Relation B:
\(D = \left \{ (-2,y) \, | \, \answer {-4} \leq y < \answer {3} \right \}\)
\(F = \left \{ (t,s) \, | \, s \geq \answer {0} \right \}\)
Relation H:
\(H = \left \{ \left (v,w\right ) \, | \, \answer {-3} < v \leq \answer {2} \right \}\)
Relation I:
\(I = \left \{ (u,v) \, | \, u \geq \answer {0}, \! v \geq \answer {0}\right \}\)
For the relations in Exercises cannotgraphfirst and cannotgraphsecond, give two examples of points which belong to the relation and two points which do not belong to the relation.
For each equation given in Exercises oldonethreefirst - oldonethreelast:
\(x\)-intercepts: \((-6, 0)\), \((2,0)\)
\(y\)-intercepts: \(\left (0, \pm 2\sqrt {3}\right )\)
\(\begin{array}{|r||c|c|} \hline x & y & (x,y) \\ \hline -6 & 0 & (-6,0) \\ \hline -4 & \pm 2 \sqrt {3} & \left (-4,\pm 2 \sqrt {3}\right ) \\ \hline -2 & \pm 4 & (-2, \pm 4) \\ \hline 0 & \pm 2 \sqrt {3} & \left (0,\pm 2 \sqrt {3}\right ) \\ \hline 2 & 0 & (2, 0) \\ \hline \end{array} \)
The graph is symmetric about the \(x\)-axis
The graph is not symmetric about the \(y\)-axis: \((-6, 0)\) is on the graph but \((6, 0)\) is not.
The graph is not symmetric about the origin: \((-6, 0)\) is on the graph but \((6, 0)\) is not.
The equation does not describe \(y\) as a function of \(x\).
The graph of the equation is the graphs of \(f_{1}(x) = \sqrt {16-(x+2)^2}\) together with \(f_{2}(x) = -\sqrt {16-(x+2)^2}\).
\(x\)-intercepts: \((-1, 0), (1, 0)\)
The graph has no \(y\)-intercepts
\(\begin{array}{|r||c|c|} \hline x & y & (x,y) \\ \hline -3 & \pm \sqrt {8} & (-3, \pm \sqrt {8}) \\ \hline -2 & \pm \sqrt {3} & (-2, \pm \sqrt {3}) \\ \hline -1 & 0 & (-1, 0) \\ \hline 1 & 0 & (1, 0) \\ \hline 2 & \pm \sqrt {3} & (2, \pm \sqrt {3}) \\ \hline 3 & \pm \sqrt {8} & (3, \pm \sqrt {8}) \\ \hline \end{array} \)
The graph is symmetric about the \(x\)-axis.
The graph is symmetric about the \(y\)-axis.
The graph is symmetric about the origin.
The equation does not describe \(y\) as a function of \(x\).
The graph of the equation is the graphs of \(f_{1}(x) = \sqrt {x^2-1}\) together with \(f_{2}(x) = -\sqrt {x^2-1}\).
Re-write as: \(y = \pm \dfrac {\sqrt {9x^2+36}}{2}\).
The graph has no \(x\)-intercepts
\(y\)-intercepts: \((0, \pm 3)\)
\(\begin{array}{|r||c|c|} \hline x & y & (x,y) \\ \hline -4 & \pm 3 \sqrt {5} & \left (-4,\pm 3 \sqrt {5}\right ) \\ \hline -2 & \pm 3 \sqrt {2} & \left (-2,\pm 3 \sqrt {2}\right ) \\ \hline 0 & \pm 3 & (0, \pm 3) \\ \hline 2 & \pm 3 \sqrt {2} & \left (2,\pm 3 \sqrt {2}\right ) \\ \hline 4 & \pm 3 \sqrt {5} & \left (4,\pm 3 \sqrt {5}\right ) \\ \hline \end{array}\)
The graph is symmetric about the \(x\)-axis.
The graph is symmetric about the \(y\)-axis.
The graph is symmetric about the origin.
The equation does not describe \(y\) as a function of \(x\).
The graph of the equation is the graphs of \(f_{1}(x) = \dfrac {\sqrt {9x^2+36}}{2}\) together with \(f_{2}(x) = - \dfrac {\sqrt {9x^2+36}}{2}\).
The graph has no \(x\)-intercepts
The graph has no \(y\)-intercepts
\(\begin{array}{|r||c|c|} \hline x & y & (x,y) \\ \hline -2 & \frac {1}{2} & (-2, \frac {1}{2}) \\ \hline -1 & 4 & (-1, 4) \\ \hline -\frac {1}{2} & 32 & (-\frac {1}{2}, 32) \\ \hline \frac {1}{2} & -32 & (\frac {1}{2}, -32)\\ \hline 1 & -4 & (1, -4) \\ \hline 2 & -\frac {1}{2} & (2, -\frac {1}{2}) \\ \hline \end{array} \)
The graph is not symmetric about the \(x\)-axis: \((1, -4)\) is on the graph but \((1, 4)\) is not.
