Find the equation of the tangent plane to the graph of f(x,y)=8x^{2}-2xy^{2} at the point (5,4). A)z=23x-15y+42 B)z=48x-80y+120 C)0=48x-80y+120 D)0=23x-15y+42

Step 1
We first find the partial derivatives
${\displaystyle f=8x^2-2x{y}^2}$
${\displaystyle f_x=16x-2y^2}$
${\displaystyle f_y=-4xy}$
At (5,4)
${\displaystyle f_x=16x-2y^2=16\left(5\right)-2{\left(4\right)}^2=48}$
${\displaystyle f_y=-4xy=-4\left(5\right)\left(4\right)=-80}$
So the direction vector of the normal is = (48,-80)
Step 2
So the equation of the tangent plane is:
${\displaystyle z=\left(8{\left(5\right)}^{2}02\left(5\right){\left(4\right)}^2\right)+48(x-5)+(-80)(y-4)}$
z=40+48x-240-80y+320
z=48x-80y+120
Answer: B