Question

Use any method to locate the zeros of y=−3x2+12x+15
𝑦
=
−
3
𝑥
2
+
12
𝑥
+
15
.(1 point)
Responses

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

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

There are no zeros.
There are no zeros.

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

Answer 1

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

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

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, there will be two real solutions (zeros).

Now we can substitute back into the quadratic formula:

\[
x = \frac{-12 \pm \sqrt{324}}{2(-3)}
\]
\[
x = \frac{-12 \pm 18}{-6}
\]

Calculating the two possible values:

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 equation are \( x = -1 \) and \( x = 5 \).

The points where \( y = 0 \) are:

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

Among the given options, the correct responses are:

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