Explanation:
The question states fine the equation of the normal line of the function
[tex]f(x)=\sqrt{x+2},at\text{ }x=-1[/tex][tex]f(-1)=y_0=\sqrt{-1+2}=\sqrt{1}=1[/tex]
The slope os the the normal line at x= x0 is the negative reciprocal of the derivative of the function, evaluated at x=x0
[tex]M(x_0)=-\frac{1}{f^{\prime}(x_0)}[/tex]
Find the derivative of f(x)
[tex]\begin{gathered} f(x)=\sqrt{x+2} \\ f^{\prime}(x)=(x+2)^{\frac{1}{2}} \\ f^{\prime}(x)=\frac{1}{2\sqrt{x+2}} \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} M(x_{0})=-\frac{1}{f^{\prime}(x_{0})} \\ M(x_0)=-2\sqrt{x_0+2} \end{gathered}[/tex]
Next, find the slope at the given point.
[tex]\begin{gathered} m=M\left(−1\right)=−2 \\ M(x_0)=-2\sqrt{x_0+2} \\ M(-1)=-2\sqrt{-1+2} \\ M(-1)=-2\sqrt{1} \\ M(-1)=-2 \end{gathered}[/tex]
Finally, the equation of the normal line is
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y-(1)=-2(x-(-1) \\ y-1=-2(x+1) \\ y-1=-2x-2 \\ y=-2x-2+1 \\ y=-2x-1 \end{gathered}[/tex]
Hence,
The final answer is
[tex]y=-2x-1[/tex]
OPTION A is the correct answer
