Question

What point is symmetrical to (−1,−3)
(
−
1
,
−
3
)
for the graph of y=x2+12x+8
𝑦
=
𝑥
2
+
12
𝑥
+
8
?(1 point)
()

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 determine the vertex of the quadratic function since it will act as the line of symmetry.

1. **Find the vertex** of the quadratic function \(y = x^2 + 12x + 8\). The formula to find the vertex \(x\) coordinate is given by \(x = -\frac{b}{2a}\).

Here, \(a = 1\) and \(b = 12\):
\[
x = -\frac{12}{2 \cdot 1} = -6
\]

2. **Find the y-coordinate** of the vertex by plugging \(x = -6\) back into the equation:
\[
y = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28
\]
So, the vertex is \((-6, -28)\).

3. Now that we have the vertex \((-6, -28)\), we can find the symmetrical point to \((-1, -3)\). The symmetry point across the vertex will have the same distance from the vertex as the original point.

4. **Calculate the horizontal distance** from the vertex \((-6, -28)\) to the point \((-1, -3)\):
\[
-1 - (-6) = -1 + 6 = 5
\]
The distance is \(5\).

5. **Find the symmetrical point** by moving 5 units to the left of the vertex:
\[
-6 - 5 = -11
\]
The y-coordinate remains the same. Hence, the y-coordinate of the symmetrical point is still \(-3\).

6. **Conclusion**: The point symmetrical to \((-1, -3)\) with respect to the vertex \((-6, -28)\) is:
\[
(-11, -3)
\]

So, the symmetrical point is \((-11, -3)\).