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"?

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 
$\frac{x_1+x_2}{2}=2$
$x_1+x_2=4$

Remark: in the x-axis, if c is the midpoint of a and b, then $c= \frac{a+b}{2}$


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