The given function
f(x)=\frac{2-x}{\sqrt{2x+2}}f(x)=2−x√2x+2
setting x=2x=2 in the given function, we get yy-coordinate of point
y=f(2)y=f(2)
=\frac{2-2}{\sqrt{2\cdot 2+2}}=0=2−2√2⋅2+2=0
Now, differentiating above equation w.r.t. xx we get slope dy/dxdydx of tangent as follows
dy/dx=d/dxf(x)dydx=ddxf(x)
=\frac{\sqrt{2x+2}(-1)-(2-x)\frac{1}{\sqrt{2x+2}}}{(\sqrt{2x+2})^2}=√2x+2(−1)−(2−x)1√2x+2(√2x+2)2
=\frac{-(2x+2)-(2-x)}{(2x+2)^{3/2}}=−(2x+2)−(2−x)(2x+2)32
=\frac{-x-4}{(2x+2)^{3/2}}=−x−4(2x+2)32
setting x=2x=2 in above equation, we get slope of tangent at (2, 0)(2,0)
=\frac{-2-4}{(2\cdot 2+2)^{3/2}}=−2−4(2⋅2+2)32
=-1/\sqrt6=−1√6
hence the slope mm of normal at the same point
=-1/(-1/\sqrt6)=\sqrt6=−1−1√6=√6
hence the equation of normal at (x_1, y_1)\equiv (2, 0) (x1,y1)≡(2,0) & having slope m=\sqrt6m=√6 is given by following formula
y-y_1=m(x-x_1)y−y1=m(x−x1)
y-0=\sqrt6(x-2)y−0=√6(x−2)
y=\sqrt6(x-2)y=√6(x−2)
