Question

PLEASE HELP!!!!!! The line of symmetry for the quadratic equation y = ax 2 - 8x - 3 is x = 2. What is the value of "a"?

A) -2
B) -1
C) 2

Answer 1

`y= ax^(2) -8x-3`

1.

the line of symmetry is x=2, means that the x coordinate of the vertex is x=2.

the point x=2 is the midpoint of the roots
`x_1` and
`x_2`.

so

`(x_1+x_2)/(2)=2`

`x_1+x_2=4`

Remark: in the x-axis, if c is the midpoint of a and b, then
`c= (a+b)/(2)`


2.
since
`x_1` and
`x_2` are roots


`a(x_1)^(2) -8(x_1)-3=0` and
`a(x_2)^(2) -8(x_2)-3=0`

3.
equalizing:


`a(x_1)^(2) -8(x_1)-3=a(x_2)^(2) -8(x_2)-3`


`a(x_1)^(2) -8(x_1)=a(x_2)^(2) -8(x_2)`


`a(x_1)^(2)-a(x_2)^(2) =8(x_1) -8(x_2)`

in the left side factorize a, in the left side factorize 8:


`a[(x_1)^(2)-(x_2)^(2)] =8(x_1 -x_2)`

in the right side use the difference of squares formula:


`a(x_1 -x_2)(x_1 +x_2) =8(x_1 -x_2)`

simplify by
`(x_1 -x_2)`


`a(x_1 +x_2) =8`

substitute
`(x_1 +x_2)` with 4:


`a*4 =8`

a=2


Answer: C)2