You find the y-coordinate of the vertex by substituting -3 in for x and evaluating. So (-3)^2 + 6(-3) + 2 = -7, so the vertex is (-3, -7). Does that help?
Plug in what you got when you did x= -b/2a into the original equation. So he got -3 so he put it into y=x^2+6x+2 and got -7 therefore his vertex was (-3, -7) (I know this is four years later sorry heh. Hope this still helps lmao)
You find the y-coordinate of the vertex by substituting -3 in for x and evaluating. So (-3)^2 + 6(-3) + 2 = 9 - 18 + 2 = -7, so the vertex is (-3, -7). Does that help?
+Micheal F You can do it a variety of ways. One way is simply to factor out the GCF, which is x. So you get y = x(-x + 6). That means either x = 0 or x = 6. That gives you your x-intercepts of (0, 0) and (6, 0). Your vertex is then in the middle, at x = 3. Substitute 3 in for x and you get y = 9, so (3, 9) would be your vertex. You can now sketch the graph.
Suzana Bacaleinick Are you asking about the (-3, 12) in the table of values in GP1? If so, you fill in -3 for x, so you get 5(-3)^2 + 10(-3) - 3. That simplifies to 5 * 9 + -30 - 3, which simplifies to 45 + -30 -3, which simplies to 12. So the ordered pair is (-3, 12).