqrt{324} \):

\[
\sqrt{324} = 18
\]

Substituting back into our formula:

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

This gives us two possible solutions:

1. Using the positive root:
\[
x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1
\]

2. Using the negative root:
\[
x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5
\]

### Step 3: Solutions

Therefore, the zeros of the equation \( y = -3x^2 + 12x + 15 \) are:

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

These are the points where the quadratic function intersects the x-axis.