find the crical points of the given func- on. Use the Second Derivave Test to determine if each crit- ical point corresponds to a relave maximum, minimum, or saddle point.

- f(x,y) = x^2 +4
*x+y^2-9*y+3*x*y

f(x) and f(y)

```
f=expression(x^2 +4*x+y^2-9*y+3*x*y)
#F(x)
D(f,'x')
```

`## 2 * x + 4 + 3 * y`

```
#F(y)
D(f,'y')
```

`## 2 * y - 9 + 3 * x`

``` To find the critical points, we will solve these equations:

- 2x + 4 +3y = 0

2x = -3y - 4

x = -1.5y - 2

Substiture x into B:

- 2y - 9 +3x = 0

2y - 9 + 3(-1.5y - 2) = 0

2y - 9 -4.5y - 6 = 0

-2.5y - 15 = 0

y=15/(-2.5)

y = -6

x = (-1.5 * -6) - 2

x = 7

Critical point is (7,-6)

```
f=expression(2 * x + 4 + 3 * y)
#F(x)
D(f,'x')
```

`## [1] 2`

```
#F(y)
D(f,'y')
```

`## [1] 3`

```
f=expression(2 * y - 9 + 3 * x)
#F(x)
D(f,'x')
```

`## [1] 3`

```
#F(y)
D(f,'y')
```

`## [1] 2`

f(xx)=2

f(xy)=3

f(yy)=2

f(xy)=3

I am not sure how to enterpret this answer since I can’t plug my critical point into the second derivative

There were a lot of valuable elements in this course, I really liked Regression module since it felt like it was the most hands on topic.