Solution:
Given:
[tex]y=-\left(x+4\right)^2+2[/tex]
To determine the maximum point of the function,
step 1: Take the first derivative of the function.
[tex]y^{\prime}=-2x-8[/tex]
Step 2: Find the critical point of the function.
At the critical point, y' equals zero.
Thus,
[tex]\begin{gathered} -2x-8=0 \\ add\text{ 8 to both sides,} \\ -2x-8+8=0+8 \\ \Rightarrow-2x=8 \\ divide\text{ both sides by -2} \\ \frac{-2x}{-2}=\frac{8}{-2} \\ \Rightarrow x=-4 \end{gathered}[/tex]
Step 3: Take the second derivative of the function.
Thus, we have
[tex]y^{\prime}^{\prime}=-2[/tex]
Since y'' is less than zero, it implies that there's a maximum value/point.
step 4:: Evaluate the maximum value of y.
This gives:
[tex]\begin{gathered} y=-(x+2)^2+2 \\ where\text{ x=-4} \\ y=-(-4+4)^2+2=2 \\ \Rightarrow y=2 \end{gathered}[/tex]
Hence, the maximum point of the function is
[tex](-4,2)[/tex]
The correct option is
