Question

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

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

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

There are no zeros.

Answer 1

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

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

To simplify, we can divide the entire equation by -3:

\[
x^2 - 4x - 5 = 0
\]

Next, we can factor the quadratic equation:

\[
(x - 5)(x + 1) = 0
\]

Setting each factor equal to zero gives us the potential solutions:

\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\]

\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]

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

So, the corresponding points where \( y = 0 \) are \( (5, 0) \) and \( (-1, 0) \).

The correct response is:

\[
(-1, 0) \quad (5, 0)
\]

Therefore, the most suitable choice from the options provided is:

\((−1,0)(5,0)\)