Quadratic graph

Suppose the vertex of the quadratic graph y = ax^2 + bx + c is ( 2, -8 ) and it cuts the line y = k at ( -1, k). Suppose the line has 2 intersecting pts of the quadratic graph, find the x-coordinate of the other intersecting pt.
sub -1, k into y = ax^2 + bx + c
k=a-b+c
we know that the vertex of quadratic equation is
(-b/2a, (4ac-b^2)/4a)
so
-b/2a=2, b=-4a
(4ac-b^2)/4a=-8
4ac-16a^2=-32a
ac-4a^2=-8a
c=4a-8
sub into
k=a-b+c
k=a+4a+4a-8=9a-8
Now
y = ax^2 + bx + c=ax^2-4ax+4a-8
sub k=9a-8
ax^2-4ax+4a-8=9a-8
ax^2-4ax-5a=0
x^2-4a-5=0
(x+1)(x-5)=0
x=-1 or 5
so the x-coordinate of the other intersecting pt is 5

y = ax² + bx + c
= a[x² + b/a*x + (b/(2a))²] + c - b²/(4a)
= a[x + b/(2a)]² + [c - b²/(4a)]
Since [x + b/(2a)]² is always greater than or equal to zero,
y is the minimum if x + b/(2a) = 0, or x = -b/(2a)
Minimum value of y= c - b²/(4a)

So (-b/(2a), c - b²/(4a)) is the turning point of y = ax² + bx + c.
The point is (2, -8).
Therefore, b/(2a) = -2
b = -4a..........(1)
and c - b²/(4a) = -8
c - 16a²/(4a) = -8
c - 4a = -8
c = 4a - 8..........(2)

The equation becomes,
y = ax² - 4ax + 4a - 8

Let ß be the x-coordinate of the other intersecting point between y = ax² + bx + c and y = k.
-1, ß are the roots of the below simultaneous equations,
y = ax² + bx + c = ax² - 4ax + 4a - 8..........(3)
y = k..........(4)

This is equivalent to solving (3) = (4),
ax² - 4ax + 4a - 8 = k
ax² - 4ax + 4a - 8 - k = 0

Sum of roots = -1 + ß = -(-4a)/a = 4
ß = 4 + 1 = 5.

Hence, the x-coordinate of the other intersecting point is 5.