Question

1 Calculus.

Answer 1

I see that in every case, we increment x by 2.

We begin drawing the slope field at or near which point?

Answer 2

the answer to that question is right in the problem statement, altho perhaps not in the form you're used to when you describe a point in the plane.

Answer 3

We would put it near 2 right? Not sure about how the answer is right in the problem statement can you explain and break it down? Then I can solve?

Answer 4

All four of the answer choices have the same initial values for x and y. Can you spot these values?

x=?

y=?

Answer 5

Yeah so when x and y=0 we have n=6 and so on

Answer 6

Please take another look at the first two columns of the first possible answer.
Which x and which y values are given there? n is nthing but a counter, like counting on your fingers.

Answer 7

my bad i put the wrong thing when n and x=0 y=6

Answer 8

Right. So, if you were to graph this slope field, your starting point would definitely be (0,6). x=0 and y = 6.

You don't have to do so, but actually graphing this might make the picture clearer for you. Should we move on, or do you want to graph this first point?

Answer 9

Ok I graphed it. Should I graph the rest of the points? If so how do i figure which is right? like will my points from a specific graph make a straight slope line?

Answer 10

You are given a formula to be used in calculating the slope at any point (x,y).

What is that formula?

Answer 11

Are you looking at what is in the question or slope point form?

Answer 12

it's either y=mx+b or
y'=-2x-4y

Answer 13

Yes, it's the d. e. Remember that (dy/dx) = slope = y' = m in this particular problem.

Remember that you are starting at (0,6), and that you are to increase x by 2.
Your job right now is to determine by how much y increases.

Note: (increase in y) = (slope of tangent line at (0,6) * (change or increase in x).

What is the slope of the tangent line to the curve at (0,6)?

Answer 14

I still need your answer for that. Note, however, that I've made a mistake; your increments in x are 0.2, not 2.0. But that doesn't affect anything (yet).

Answer 15

At the starting point, (0,6), x=0 and y=6. The slope of the tangent line to the curve y at that point is given by dy/dx = -2x - 4y.

Evaluate that slope at (0,6), please.

Answer 16

How exactly do I do that? (0,6) isn't on the curve...

Answer 17

Actually, (0,6) MUST be on your graph, and your curve will go through that point.

Answer 18

Again, x=0 and y=6. The slope is dy/dx = -2x -4y.

Evaluate the slope for x=0 and y=6.

Answer 19

Let x=0 in the slope formula.

Answer 20

ohhh i think I see I got -24

Answer 21

plugging in x=0 adn y=6 to the original equation. Now I'll plug in for point slope form.

Answer 22

y=m0+6?

Answer 23

That is the slope. Becasuse the slope is negative, the curve y is decreasing at (0,6).
The change in y, from the current y=6, is (slope of TL)(increment in x).

Can you figure that out?

Use any method you like. However, your "m0" is not correct.

Calculate the slope first (you already have) and then multi. it by the change in x.

Show your work, pls.

Answer 24

so slope is -24 and we are looking to find x=.2 and solve for y?

Answer 25

Hold on to that y=6 (initial value of the function). Calculate the change in y from the formula I gave you above: (slope of TL) * (increment in x).

Answer 26

so dy/dx=-2(x)-4(6)? Are we using slope point or the function we were given?????

Answer 27

remember that x=0. Your calculated slope must reflect that:

slope of tl to graph at (0,6) = (slope of tl at (0,6) times (increment in x = 0.2 )

The slope is -2(0) - 4(6) = -24; the increment in x is 0.2.

By how much does y change from its initial value, 6?

Answer 28

We can rule out A and D because the y value is increasing and we have a negative slope.

Answer 29

right I understand that, but what equation do we use for it?

Answer 30

calchater, I'd appreciate your answering my questions rather than jumpting ahead to select an answer. I promise to answer your questions at a suitable time.

Answer 31

I know you want me to find the rate of change for y.... I am asking what equation to use... As far as the ruling out... I was just eliminating the choices to narrow the answer.

Answer 32

The initial value of y is 6; as we move from x= 0 to x=0.2, y decreases by (how much?) to (what value ?)

Answer 33

so plug x=.2 into what equation the one we are given????? to find the change in y?

Answer 34

So -2(.2)-4y=what?

Answer 35

You found that the slope at (0,6) is -24. Your increment is 0.2.

Therefore, y changes by -24(0.2) = - 4.8. Notice how I subst. x=0 and y = 6?

Answer 36

The new y-value, at x=0.2, is therefore 6 - 4.8 = ??

Answer 37

1.2 So we are using the slope to approximate rate of change?

Answer 38

Yes, indeed we are!

Now, to answer your question: Here is one (of several forms) of the equation we're using to approximate y = f(x):\[y=f(x)=f(x _{0})+f '(x _{o})(x-x _{0})\]

Answer 39

so 6+6(x-0) plugging in?

Answer 40

Now let's quickly illustrate how this works.

Your initial y value is 6. that's your f(x_0). Your slope is -24. That's your f '(x_0).
Your x - x_0 is 0.2 - 0 = 0.2. Hope this makes sense.

Answer 41

ok so then 6+-24*.2 or to simplify 1.2

Answer 42

"so 6+6(x-0) plugging in?" The initial 6 is fine. That's your starting y value.

That +6 is not correct; the slope is -24, as we found earlier.
(x-0) is fine because the starting value of x was 0.

Try to reconcile what you've done with what I've done.

Answer 43

any questions?

Answer 44

Not so far (I corrected that -24 above) so then y should be 1.2 for the second column leaving us with d?

Answer 45

and if we needed to continue when x=.4 y=.16?

Answer 46

You have four answer choices; your criteria for selection of the correct one has to be whether the choice matches your actual calculations or not.

The first point is (0,6). The second is (0.2, 1.2) First, do you agree? Secondly, does any of the answer choices reflect these calculations?

Answer 47

I agree and My choice was D.