Let's write the equation for every function:
Since f(x) passes through (0,-1) and (5,14), the equation will be given by:
[tex]\begin{gathered} (x1,y1)=(0,-1) \\ (x2,y2)=(5,14) \\ m=\frac{14-(-1)}{5-0}=\frac{15}{5} \\ m=3 \end{gathered}[/tex]Using the point-slope equation:
[tex]\begin{gathered} y-y1=m(x-x1) \\ y-(-1)=3(x-0) \\ y+1=3x \\ y=3x-1 \\ f(x)=3x-1 \end{gathered}[/tex]Using the same procedure for g(x):
[tex]\begin{gathered} (x1,y1)=(-6,-1) \\ (x2,y2)=(-1,14) \\ m=\frac{14-(-1)}{-1-(-6)}=\frac{15}{5}^{} \\ m=3 \\ \end{gathered}[/tex][tex]\begin{gathered} y-y1=m(x-x1) \\ y-(-1)=3(x-(-6)) \\ y+1=3x+18 \\ y=3x+17 \\ g(x)=3x+17 \end{gathered}[/tex]Now we have our functions:
[tex]\begin{gathered} f(x)=3x-1 \\ g(x)=3x+17 \end{gathered}[/tex]Part A:
We can translated f(x) k units up in order to get g(x) ( A vertical translation)
[tex]f(x)=3x-1+k[/tex]Or, we can translated f(x) k units to the left in order to get g(x) ( A horizontal translation)
[tex]f(x)=3(x+k)-1[/tex]We can see a graph of the functions:
Where the red function is f(x) and the blue function is g(x).
Part B:
Since both functions must be equal:
[tex]\begin{gathered} f(x)=g(x) \\ so\colon \\ 3x-1+k=3x+17 \end{gathered}[/tex]Solve for k:
[tex]\begin{gathered} k=3x+17-3x+1 \\ k=18 \end{gathered}[/tex]-----------------------
For the other case, let's use the same procedure:
[tex]\begin{gathered} 3(x+k)-1=3x+17 \\ 3x+3k-1=3x+17 \\ 3k=3x+17-3x+1 \\ k=\frac{18}{3} \\ k=6 \end{gathered}[/tex]Part C:
For the vertical translation:
[tex]3x-1+18[/tex]For the horizontal translation:
[tex]3(x+6)-1[/tex]
