Given:
The table of values is
x y
-4 2
-3 5
-2 8
-1 11
To find:
The slope of the line that contains these points.
Solution:
From the given table consider any two point.
Let the line passes through the points (-4,2) and (-3,5). So, the equation of the line is
![y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%28x-x_1%29)
![y-2=\dfrac{5-2}{-3-(-4)}(x-(-4))](https://tex.z-dn.net/?f=y-2%3D%5Cdfrac%7B5-2%7D%7B-3-%28-4%29%7D%28x-%28-4%29%29)
![y-2=\dfrac{3}{-3+4}(x+4)](https://tex.z-dn.net/?f=y-2%3D%5Cdfrac%7B3%7D%7B-3%2B4%7D%28x%2B4%29)
![y-2=\dfrac{3}{1}(x+4)](https://tex.z-dn.net/?f=y-2%3D%5Cdfrac%7B3%7D%7B1%7D%28x%2B4%29)
Using distributive property, we get
![y-2=3x+12](https://tex.z-dn.net/?f=y-2%3D3x%2B12)
![y=3x+12+2](https://tex.z-dn.net/?f=y%3D3x%2B12%2B2)
![y=3x+14](https://tex.z-dn.net/?f=y%3D3x%2B14)
Therefore, the required equation of line is
.
I believe the answer would be D.
Answer: i took a screen shot hope it helps
Step-by-step explanation:
Answer:
See steps below
Step-by-step explanation:
The function
is a particular case of the general Rosenbrock’s Function.
a)
Since
for all the values of x and equals 0 when x=1 and
for all the values of y and equals 0 only when y=1, we conclude that (1,1) is a minimum.
On the other hand,
f(x,y)>0 for (x,y) ≠ (1,1) so (1,1) is a global minimum.
b)
To confirm that this function is not convex, we will be using the following characterization of convexity
“f is convex if, and only if, the Laplace operator of f
for every (x,y) in the domain of f”
Given that the domain of f is the whole plane XY, in order to prove that f is not convex, we must find a point (x,y) at where the Laplace operator is < 0.
The Laplace operator is given by
Let us compute the partial derivatives
and
we have then
if we take (x,y) = (0,1)
hence f is not convex.