Question

Use any method to locate the zeros of y=−3x^2 + 12x + 15
.(1 point)

Answer 1

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

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

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

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

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

Now, we find the discriminant:

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

Since the discriminant is positive, there will be two distinct real roots. We can now substitute these values into the quadratic formula:

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

Calculating \( \sqrt{324} \):

\[
\sqrt{324} = 18
\]

Now we plug this into the formula:

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

Calculating the two possible values for \( x \):

1. For the plus sign:

\[
x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1
\]

2. For the minus sign:

\[
x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5
\]

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

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

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