Question

Use the method to locate the zeros of y = -3x squared + 12x + 15

Answer 1

To locate the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we need to set the equation to zero and solve for \( x \):

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

Since it's a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = -3 \), \( b = 12 \), and \( c = 15 \), we can either factor it, complete the square, or use the quadratic formula. Let's use the quadratic formula, which is:

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

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

2. **Apply the quadratic formula**:
\[
x = \frac{-12 \pm \sqrt{324}}{2 \cdot -3}
\]
\[
\sqrt{324} = 18
\]

So the equation becomes:
\[
x = \frac{-12 \pm 18}{-6}
\]

3. **Calculate the two possible values for \( x \)**:
- For the positive root:
\[
x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1
\]

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

Thus, the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \) are:

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