Question

What point is symmetrical to (−1,−3)
for the graph of y=x2+12x+8
?(1 point)
(
)
Remaining Attempts : 3

Answer 1

To find the point that is symmetrical to the point \((-1, -3)\) with respect to the graph of the function \(y = x^2 + 12x + 8\), we first need to identify the vertex of the parabola described by the function.

The given quadratic function can be expressed in standard form as follows:

1. Complete the square or use the vertex formula \(x = -\frac{b}{2a}\).

The function is \(y = x^2 + 12x + 8\), where \(a = 1\) and \(b = 12\).

2. Calculate the x-coordinate of the vertex:

\[
x = -\frac{b}{2a} = -\frac{12}{2(1)} = -6.
\]

3. Substitute \(x = -6\) back into the equation to find the y-coordinate of the vertex:

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

So the vertex of the parabola is \((-6, -28)\).

Now, to find the reflected point, we need to calculate the distance from the point \((-1, -3)\) to the vertex \((-6, -28)\):

1. The x-coordinate distance:

\[
\text{Distance in x} = -1 - (-6) = -1 + 6 = 5.
\]

2. The y-coordinate distance:

\[
\text{Distance in y} = -3 - (-28) = -3 + 28 = 25.
\]

Now, to reflect the point \((-1, -3)\) across the vertex \((-6, -28)\), we'll move the same distance from the vertex in the opposite direction:

1. For the x-coordinate:

\[
x' = -6 - 5 = -11.
\]

2. For the y-coordinate:

\[
y' = -28 - 25 = -53.
\]

Thus, the point that is symmetrical to \((-1, -3)\) with respect to the graph is:

\[
\boxed{(-11, -53)}.
\]