The graph is not symmetric about the \(y\)-axis: \((1, -4)\) is on the graph but \((-1, -4)\) is not.
The graph is symmetric about the origin.
The equation does describe \(y\) as a function of \(x\), namely \(y=f(x) = - 4x^{-3}\).
\(v\)-intercept: \((4,0)\)
\(w\)-intercepts: \(\left (0, \pm 2 \right )\)
\(\begin{array}{|r||c|c|} \hline v & w & (v,w) \\ \hline -5 & \pm 3 & (-5,\pm 3) \\ \hline -2 & \pm \sqrt {6} & \left (-2,\pm \sqrt {6}\right ) \\ \hline 0 & \pm 2 & (0, \pm 2) \\ \hline 2 & \pm \sqrt {2} & \left (1,\pm \sqrt {3}\right ) \\ \hline 4 & 0 & (4, 0) \\ \hline \end{array} \)
The graph is symmetric about the \(v\)-axis
The graph is not symmetric about the \(w\)-axis: \((4, 0)\) is on the graph but \((-4, 0)\) is not.
The graph is not symmetric about the origin: \((4, 0)\) is on the graph but \((-4, 0)\) is not.
The equation does not describe \(w\) as a function of \(v\).
The graph of the equation is the graphs of \(f_{1}(v) = \sqrt {4-v}\) together with \(f_{2}(v) = -\sqrt {4-v}\).
\(v\)-intercept: \((2,0)\)
\(w\)-intercept: \((0,2)\)
\(\begin{array}{|r||c|c|} \hline v & w & (v,w) \\ \hline -3 & \sqrt [3]{35} & (-3, \sqrt [3]{35}) \\ \hline -1 & \sqrt [3]{9} & (-1, \sqrt [3]{9}) \\ \hline 0 & 2 & (0, 2) \\ \hline 1 & \sqrt [3]{7} & (1, \sqrt [3]{7}) \\ \hline 2 & 0 & (2, 0) \\ \hline 3 & -\sqrt [3]{19} & (3, -\sqrt [3]{19}) \\ \hline \end{array} \)
The graph is not symmetric about the \(v\)-axis: \((0,2)\) is on the graph but \((0,-2)\) is not.
The graph is not symmetric about the \(w\)-axis: \((2, 0)\) is on the graph but \((-2, 0)\) is not.
The graph is not symmetric about the origin: \((0, 2)\) is on the graph but \((0, -2)\) is not.
The equation does describe \(w\) as a function of \(v\), namely \(w=f(v) = \sqrt [3]{8-v^3}\).
The graph has no \(v\)-intercepts.
The graph has no \(w\)-intercepts.
\(\begin{array}{|r||c|c|} \hline v & w & (v,w) \\ \hline -8 & \frac {1}{2} & \left (-8, \frac {1}{2} \right ) \\ \hline -1 & 2 & \left (-1, 2 \right ) \\ \hline -\frac {1}{8} & 8 & \left (-\frac {1}{8}, 8 \right ) \\ \hline \frac {1}{8} & 8 & \left (\frac {1}{8}, 8 \right ) \\ \hline 1 & 2 & \left (1, 2 \right ) \\ \hline 8 & \frac {1}{2} & \left (8, \frac {1}{2} \right ) \\ \hline \end{array} \)
The graph is not symmetric about the \(v\)-axis: \((-1,2)\) is on the graph but \((-1,-2)\) is not.
The graph is symmetric about the \(w\)-axis.
The graph is not symmetric about the origin: \((-1,2)\) is on the graph but \((-1,-2)\) is not.
The equation does describe \(w\) as a function of \(v\), namely \(w=f(v) = 2 v^{-\frac {2}{3}}\).
HINT: \(v^4 - 2v^2 w + w^2 = \left (v^2 - w \right )^2\) …
Extracting square roots gives: \(w = v^2 + 4\) and \(w = v^2-4\)
\(v\)-intercepts: \((-2,0), (2,0)\)
\(w\)-intercepts: \((0,-4), (0,4)\)
\(\begin{array}{|r||c|c|} \hline v & w & (v,w) \\ \hline -2 & 8 & (-2,8) \\ \hline -2 & 0 & (-2,0) \\ \hline -1 & 5 & (-1, 5) \\ \hline -1 & -3 & (-1, -3) \\ \hline 0 & \pm 4 & (0, \pm 4) \\ \hline 1 & 5 & (1, 5) \\ \hline 1 & -3 & (1, -3) \\ \hline 2 & 8 & (2,8) \\ \hline 2 & 0 & (2,0) \\ \hline \end{array}\)
The graph is not symmetric about the \(v\)-axis: \((1,5)\) is on the graph but \((1,-5)\) is not.
The graph is symmetric about the \(w\)-axis.
The graph is not symmetric about the origin: \((1,5)\) is on the graph but \((-1, -5)\) is not.
The equation does not describe \(w\) as a function of \(v\).
The graph of the equation is the graphs of \(f_{1}(v) = v^2+4\) together with \(f_{2}(v) = v^2-4\).