What is the equation of a vertical line passing through (-5, -2)?

Mathematics

1 answer:

elena-s [515]3 years ago

8 0

Answer:

x=-5

Step-by-step explanation:

if point the line is passing through is -5, -2 and the line is vertical than x would have to stay the same no matter where the y is because it's vertical.

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Darina [25.2K]

Answer: B

0.07274

Step-by-step explanation:

Length

72.74

Millimeter

0.07274

Meter

Formula

divide the length value by 1000

5 0

2 years ago

Read 2 more answers

(a)5%)Let fx, y) = x^4 + y^4 - 4xy + 1. and classify each critical point Find all critical points of fx,y) as a local minimum, l

lara31 [8.8K]

$f(x,y)=x^4+y^4-4xy+1$

has critical points wherever the partial derivatives vanish:

$f_x=4x^3-4y=0\implies x^3=y$

$f_y=4y^3-4x=0\implies y^3=x$

Then

$x^3=y\implies x^9=x\implies x(x^8-1)=0\implies x=0\text{ or }x=\pm1$

If $x=0$ , then $y=0$ ; critical point at (0, 0)
If $x=1$ , then $y=1$ ; critical point at (1, 1)
If $x=-1$ , then $y=-1$ ; critical point at (-1, -1)

$f(x,y)$ has Hessian matrix

$H(x,y)=\begin{bmatrix}12x^2&-4\\-4&12y^2\end{bmatrix}$

with determinant

$\det H(x,y)=144x^2y^2-16$

At (0, 0), the Hessian determinant is -16, which indicates a saddle point.
At (1, 1), the determinant is 128, and $f_{xx}(1,1)=12$ , which indicates a local minimum.
At (-1, -1), the determinant is again 128, and $f_{xx}(-1,-1)=12$ , which indicates another local minimum.