Question

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

Answer 1

To find the point that is symmetrical to \((-1, -3)\) for the graph of the function \(y = x^2 + 12x + 8\), we need to identify the axis of symmetry of the parabola defined by this quadratic equation.

1. **Find the vertex of the parabola** using the formula \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = 12\):

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

Plugging \(x = -6\) back into the equation to find the \(y\)-coordinate:

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

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

2. **Find the symmetrical point**: Since the axis of symmetry is a vertical line at \(x = -6\), the symmetrical point to \((-1, -3)\) will have the same \(y\)-coordinate but will be equidistant from \(-6\).

- Calculate the distance from \(-1\) to the axis of symmetry \(-6\):

\[
|(-1) - (-6)| = |-1 + 6| = |5| = 5
\]

Then, move to the left from -6 by 5 units to find the symmetrical point:

\[
-6 - 5 = -11
\]

Thus, the symmetrical point \( (-11, -3) \) retains the same \(y\)-coordinate of \(-3\).

3. **Final Result**: The point symmetrical to \((-1, -3)\) with respect to the graph of \(y = x^2 + 12x + 8\) is

\[
\boxed{(-11, -3)}.
\]