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

Question

.(1 point)
Responses

(0,15)
left parenthesis 0 comma 15 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

(−1,0)(5,0)

Answer 1

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

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

Dividing the entire equation by -3 to simplify it:

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

Next, we can factor this equation. We are looking for two numbers that multiply to \(-5\) (the constant term) and add to \(-4\) (the coefficient of the \(x\) term). These numbers are \(-5\) and \(1\):

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

Setting each factor to zero gives us the solutions:

1. \( x - 5 = 0 \) \( \Rightarrow x = 5 \)
2. \( x + 1 = 0 \) \( \Rightarrow x = -1 \)

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

Therefore, the coordinates of the zeros are:

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

The correct response is:

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

Answer 2

Complete the square to identify the vertex of y=x2−16x+14
.(1 point)
Responses

(8,78)
left parenthesis 8 comma 78 right parenthesis

(8,14)
left parenthesis 8 comma 14 right parenthesis

(8,−50)
left parenthesis 8 comma negative 50 right parenthesis

(8,30)