Question

Use any method to locate the zeros of y=−3x2+12x+15 .

Answer 1

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

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

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

There are no zeros.

Answer 2

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

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

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

First, we need to compute the discriminant (\( b^2 - 4ac \)):

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

Since the discriminant is positive, we have two real and distinct solutions. Now we can substitute the values into the quadratic formula:

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

Calculating \( \sqrt{324} \):

\[
\sqrt{324} = 18
\]

Now substituting back into the formula:

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

This gives us two cases to solve:

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

Thus, the zeros of the equation \( y = -3x^2 + 12x + 15 \) are:

\[
x = -1 \quad \text{and} \quad x = 5
\]

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

Answer 3

To find the zeros of the equation \( y = -3x^2 + 12x + 15 \) again, we confirm that we previously found the zeros, which are indeed \( x = -1 \) and \( x = 5 \).

In terms of ordered pairs (x, y), the zeros of the function occur where \( y = 0 \), or specifically, at the points:

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

From the options provided:

1. \( (1,0)(-5,0) \)
2. \( (-1,0)(5,0) \)
3. \( (0,15) \)
4. There are no zeros.

The correct response is:

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