The axis of symmetry of a quadratic is x= -b/(2a) given the form ax^2+bx+c.
In this case -b/(2a)=1 and b=2 so:
--2/(2a)=1
2/(2a)=1
1/a=1
a=1
The way to do it without remembering that the vertex of any parabola is at the point (-b/(2a), (4ac-b^2)/(4a)) is to actually find the vertex form of the parabola...which is of the form y=a(x-h)^2+k, where (h,k) is the vertex...
y=ax^2-2x-3
y+3=ax^2-2x
(y+3)/a=x^2-2x/a
(y+3)/a+1/a^2=x^2-2x/a+1/a^2
(y+3)/a+1/a^2=(x-1/a)^2
(y-3)+1/a=a(x-1/a)^2
y-3=a(x-1/a)^2+1/a
y=a(x-1/a)^2+1/a+3
So we know the minimum point and vertex occur when the squared term is minimized, so x=1/a is the axis of symmetry. We are told that this occurs when x=1, we know that x=x so we can then say:
1/a=1
a=1
