Question

What point is symmetrical to (−1,−3)
for the graph of y=x^2+12x+8
?(1 point)
(
)

Answer 1

To find the point that is symmetrical to \((-1, -3)\) for the graph of the quadratic function \(y = x^2 + 12x + 8\), we need to first determine the vertex of the quadratic. The vertex of a parabola given by the equation \(y = ax^2 + bx + c\) is located at the x-coordinate:

\[
x = -\frac{b}{2a}
\]

In this case, \(a = 1\) and \(b = 12\):

\[
x = -\frac{12}{2 \cdot 1} = -6
\]

Next, we need to find the y-coordinate of the vertex by plugging \(x = -6\) back into the original equation:

\[
y = (-6)^2 + 12(-6) + 8
\]
\[
y = 36 - 72 + 8
\]
\[
y = -28
\]

Thus, the vertex of the parabola \(y = x^2 + 12x + 8\) is \((-6, -28)\).

Now, since the original point is \((-1, -3)\), we need to find the reflection of this point across the vertex \((-6, -28)\). To find the point that is symmetrical, we can compute the distance from the point to the vertex and then move the same distance in the opposite direction.

First, calculate the distance between \((-1, -3)\) and \((-6, -28)\):

The x-coordinate distance is:
\[
-1 - (-6) = -1 + 6 = 5
\]

The y-coordinate distance is:
\[
-3 - (-28) = -3 + 28 = 25
\]

Thus, the distance is:
\[
d_x = 5 \quad \text{and} \quad d_y = 25
\]

To find the symmetrical point \(P\), we move from the vertex \((-6, -28)\) by the same distance but in the opposite direction:

For the x-coordinate:
\[
-6 - 5 = -11
\]

For the y-coordinate:
\[
-28 - 25 = -53
\]

Thus, the point symmetrical to \((-1, -3)\) across the vertex \((-6, -28)\) is \((-11, -53)\).

The final answer is:
\[
\boxed{(-11, -53)}
\]