Question

Use any method to locate the zeros of y=−3x2+12x+15.(1 point)
Responses

(1,0)(−5,0)
left parenthesis 1 comma 0 right parenthesis left parenthesis negative 5 comma 0 right parenthesis

(0,15)
left parenthesis 0 comma 15 right parenthesis

(−1,0)(5,0)
left parenthesis negative 1 comma 0 right parenthesis left parenthesis 5 comma 0 right parenthesis

There are no zeros.

Answer 1

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

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

where \(a = -3\), \(b = 12\), and \(c = 15\).

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

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

Since the discriminant is positive, it indicates that there are two real zeros. Now we can substitute back into the quadratic formula:

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

Calculating \(\sqrt{324}\):

\[
\sqrt{324} = 18
\]

Now substituting back:

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

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

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

The zeros of the function are:

\[
(-1, 0) \quad \text{and} \quad (5, 0)
\]

Thus, the correct response is:

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