4. A triangle has vertices at A(-3,-1),B(-6,-5), C(-1,- 4). Which transformation would produce an image with
vertices A'(3,-1),B'(6,-5), C'(1, - 4)?
A. a reflection over the x-axis
B. a reflection over the y-axis
C. a rotation 90° clockwise
D. a rotation 90° counterclockwise

Idon't see any graph, however the graph of the function : f(x) = x²-2x+3 is a parabola: 1) open upward (since a> 0)2) it intersects y-axis in 3 (y=3)3) its axis of symmetry is x = +1 ( -b/2a)4) it's minimum is y = -2, so its vertex is (1,-2)5) it doesn't intersect the x-axis since discriminant < 0