Question

g

Answer 1

is there a question, or just the function?

Answer 2

u need to find the domain ??

Answer 3

the asymptotes horizontal and vertical

Answer 4

For the vertical asymptotes, set the denominator equal to 0, like so:

(x^2-25) = 0

Then solve for the two x values.

Answer 5

The horizontal asymptote is just 0, as the degree of x is lower in the numerator than the denominator.

Answer 6

Yes, Erak, is right. And the limit as x approaches zero can be found to zero by L'Hosp

Answer 7

just post it and somebody will get to it

Answer 8

compare the graph of h(x)=(x-3)^2+3 to the graph of f(x)=x^2

Answer 9

I assume that's just listing the transformations?

Answer 10

yes

Answer 11

So in h(x)=(x-3)^2+3, what do the "-3" and "+3" parts of the equation represent?

Answer 12

down and up or postive and negative

Answer 13

Okay for the "-3", when it's in the bracket with x, that represents a horizontal shift, so left to right. If you set that bracket to 0, what value of x do you get?

x-3 = 0

Answer 14

so -3?

Answer 15

well if you move the "-3" over to the right side, like so:

x-3 = 0
x = 3

You see that it's actually positive 3, meaning it has a horizontal shift of 3 units right. Note how it's negative 3 in the bracket, but it's actually moving right.

Answer 16

h(x)=(x-3)^2+3 <--- now for that 3 at the end of the equation

That's your vertical shift, and there's nothing interesting about it. If it says +3, it literally means a vertical shift of 3 units upwards.

Answer 17

okay so its h(x) shifts 3 units down and 3 units up?

Answer 18

3 units right, and 3 units up

Answer 19

ok can you help me on another one

Answer 20

Go ahead

Answer 21

find the inimum value of c=4x-3y using the following constraints
x>/0
y >/ 0
4x+2y</12
2y</x+2

Answer 22

i tried to do the line under the least and greater sign thats why i put the /

Answer 23

c=4x-3y is your equation?

Answer 24

yes

Answer 25

hello?

Answer 26

Start with a blank xy axis

Answer 27

The fact that \(\Large x \ge 0\) and \(\Large y \ge 0\) means we're restricting ourselves to just the first quadrant. This shaded region in blue.

Answer 28

Do you agree with that statement @ashontae19 ?

Answer 29

yes

Answer 30

Next we graph 4x+2y<=12. This means we graph the line 4x+2y=12 and then shade below the boundary line. The boundary line is solid. The graph is shown in red.

Answer 31

can you do the work all in one step please

Answer 32

Because we're restricted to just the first quadrant, this means we can ignore the stuff in Q2,Q3,Q4

so we can shrink the red region down to this

Answer 33

Now let's graph 2y<=x+2. You would graph 2y = x+2 to form the solid boundary line. Then you'd shade below the boundary line. Graph is shown in blue.

Answer 34

ok i got that thanks how would i plug into the objective function

Answer 35

Notice how the red region and blue region overlap to form the purple region. The purple region in just Q1 looks like this.

Answer 36

yea so how would i plug into the objective function

Answer 37

Focus on just the purple region. Erase the other colors if needed.

Answer 38

Then plot the points A,B,C,D as the corners of the purple region.

Answer 39

Are you able to determine the coordinates of A,B,C, and D?

Answer 40

can you write the numbers my computer going slow i have answer choices
2
0
-3
-9

Answer 41

where is point A located in terms of (x,y) coordinates?

Answer 42

0,0

Answer 43

yep, so (x,y) = (0,0) meaning that
x = 0 and y = 0

Answer 44

plug x = 0 and y = 0 into c=4x-3y

Answer 45

if x = 0 and y = 0
then c = ??

Answer 46

0

Answer 47

good

Answer 48

where is point B located in terms of (x,y) coordinates?

Answer 49

3,1

Answer 50

not (3,1) but you're close

Answer 51

try again

Answer 52

3,0

Answer 53

yes

if x = 3 and y = 0
then c = ??

Answer 54

is 0 ?

Answer 55

c = 4x-3y
c = 4*3-3*0 ... plug in (x,y) = (3,0)
c = 12-0
c = 12

Answer 56

You replace every x with 3. You replace every y with 0.

Answer 57

my answer choices are
2
0
-3
-9

Answer 58

so which one would it be

Answer 59

where is point C located in terms of (x,y) coordinates?

Answer 60

2,2

Answer 61

yes

if x = 2 and y = 2
then c = ??

Answer 62

2

Answer 63

yes

Answer 64

where is point D located in terms of (x,y) coordinates?

Answer 65

0,1

Answer 66

ok so the answer is 2? the final answer???

Answer 67

if x = 0 and y = 1
then c = ??

Answer 68

1

Answer 69

so is the final answer 2???

Answer 70

So to recap we have these four points
A = (0,0)
B = (3,0)
C = (2,2)
D = (0,1)
which are the boundary of the feasible region


If we plug the coordinates of point A into the objective function, then we get c = 0


If we plug the coordinates of point B into the objective function, then we get c = 12


If we plug the coordinates of point C into the objective function, then we get c = 2


If we plug the coordinates of point D into the objective function, then we get c = -3


Which is the smallest value of c?

Answer 71

-3

Answer 72

yes which happens at point D

so if x = 0 and y = 1, then the min value is c = -3

Answer 73

so -3 is the final answer

Answer 74

ok thanxs lasr one find the maximum value of c=-2x+4y using the following constraints

x≤2
x≥-1
y≤5
y≥-3

Answer 75

you'll follow the same steps as before.

step 1) Plot each inequality on the same xy grid

step 2) figure out where ALL four regions overlap. Mark the boundary and specifically mark the corner points of the boundary of this feasible region

step 3) Plug each point into the objective function c=-2x+4y

step 4) The point that produces the largest c value is going to lead you to the answer

Answer 76

Give those steps a shot and tell me what you get. Tell me what the corner points are. Don't just tell me the final answer.

Answer 77

can you just work out the final answer real fast because im getting tired because i got to hurry up

Answer 78

Sorry I can't

Answer 79

well do you know what the subject to would be

Answer 80

I'm not sure what you mean?

Answer 81

This post has grown much, much too long. Would you please post no more than one or two questions at a time.