First equation
[tex]f(x)=-(x-1)^2+4[/tex]The general form of a parabola is given as
[tex]y=a(x-h)+k[/tex]Comparing the general form with the first equation
[tex]a=-1,h=1,k=4[/tex]The vertex of a parabola is given as
[tex]Vertex=(h,k)[/tex]Therefore, the vertex of the first parabola is (1, 4)
The focus of a parabola is given as
[tex]Focus=(h,k+\frac{1}{4a})[/tex]Substitute the values of h, k and a into the formula for focus
This gives
[tex]\begin{gathered} \text{Focus}=(1,4+\frac{1}{4(-1)}) \\ \text{Focus}=(1,4+\frac{1}{-4}) \\ \text{Focus}=(1,\frac{15}{4}) \\ \text{Focus}=(1,3\frac{3}{4}) \end{gathered}[/tex]Therefore, the focus of the first equation is
[tex]\text{Focus=}(1,3\frac{3}{4})[/tex]The directrix of the parabola will have the equation
[tex]y=k-\frac{1}{4a}[/tex]Substituting values gives
[tex]\begin{gathered} y=4-\frac{1}{4(-1)} \\ y=4-\frac{1}{-4} \\ y=4+\frac{1}{4} \\ y=\frac{17}{4} \\ y=4\frac{1}{4} \end{gathered}[/tex]The directrix of the first equation has the eqaution
[tex]y=4\frac{1}{4}[/tex]Therefore, all the attribute of the first equation are
[tex]\begin{gathered} \text{vertex}=(1,4) \\ \text{focus}=(1,3\frac{3}{4}) \\ \text{directrix}\Rightarrow y=4\frac{1}{4} \end{gathered}[/tex]For the second equation
[tex]y=2(x+1)^2+4[/tex]Comparing with the general form of a parabola
It follows
[tex]a=2,h=-1,k=4[/tex]Following the calculations as above
Then
All the attributes of the second equations are
[tex]\begin{gathered} \text{vertex}=(-1,4) \\ \text{focus}=(-1,4\frac{1}{8}) \\ \text{directrix}\Rightarrow y=3\frac{7}{8} \end{gathered}[/tex]