To find the point that is symmetrical to \((-1, -17)\) with respect to the graph of the equation \(y = -x^2 - 16\), we need to reflect the point across the vertex of the parabola represented by this equation.
First, let’s find the vertex of the parabola. The given equation is:
\[ y = -x^2 - 16 \]
This is a downward-opening parabola. The vertex form of a parabola \(y = a(x - h)^2 + k\) reveals that the vertex corresponds to the values \(h\) and \(k\). For our equation in standard form \(y = ax^2 + bx + c\), we can find the vertex using \(h = -\frac{b}{2a}\).
In our equation \(y = -x^2 - 16\):
- \(a = -1\)
- \(b = 0\)
- \(c = -16\)
Calculating the vertex:
\[ h = -\frac{0}{2(-1)} = 0 \]
Then, we substitute \(h\) back to find \(k\):
\[ k = -0^2 - 16 = -16 \]
So, the vertex is located at the point \((0, -16)\).
Next, we find the symmetrical point to \((-1, -17)\). The reflection can be found by using the relationship of distances. The distance from \((-1, -17)\) to the vertex \((0, -16)\) needs to be the same on the opposite side.
- Calculate the distance from \((-1, -17)\) to the vertex \((0, -16)\):
\[ \text{Distance} = \sqrt{(0 - (-1))^2 + (-16 - (-17))^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{2} \]
- Since the vertex is \((0, -16)\), to find the symmetrical point, we move the same distance in the opposite direction along the horizontal axis. Given that our initial point \((-1, -17)\) is to the left of the vertex, the symmetrical point will be \(1\) unit to the right (the horizontal distance is \(1\) unit).
Thus, the symmetrical point can be calculated as follows:
- The \(x\)-coordinate of the symmetrical point:
\[ 0 + 1 = 1 \]
- The \(y\)-coordinate of the symmetrical point stays the same because it will also be symmetrical vertically:
The original point \(y\)-coordinate is \(-17\) which reflects over the vertex value of \(y = -16\).
So the \(y\)-coordinate becomes:
\[ y = -16 + (-16 + (-17)) = -16 + 1 = -15 \]
Thus, the symmetrical point with respect to the vertex is:
\[ (1, -15) \]
So, the point that is symmetrical to \((-1, -17)\) with respect to the graph of \(y = -x^2 - 16\) is \(\boxed{(1, -15)}\).
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