since the vertex occurs at x = -b/2a,
y = a(-b/2a)^2 + b(-b/2a) + c
= b^2/4a - b^2/2a + b
= -b^2/4a + c
For the graph y= ax^2 + bx + c, show that the y-coordinate of the vertex is
-b^2/4a + c.
2 answers
thanks