What are the coordinates of the vertex for f(x) =3x^2+6x+3

The x-coordinate of the vertex can be found by -b/2a where the equation is ax^2 + bx + c

3x^2 + 6x + 3

b is 6, and a is 3

-b/2a = -6/2(3) = -6/6 = -1

The x-coordinate of the vertex is -1. To find the y-coordinate plug the x-coordinate into the equation and solve for y.

f(-1) = 3(-1)^2 + 6(-1) + 3
= 3(1) - 6 + 3
= 3 - 6 + 3
= 0

The y-coordinate is 0

The vertex is (-1, 0)

40 & 50 in the right corner are the complementary angles as when placed in a sum (added together) they make 90°