Question

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

(−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.

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

(0,15)

Answer 1

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

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

We can solve this quadratic equation using the quadratic formula, which is:

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

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

Now, let's 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 and distinct solutions.

Now, let's find the roots using the quadratic formula:

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

Calculating \(\sqrt{324}\):

\(\sqrt{324} = 18\)

Now substitute back into the formula:

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

This gives us two cases to solve:

1. **Case 1**:
\[
x = \frac{{-12 + 18}}{{-6}} = \frac{6}{{-6}} = -1
\]

2. **Case 2**:
\[
x = \frac{{-12 - 18}}{{-6}} = \frac{-30}{{-6}} = 5
\]

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

This means the zeros of \(y = -3x^2 + 12x + 15\) are:

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

The correct response is:

\((−1,0)(5,0)\)