Answer
(a) $z=8x+4y+1$
(b) $\frac{x-1}{8}=\frac{y+2}{4}=\frac{z-1}{-1}$
Work Step by Step
$z=3x^2-y^2+2x$ and $(1,-2,1)$
(a) Equation for a tangent plane is given as:
$z-1=8(x-1)+4(y+2)$
$z-1=8x+4y$
$z=8x+4y+1$
(b) $8x+4y-z=-1$
Equation of normal line is:
$\frac{x-1}{8}=\frac{y+2}{4}=\frac{z-1}{-1}$
