the trick here is to substitute points given
[tex]f(x) = a \times {b}^{x} [/tex]
substituting point A x-value =b
[tex]f(b) = a \times {b}^{b} [/tex]
turns out false since (b,a*b^b) !=(b,0)
substituting point B x-value=0
[tex]f(0) = a \times {b}^{0} = a[/tex]
turns out true since (0,a)=(0,a)
we already got f(x) at x =0 which is not the point (0,0)
so c is false
substituting D x-value=a
[tex]f(a) = a \times {b}^{a} [/tex]
hence B is the correct answers
