Question

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

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

B) There are no zeros.

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

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

Answer 1

To find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we can set \( y \) to 0 and solve for \( x \):

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

We can start by using the quadratic formula, which states:

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

In this case, \( a = -3 \), \( b = 12 \), and \( c = 15 \).

Now, calculate the discriminant:

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

Since the discriminant is positive, there are two distinct real roots. Now we will substitute the values into the quadratic formula:

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

Calculating \( \sqrt{324} \):

\(\sqrt{324} = 18\)

Substituting back into the formula gives:

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

This will give us two possible solutions:

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

2. Using \( - \):
\[
x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5
\]

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

The points in ordered pair form are:

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

So, the correct response is:

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