anonymous
  • anonymous
.hjh
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Well, the average rate of change between two points is simply the slope between them. So I assume you know slope to be determined by \[\frac{ y_{2} - y_{1} }{ x_{2}-x_{1} }\] So in the first statement, if the average rate of change, the slope, is 0, what must be true about \[\frac{ y_{2}-y_{1} }{ x_{2}-x_{1} }\]?
anonymous
  • anonymous
I don't know xc ... maybe between -3 and 3 there are solutions?
anonymous
  • anonymous
Well, if the average rate of change is 0 then the slope is 0. Which means \[\frac{ y_{2} - y_{1} }{ x_{2}-x_{1} } = 0\] Basically, how can we make that fraction for slope =0?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
would the demominaters be -3 and 3? 3-(-3)
anonymous
  • anonymous
Yes, they would be. So you would have \[\frac{ y_{2}-y_{1} }{ 6 } = 0\] So what does that say about y_2 and y_1?
anonymous
  • anonymous
That there not identified yet?
misty1212
  • misty1212
|dw:1433728163730:dw|
anonymous
  • anonymous
Sorry, not trying to tease or anything, just wanted you to see it. Well, the only way a fraction can be 0 is if the numerator is 0. Which means \(y_{2} - y_{1} = 0\) \(y_{1} = y_{2}\) The idea is that the y coordinate of both points has to be the same in order for the average rate of change to be 0. That make sense?
anonymous
  • anonymous
Kinda, it makes the reasoning for why its 0
anonymous
  • anonymous
would that explain tuckers part on why he's correct? or is there more to tuckers reasoning :o
anonymous
  • anonymous
Yeah, exactly. So the first person is essentialy saying the y-coordinates for both points are the same. It doesnt matter what they are, we just want them to be the same. So, for example, lets say our y-value is 0. So we would have the points (-3,0) and (3,0).
anonymous
  • anonymous
As for the 2nd person, considering the graph between -3 and 3 we know it'll go up and then back down. But at the same time, we also want the graph to start and begin at the same y-value. So we want something like this. |dw:1433728476305:dw| So, the best example of how this is possible is if the graph is a parabola. But really, any continuous function that has the same y-coordinate at x = -3 and x = 3 and is not a line will make them both correct.
anonymous
  • anonymous
Ok! That makes since, and yea i thought it would be a parabola, thanks again! cx xD
anonymous
  • anonymous
Well, I guess it also has to be going up in between -3 and 3, lol. So I guess it cant be EVERYTHING with x-intercepts at -3 and 3, but you get the idea.
anonymous
  • anonymous
Yea lol, thats all i think of when i hear " up and down", and is it ok if i write out the answers and you read to see if it souds right?
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
* Tucker says that for the function, between x = -3 and x = 3, the average rate of change is 0* Here Tucker says that between -3 and 3 x coordinates that the average rate of change is 0.. the slope is y2−y1/x2−x1 and if you plug in the x coordinates you get y2−y1/6=0.... whats true is that the x-coordinates are -3 and 3, whats true about y is y2−y1=0 so y1=y2, the y coordinate of both points has to be the same in order for the average rate of change to be 0
anonymous
  • anonymous
sorry that took forever, but thats tuckers reasoning
anonymous
  • anonymous
It doesnt need to be that wordy, that was just me trying to explain the idea :P Really, all you need to say is something like this: "Tucker can be correct as long as the average of change is 0. As in the slope between the points (-3,y1) and (3,y2) is 0. This is only true if y1 = y2." I dont think you really need to go into any more detail than that. I would think any detail left out from that is stuff youre assumed to know and dont really need to explain.
anonymous
  • anonymous
Ok lol, yea i was writing it like " wait... should i include this O.o" lol and thanks! ill do the other person, and less wordy lol cx
anonymous
  • anonymous
Yeah, pretty much any minimal conditions that would make the statement true :P In the end, we dont even care about the x-coordinates. The x-coordinates could be anything, but the fact that we know the rate of change has to be 0 just tells us the y-coordinates have to be equal.
anonymous
  • anonymous
Karly is correct also because if you draw a graph between -3 and 3 we know it'll go up and then back down, example parabola
anonymous
  • anonymous
example *a parobla
anonymous
  • anonymous
Well, have to phrase that a little differently, just be careful. The way you phrased it, you said if we draw a graph between -3 and 3. Just by drawing that graph, that doesnt mean it will go up and then down. Just trying to be careful with the wording. We want to state a condition that makes both of them correct. "Karly claims the goes up through a turning point and then comes back down. This won't contradict Tucker's statement as long as the graph has the same y-coordinate at x = -3 and x = 3. An example where this could easily be true is a parabola." Not trying to be super picky or mean or anything, I guess Im almost making sure the English is good now and not the math, haha.
anonymous
  • anonymous
No it's ok, I don't want to do bad Cx and thank you so much, your really a big help! cx
anonymous
  • anonymous
You're welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.