If the image of the function y = a ^ (2x + b) + 1 (a > 0, a ≠ 1, B ∈ R) passes the fixed point (1,2), then what is B?

If the image of the function y = a ^ (2x + b) + 1 (a > 0, a ≠ 1, B ∈ R) passes the fixed point (1,2), then what is B?

Substituting (1,2) into the analytic expression of the function, we get
2=a^(2+b)+1
That is, a ^ (2 + b) = 1
If a ＞ 0 and a ≠ 1, then B + 2 = 0
The solution is b = - 2

If the image of the function f (x) = (2x + 1) \ (x + a) is symmetric with respect to the line y = x, then the real number a is --; if the image of the function f (x) = (2x + 1) \ (x + a) is mutually symmetric with respect to the line y = x, then the real number a is ---

F (x) = (2x + 1) / (x + a) is: x = (1-A * f (x)) / (f (x) - 2), so the inverse function of F (x) = (2x + 1) / (x + a) is: G (x) = (1-A * x) / (X-2) because it is symmetric with y = X. if f (x) = g (x), we can get: (2x + 1) / (x + a) = (1-A * x) / (X-2) - ax ^ 2 + (- A ^ 2 + 1) x + a = 2x ^ 2-3x-2 for any x equality

1 in the following points, the point not on the image of function y = - 2x + 3 is () a (- 5,13) B (0.5,2) C (3,0) d (1,1)

c