To determine which points lie on the line with a slope of [tex]\(\frac{2}{3}\)[/tex] passing through the point [tex]\((0, 2)\)[/tex], we can use the point-slope form of the equation of the line. The point-slope form of a line's equation is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\(m\)[/tex] is the slope. Here, [tex]\((x_1, y_1) = (0, 2)\)[/tex] and [tex]\(m = \frac{2}{3}\)[/tex]. Plugging these values into the point-slope equation, we get:
[tex]\[ y - 2 = \frac{2}{3}(x - 0) \][/tex]
[tex]\[ y - 2 = \frac{2}{3}x \][/tex]
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
Now we will check each of the given points to see if they satisfy the equation [tex]\(y = \frac{2}{3}x + 2\)[/tex].
1. For the point [tex]\((-3, 0)\)[/tex]:
[tex]\[ y = \frac{2}{3}(-3) + 2 \][/tex]
[tex]\[ y = -2 + 2 \][/tex]
[tex]\[ y = 0 \][/tex]
Point [tex]\((-3, 0)\)[/tex] satisfies the equation.
2. For the point [tex]\((-2, -3)\)[/tex]:
[tex]\[ y = \frac{2}{3}(-2) + 2 \][/tex]
[tex]\[ y = -\frac{4}{3} + 2 \][/tex]
[tex]\[ y = -\frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{2}{3} \][/tex]
Point [tex]\((-2, -3)\)[/tex] does not satisfy the equation.
3. For the point [tex]\((2, 5)\)[/tex]:
[tex]\[ y = \frac{2}{3}(2) + 2 \][/tex]
[tex]\[ y = \frac{4}{3} + 2 \][/tex]
[tex]\[ y = \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{10}{3} \][/tex]
Point [tex]\((2, 5)\)[/tex] does not satisfy the equation.
4. For the point [tex]\((3, 4)\)[/tex]:
[tex]\[ y = \frac{2}{3}(3) + 2 \][/tex]
[tex]\[ y = 2 + 2 \][/tex]
[tex]\[ y = 4 \][/tex]
Point [tex]\((3, 4)\)[/tex] satisfies the equation.
5. For the point [tex]\((6, 6)\)[/tex]:
[tex]\[ y = \frac{2}{3}(6) + 2 \][/tex]
[tex]\[ y = 4 + 2 \][/tex]
[tex]\[ y = 6 \][/tex]
Point [tex]\((6, 6)\)[/tex] satisfies the equation.
So, the three points that Vera could use to graph the line are:
[tex]\[
\boxed{(-3, 0), (3, 4), (6, 6)}
\][/tex]
