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#1 Re: Help Me ! » Cost of Transporting Goods » Yesterday 21:22:54

Bob

I'm like the Canadian Mounties; I never give up. I'll post more help if you want.  Just say.

Bob

#2 Re: Help Me ! » Function Decreasing Over Interval » Yesterday 21:22:09

Bob

Both new functions reflect the original in an axis. So it's the same for both: increasing from -2 to + 7.

Bob

#3 Re: Help Me ! » Function Decreasing Over Interval » Yesterday 07:21:13

Bob

I think so.

It would have been better if the questioner had said "For each function identify if it is decreasing, increasing or neither including where in the domain this occurs." But they possibly thought it was obvious.

Bob

#4 Re: Help Me ! » Function Decreasing Over Interval » 2024-04-20 19:37:58

Bob

When the red graph is decreasing, the blue graph is increasing.

When the blue graph is increasing, the red graph is decreasing.

Correct.

The graph of y = -f(x) is decreasing at the point (7, 0).

The graph of y = f(x) is increasing at the point (7, 0).

The graph I made up had zero gradient at the end points of the interval.

The original question describes the interval as (-2, 7) . Don't get confused here with coordinates. This doesn't mean x=-2, y = 7; but rather all x values from -2 to +7. As the brackets are round not [ ] the end points are not included so it's ok for me to make up a graph that stops decreasing at the endpoints.  If it's not decreasing it must either have an instantaneous change to positive or go through zero gradient at a local maximum or minimum. I chose the latter as it's easier to make up an equation for this.

Bob

#5 Re: Help Me ! » Cost of Transporting Goods » 2024-04-20 19:27:25

Bob

Is this right? No. You're charging too much.

For part 2, 100 miles has already been charged so 0.4 only applies for 200 miles (300 minus 100)

Same for the remaining parts. Subtract the miles that have already been charged to determine how many more miles to charge at the next rate.

A step function graph looks like a staircase, flat parts getting higher.

This graph has sloping lines with gradients 0.5, 0.4, 0.25, 0 so the sections go up in sloping lines except for the last which is flat.

Bob

#6 Re: Help Me ! » Function Decreasing Over Interval » 2024-04-20 06:06:18

Bob

I have made up a function, shown in red, that is decreasing from x = -2 to x = +7

-f(x) is a reflection in the x axis. I have shown that graph in blue.

DvjpEby.gif

From x = -2 to x = + 7 the blue graph is increasing.

Bob

#7 Re: Help Me ! » Graph Involving Absolute Value » 2024-04-19 23:59:04

Bob

A = (-2, -1)

B = (-1, -1)

C = (1, 1)

D = (2, 0).

A. Draw the graph of y = | f(x) |.  You have to work out f(x) and then make any negative values into positives. Some are already positive and so those are unchanged.

B. Draw the graph of y = f(| x |) You have to change any negative x values into positives; then use the function to work out what y values you get.  I found that the original 4 points become just 2 points repeated.

What is the basic difference? Neither new graph looks anything like the original; nor do they share any similarities. Hard to see what the questioner is searching for here.

What you can say is that  y = | f(x) | has no negative y values and  y = f(| x |) has no negative x values.  Is that what is wanted?

Bob

#9 Re: Help Me ! » Function Decreasing Over Interval » 2024-04-19 23:47:27

Bob

Part A fully correct.

Looks like a typo crept in for B . (3,12) was correct but then you put (3,10)

Bob

#10 Re: Help Me ! » Function Decreasing Over Interval » 2024-04-19 23:43:41

Bob

You did the hard bit in post 2. 

A graph that is decreasing means it slopes downwards as you go from left to right x=2 to x=7.

If you reflect it what does that do to  the slope?

Bob

#11 Re: Help Me ! » Cost of Transporting Goods » 2024-04-19 20:45:07

Bob

I've added an extra column to my table:

0 < x ≤ 100                      cost per mile 0.5  max cost = 100 x 0.5 = 50
100 < x ≤ 300                   cpm 0.4               max cost = 200 x 0.4 = 80
300 < x ≤ 400                    cpm 0.25              max cost = 100 x 0.25 = 25
400 < ≤ 960                       cpm no charge

C(x) = {0.50x, if 0 ≤ x ≤ 100....Part 1

This looks ok.

