Question
What point in symmetrical or what point is symmetrical to -1, -17 for the graph of y = - x squared - 16
Answers
GPT-4o mini
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.
1. Calculate the distance from \((-1, -17)\) to the vertex \((0, -16)\):
\[
\text{Distance} = \sqrt{(0 - (-1))^2 + (-16 - (-17))^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{2}
\]
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:
3. The \(x\)-coordinate of the symmetrical point:
\[
0 + 1 = 1
\]
4. 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 \(\boxed{(1, -15)}\).
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.
1. Calculate the distance from \((-1, -17)\) to the vertex \((0, -16)\):
\[
\text{Distance} = \sqrt{(0 - (-1))^2 + (-16 - (-17))^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{2}
\]
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:
3. The \(x\)-coordinate of the symmetrical point:
\[
0 + 1 = 1
\]
4. 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 \(\boxed{(1, -15)}\).