# Math Help - Equation of tangent plane

1. ## Equation of tangent plane

Find the equation of the plane tangent to the graph of the function

z(x,y) = 7(x^2) − 6(y^2) − 7x + 4y + 7 at the point (x,y)=(9,9)

the z(x,y) is throwing me off since it isnt f(x,y). Is there any difference?

I've done this problem about a million time, and am still not getting the right answer.

my solution:
$z_x = 14x - 7$
$z_x (9,9) = 14(9) - 9 = 119$

$z_y = -12y + 4$
$z_y = -12(9) + 4 = -104$

there is no value for z though which confuses me.

I know that an equation of the tangent plane to the surface z = f(x,y) at the point $P (x_0, y_0, z_0)$ is
$z-z_0=f_x (x_0 ,y_0 )(x-x_0 ) + f_y (x_0, y_0)(y-y_0)$

but when I plug in everything into this equation I get $z = 119(x-9) - 104(y-9)$ which is wrong. Any help would be appreciated.

2. Originally Posted by rain21
Find the equation of the plane tangent to the graph of the function

z(x,y) = 7(x^2) − 6(y^2) − 7x + 4y + 7 at the point (x,y)=(9,9)

the z(x,y) is throwing me off since it isnt f(x,y). Is there any difference?

I've done this problem about a million time, and am still not getting the right answer.

my solution:
$z_x = 14x - 7$
$z_x (9,9) = 14(9) - 9 = 119$

$z_y = -12y + 4$
$z_y = -12(9) + 4 = -104$

there is no value for z though which confuses me.

I know that an equation of the tangent plane to the surface z = f(x,y) at the point $P (x_0, y_0, z_0)$ is
$z-z_0=f_x (x_0 ,y_0 )(x-x_0 ) + f_y (x_0, y_0)(y-y_0)$

but when I plug in everything into this equation I get $z = 119(x-9) - 104(y-9)$ which is wrong. Any help would be appreciated.
$z(9,9)$ is the value of z.

3. Originally Posted by General
$z(9,9)$ is the value of z.
nvm solved it