The critical points of <em>h(x,y)</em> occur wherever its partial derivatives
and
vanish simultaneously. We have

Substitute <em>y</em> in the second equation and solve for <em>x</em>, then for <em>y</em> :

This is to say there are two critical points,

To classify these critical points, we carry out the second partial derivative test. <em>h(x,y)</em> has Hessian

whose determinant is
. Now,
• if the Hessian determinant is negative at a given critical point, then you have a saddle point
• if both the determinant and
are positive at the point, then it's a local minimum
• if the determinant is positive and
is negative, then it's a local maximum
• otherwise the test fails
We have

while

So, we end up with

Nope. its y=3x-1. the y intercept is -1, so we put that at the end. And if we take 2 points on the line and use y/x, we get 3/1 or 3, and this is our slope. Also the line is going up so its not negative
Answer:
14, 7, 28
Step-by-step explanation:
The correct answer is 8 because 8+4 equals 12
Step-by-step explanation:
L ----------V-------------U
LV = VU
UV = 6-3x
LU = 14x + 2 = 2UV
14X +2 = 12 - 6X
20X = 10
X = 0.5
UV = 6 - 3(0.5)
= 4.5
LU = 9