The line of symmetry for the quadratic equation y=ax^2-2x-3 is x=1. what is the value of "a"?

a. -1

b. 1

c. 2

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