Trig The point (-9,5) is on the graph of h(x). What other point must be a part of h(x) for the function to be

Reflection symmetric over x = -1?

I got this wrong on the homework. Could someone please help me out with this? I really don't understand... appreciate some help here

You want your function symmetric with respect to the vertical line x = -1, this means if (x1, y) belongs to the graph then so does (x2, y) where (x2, y) results from reflecting (x1, y) about x = -1.
Therefore x = -1 is the midpoint of x1 and x2 , i.e. -1 = (x1 + x2) / 2, so x2 = - 2 - x1

We the point (-9, 5) so x1 = - 9, Therefore x2 = - 2 - (-9) = - 2 + 9 = 7, so the other point is (7, 5).

(-9, 5) is 8 units to the left of x = -1
(7, 5) is 8 units to the right of x =-1