C(x) = {50 + 0.40(x - 100), if 100 < x ≤ 400...Part 2

Not  x ≤ 400. The upper limit for part 2 is 300.

C(x) = {170 + 0.25(x - 400), if 400 < x ≤ 800...Part 3

Where did 170 come from?

Bob

#13 Re: Help Me ! » First-class Mail » 2024-04-18 20:15:00

Bob

Mathematically, this is similar to the truck charge question.  First summarise the info.

0 < weight ≤ 1              fixed charge of 1.17
1 < x ≤ 13                    0.17 per ounce

So between x=0 and 1 the graph will be a horizontal line as the charge is fixed and doesn't vary with x.

After that the additional cost starts to go up in a straight line with gradient 0.17 .

The graph stops when x=13

Bob

#14 Re: Help Me ! » Cost of Transporting Goods » 2024-04-18 20:09:22

Bob

I like to try and summarise the written information using (in this case) algebra.

0 < x ≤ 100                      cost per mile 0.5
100 < x ≤ 300                   cpm 0.4
300 < x ≤ 400                    cpm 0.25
400 < ≤ 960                       cpm no charge

But, beware. This doesn't show fully what a charge will be, nor does it give you the points for a graph.  Foe example, if the distance for a package is, say, 150 miles then the charge would be 100 x 0.5 + 50 x 0.4

So, to get the function you need to include charges for a previous stage and account for how many more miles have been travelled. I'll show what I mean for distances in the 100 - 300 group.

100 < x ≤ 300                   total charge = 100 x 0.5 + (x-100) x 0.4

I'll leave you to deal with the other groups similarly. That should enable you to make the graph.

Bob

#19 Re: Help Me ! » Find x-intercepts of Two Graphs » 2024-04-17 20:09:11

Bob

Part A

Set x + 4 = 0 and solve for x.

No. You want x + 4 = -8

Part B is correct.

Bob

#21 Re: Help Me ! » Find x-intercepts of Two Graphs » 2024-04-16 17:45:18

Bob

Oh I see what happened. I read post 2 not noticing post 1.

Part B correct not A. It's to do with where the 2 is . Inside the bracket, ok. Outside it doubles the y coordinate and does nothing to the x.

Bob

#22 Re: Help Me ! » Find Point For Three Graphs » 2024-04-15 21:01:35

Bob

A. (x+1) inside the function bracket will move the graph one place left.  The -3 outside the bracket moves the point down 3.

B. (x-4) moves the point 4 places right. That gives you g(1-4) which evaluates to 5. The times by -3 and add 3 to get the new y coordinate.

C. I know g(-3) so I put 3x+9 = -3  => 3x = -12 => x = -4.  So we then have g(3.(-4)+9) = g(-12+9) = g(-3) = 5. The y cordinate remains at 5.

Bob

#23 Re: Help Me ! » Graph Transformation » 2024-04-15 20:49:14

Bob

Again refer to the graph on a previous thread.

To get h(x) 2x inside the bracket is the same as squeezing the points closer to the y axis by a factor 1/2. The y coordinates stay the same.

eg (-4,-2) moves to (-2,-2)

g(x) reflects the points in the y axis so (-4,-2) becomes (+4,-2)

Comment on the number of posts.

I don't mind lots of posts if it means I'm helping you.  But it does get hard to keep track when so many appear at once.

Two suggestions that would help me:

Either (1) number your posts in the title so I can keep track of which ones I've viewed and dealt with,

Or (2) post the next query in the same thread you are already using so we get a chain of items all in one longer thread.  I did it that way some years back for an English lady who was studying for GCSE (late entrant). Her thread got to thousands of individual posts but it was easy for both of us to keep track.

Bob

#24 Re: Help Me ! » Graph Transformation » 2024-04-15 20:36:55

Bob

QQNHtmb.gif

The original is in red and F9x) in blue.

Bob

#25 Re: Help Me ! » Function Decreasing Over Interval » 2024-04-15 19:44:26

Bob

They want you to say if the transformed graph has a decreasing or increasing section and if so where.

Bob

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