Question

(kx^2)+m, x<-2
g(x)=e^(ln(2x+3))-2<x<3
kx-m, x>3

Find values of k and m

Answer 1

I assume you want k and m so that g(x) is continuous?

Just to make things easier, recall that

e^(ln u) = u
so, e^(ln(2x+3)) = 2x+3

If so, then you want the endpoints to match:

(k(-2)^2)+m = 2(-2)+3
k(3)-m = 2(3)+3

or,

4k+m = -1
3k-m = 9

I guess you can handle that, yeah?
As always, check my arithmetic.