Let's determine which points are on the graph of the equation [tex]\( y = \left(\frac{1}{2}\right)^x \)[/tex].
1. Point (2, [tex]\(\frac{1}{4}\)[/tex]):
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \][/tex]
- The y-value matches [tex]\(\frac{1}{4}\)[/tex] given in the point.
- Therefore, (2, [tex]\(\frac{1}{4}\)[/tex]) is a point on the graph.
2. Point [tex]\((0, \frac{1}{2})\)[/tex]:
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \left( \frac{1}{2} \right)^0 = 1 \][/tex]
- The y-value is 1, but the point has a y-value of [tex]\(\frac{1}{2}\)[/tex].
- Therefore, [tex]\((0, \frac{1}{2})\)[/tex] is not a point on the graph.
3. Point [tex]\((2, 1)\)[/tex]:
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \][/tex]
- The y-value is [tex]\(\frac{1}{4}\)[/tex], but the point has a y-value of 1.
- Therefore, [tex]\((2, 1)\)[/tex] is not a point on the graph.
4. Point [tex]\((0, 0)\)[/tex]:
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \left( \frac{1}{2} \right)^0 = 1 \][/tex]
- The y-value is 1, but the point has a y-value of 0.
- Therefore, [tex]\((0, 0)\)[/tex] is not a point on the graph.
So the point that is on the graph of [tex]\( y = \left(\frac{1}{2}\right)^x \)[/tex] is:
[tex]\[
(2, \frac{1}{4})
\][/tex]
