Question

Write the equation of a parabola that opens downward with no stretch, given the axis of symmetry is x=3
and the vertex is (3,11)
.(1 point)
Responses

y=−(x−3)2−11
y is equal to negative left parenthesis x minus 3 right parenthesis squared minus 11

y=−(x+3)2+11
y is equal to negative left parenthesis x plus 3 right parenthesis squared plus 11

y=−(x−3)2+11
y is equal to negative left parenthesis x minus 3 right parenthesis squared plus 11

y=−(x+11)2−3
y is equal to negative left parenthesis x plus 11 right parenthesis squared minus 3

Answer 1

To write the equation of a parabola that opens downward with its vertex at (3, 11) and axis of symmetry at \( x = 3 \), we use the vertex form of a parabola's equation:

\[
y = a(x - h)^2 + k
\]

where \( (h, k) \) is the vertex of the parabola. In this case, \( h = 3 \) and \( k = 11 \).

Since the parabola opens downward, \( a \) should be negative. Given that there is no stretch, we will use \( a = -1 \).

Substituting these values into the vertex form gives:

\[
y = -1(x - 3)^2 + 11
\]

This simplifies to:

\[
y = -(x - 3)^2 + 11
\]

Therefore, the correct equation of the parabola is:

\[
\boxed{y = -(x - 3)^2 + 11}
\]

So the correct response is:

**y = −(x−3)² + 11**.