# 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

Find the equation of the tangent plane to the graph of $f\left(x,y\right)=8{x}^{2}-2x{y}^{2}$ 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
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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