This is a post from CarissaF. Originally Posted by CarissaF if f(x,y)=4x^2+5xy+2y^2 Then an implicit equation for the tangent plane to the graph at point (2,-3) is: I got an answer of x-2y=0 but it is wrong... I can't figure out why. This is my work. 0=f(2,-3)+f_{x}(2,-3)(x-2)+f_{y}(2,-3)(y--3) f(2,-3)=4 f_{x}(2,-3)=1 f_{y}(2,-3)=-2 0=4+1(x-2)-2(y+3) 0=4+x-2-2y-6 4=x-2y Can anyone spot my mistake?
Follow Math Help Forum on Facebook and Google+
Originally Posted by topsquark This is a post from CarissaF. Consider this $0=f_x(2,-3)(x-2)+f_y(2,-3)(y+3)$