Question

Use any method to locate the zeros of y = -3x2 + 12x + 15. (1 point)
• (-1, 0) (5, 0)
• (0,15)
• There are no zeros.
• (1, 0) (-5, 0)

Answer 1

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

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

We can first multiply the entire equation by -1 (to make calculations simpler):

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

Next, we can use the quadratic formula to find the zeros of the equation, where \( a = 3 \), \( b = -12 \), and \( c = -15 \):

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

Calculating the discriminant (\( b^2 - 4ac \)):

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

Now apply 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 solutions:

1. \( x = \frac{12 + 18}{6} = \frac{30}{6} = 5 \)
2. \( x = \frac{12 - 18}{6} = \frac{-6}{6} = -1 \)

So the zeros of the function are \( x = 5 \) and \( x = -1 \).

The respective points where these zeros occur on the graph are:

\((-1, 0)\) and \((5, 0)\).

Therefore, the correct answer is:

**(-1, 0) and (5, 0)**.