Question

Convert the equation from vertex form to standard form y=−4(x+5)^2−3
.(1 point)
Responses

y=16x^2+160x+397
y equals 16 x squared plus 160 x plus 397

y=−4x^2−40x−103
y equals negative 4 x squared minus 40 x minus 103

y=−4x^2−25x−3
y equals negative 4 x squared minus 25 x minus 3

y=16x^2+160x+412

Answer 1

To convert the equation from vertex form \( y = -4(x + 5)^2 - 3 \) to standard form, we will expand it.

1. Start with the vertex form:

\[
y = -4(x + 5)^2 - 3
\]

2. Expand \( (x + 5)^2 \):

\[
(x + 5)^2 = x^2 + 10x + 25
\]

3. Substitute this back into the equation:

\[
y = -4(x^2 + 10x + 25) - 3
\]

4. Distribute the \(-4\):

\[
y = -4x^2 - 40x - 100 - 3
\]

5. Combine like terms:

\[
y = -4x^2 - 40x - 103
\]

So, the standard form of the equation is:

\[
\boxed{y = -4x^2 - 40x - 103}
\]

Therefore, the correct response is:

**y = −4x^2 − 40x − 103**.