Notice that over the interval [-1 , 1.5], the function f seems to be equal to x^2.
Furthermore, over the interval [-3 , -1], the function f seems to be a straight line of slope equal to 2 and y-intercept equal to 3.
Therefore, we can write down:
[tex]f(x)=\mleft\{\begin{aligned}2x+3\text{ if }-3\leq x<-1 \\ x^2\text{ if }-1\leq x\leq1.5\end{aligned}\mright.[/tex]On the other hand, the function g seems to be a straight line of slope 4 and y-intercept equal to -6 from x=0 to x=2, and something similar to 2x^2 from X=2 to x=4.5, but with its vertex at x=3
[tex]g(x)=\mleft\{\begin{aligned}4x-6\text{ if }0\leq x<2 \\ 2(x-3)^2\text{ if }2\leq x\leq4.5\end{aligned}\mright.[/tex]Notice that the vertical lengths of g seem to be twice those of f, so out first guess may be to write:
[tex]g(x)=2\cdot f(x)[/tex]Additionally, the function seems to be displaced 3 units to the right, so:
[tex]g(x)=2\cdot f(x-3)[/tex]Observe that since the domain of f is equal to [-3 , 1.5], then x has to be in the interval [0 , 4.5] for f(x-3) to be well defined. Also, a change in the correspondence rule of f happens at x=-1, and for g it happens at x=2.
In terms of f, this should happen at x-3=-1, which is equivalent to x=2.
Finally, observe that:
[tex]\begin{gathered} 2\cdot f(x-3)=\mleft\{\begin{aligned}2(2(x-3)+3)\text{ if }-3\leq(x-3)<-1 \\ 2(x-3)^2\text{ if }-1\leq x-3\leq1.5\end{aligned}\mright. \\ =\mleft\{\begin{aligned}2(2x-6+3)\text{ if }-3+3\leq x<-1+3 \\ 2(x-3)^2\text{ if }-1+3\leq x<1.5+3\end{aligned}\mright. \\ =\mleft\{\begin{aligned}4x-6\text{ if }0\leq x<2 \\ 2(x-3)^2\text{ if }2\leq x<4.5\end{aligned}\mright. \\ =g(x) \end{gathered}[/tex]Therefore:
[tex]g(x)=2\cdot f(x-3)[/tex]