Question

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

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

There are no zeros.
There are no zeros.

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

(−1,0)(5,0)

Answer 1

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

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

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

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

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

Now we can plug the values into the quadratic formula:

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

Calculate \( \sqrt{324} \):

\[
\sqrt{324} = 18
\]

Then we have:

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

Calculating the two potential solutions:

1. For \( x = \frac{-12 + 18}{-6} \):

\[
x = \frac{6}{-6} = -1
\]

2. For \( x = \frac{-12 - 18}{-6} \):

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

Thus, the zeros of the equation are \( x = -1 \) and \( x = 5 \).

The zeros of the function are \( (-1, 0) \) and \( (5, 0) \).

So the correct response is:
\[
(-1, 0)(5, 0)
\]