Question

Helppppppppppppppppppppplp

Answer 1

the trick here is to substitute points given
$f(x) = a \times {b}^{x}$
substituting point A x-value =b
$f(b) = a \times {b}^{b}$
turns out false since (b,a*b^b) !=(b,0)
substituting point B x-value=0
$f(0) = a \times {b}^{0} = a$
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
$f(a) = a \times {b}^{a}$

hence B is the correct answers