Question help

[pope4]

I was doing this question in my textbook and I was confused as to point a. When I looked up the answer on the internet, it made even less sense. Why is the graph centred on the grid if you can't have negative distance, shouldn't x be on the positive side of the axis? How does h=5 if the graph is centred? In what way is it flipped?

 

[Dr.Peterson]

I was doing this question in my textbook and I was confused as to point a. When I looked up the answer on the internet, it made even less sense. Why is the graph centred on the grid if you can't have negative distance, shouldn't x be on the positive side of the axis? How does h=5 if the graph is centred? In what way is it flipped? View attachment 34161View attachment 34162

That's a good observation. I think what they really mean is that x is the signed distance from the center. The actual distance would be its absolute value; they are using x not as an actual distance, but as a coordinate. It is easy to overlook the difference!

 

[pope4]

That's a good observation. I think what they really mean is that x is the signed distance from the center. The actual distance would be its absolute value; they are using x not as an actual distance, but as a coordinate. It is easy to overlook the difference!

How do you tell which one they mean?

The answer said that the graph is flipped as well. Would you be able to explain why that is? Thank you for your help!

 

[Dr.Peterson]

How do you tell which one they mean?

The answer said that the graph is flipped as well. Would you be able to explain why that is? Thank you for your help!

Actually, now that I look closer at everything they say, I see that I totally misread it, looking only at the graph, and not looking at the details of the problem and solution myself, but thinking about other problems that look similar. I was wrong.

In their answer, they do take x as only positive; the only place they allow it to be negative is in their graph. The first line of the answer says they are considering only the right half.

Now, having paid closer attention, I see that the graph isn't of the function they found at all! Their function was [imath]y=\sqrt{5}\sqrt{-(x-5)}=\sqrt{25-5x}[/imath]. The graph shows [imath]y=\sqrt{-(x^2-25)}=\sqrt{25-x^2}[/imath], which is a circle (though it looks like an ellipse because of different scales on the axis).

So the correct answer to the problem is [imath]y=\sqrt{5}\sqrt{-(2-5)}=\sqrt{15}=3.87[/imath].

In order to graph a cross-section of the roof, I would use [imath]y=\sqrt{5(5-|x|)}[/imath]:



Here I've added in their wrong graph (blue) and their wrong answer:


But where does it say the graph is flipped? Please explain.

Anyway, all those people who liked my answer should join me in the corner. We can have a big party in honor of the author of the problem solution, who should be there.

 

[pope4]

Actually, now that I look closer at everything they say, I see that I totally misread it, looking only at the graph, and not looking at the details of the problem and solution myself, but thinking about other problems that look similar. I was wrong.

In their answer, they do take x as only positive; the only place they allow it to be negative is in their graph. The first line of the answer says they are considering only the right half.

Now, having paid closer attention, I see that the graph isn't of the function they found at all! Their function was [imath]y=\sqrt{5}\sqrt{-(x-5)}=\sqrt{25-5x}[/imath]. The graph shows [imath]y=\sqrt{-(x^2-25)}=\sqrt{25-x^2}[/imath], which is a circle (though it looks like an ellipse because of different scales on the axis).

So the correct answer to the problem is [imath]y=\sqrt{5}\sqrt{-(2-5)}=\sqrt{15}=3.87[/imath].

In order to graph a cross-section of the roof, I would use [imath]y=\sqrt{5(5-|x|)}[/imath]:
View attachment 34195


Here I've added in their wrong graph (blue) and their wrong answer:
View attachment 34196

But where does it say the graph is flipped? Please explain.

Anyway, all those people who liked my answer should join me in the corner. We can have a big party in honor of the author of the problem solution, who should be there.

That makes more sense, thank you!

In the solution, it says, and I quote, "The endpoint is (5, 0), so h = 5 and k = 0. The function is reflected in the y-axis, so b =-1." I'm not sure why it says the graph is reflected, and I'm assuming h=5 because it's entirety was on the right side and it was moved to make it symmetrical so they can take only the right side under consideration. Please let me know if my reasoning is correct!

 

[jonah2.0]

Beer induced query follows.

I was doing this question in my textbook and I was confused as to point a. When I looked up the answer on the internet, it made even less sense. Why is the graph centred on the grid if you can't have negative distance, shouldn't x be on the positive side of the axis? How does h=5 if the graph is centred? In what way is it flipped? View attachment 34161View attachment 34162

What is the ISBN of your textbook?

 

[Dr.Peterson]

That makes more sense, thank you!

In the solution, it says, and I quote, "The endpoint is (5, 0), so h = 5 and k = 0. The function is reflected in the y-axis, so b =-1." I'm not sure why it says the graph is reflected, and I'm assuming h=5 because it's entirety was on the right side and it was moved to make it symmetrical so they can take only the right side under consideration. Please let me know if my reasoning is correct!

Ah! That's what you mean by "flipped".

Here's the whole statement:

​

They are transforming the function [imath]f(x)=\sqrt{x}[/imath]. The vertex of f would be at (0,0), but for our graph it is at (5,0), so the graph is shifted right 5 units (meters) and up 0, accounting for the h and k in their [imath]a\sqrt{b(x-h)}+k[/imath]. But the basic square root graph opens to the right (that is, the graph is to the right of the vertex), whereas we want the vertex to be on the right. So it has to be reflected first, like this, where f is in blue, the reflection is in green, and the final shifted function (except for the vertical stretch (or rescaling) is in red:

​

The final step is to stretch it vertically to pass through (0,5), which they explain in more detail.

What you describe (making it symmetrical) sounds like what I did by using the absolute value, in effect making a reflected copy of the right half on the left. They didn't actually do that (since they made the wrong graph); and what they could have done instead would be to graph only over the stated domain (just the right half), taking it (as they originally said) not as a cross-section, but as showing the height as a function of distance from the center (which in reality is rotated all the way around).

So, no, I don't think your reasoning is quite correct.

 

[pope4]

Ah! That's what you mean by "flipped".

Here's the whole statement:

View attachment 34197​

They are transforming the function [imath]f(x)=\sqrt{x}[/imath]. The vertex of f would be at (0,0), but for our graph it is at (5,0), so the graph is shifted right 5 units (meters) and up 0, accounting for the h and k in their [imath]a\sqrt{b(x-h)}+k[/imath]. But the basic square root graph opens to the right (that is, the graph is to the right of the vertex), whereas we want the vertex to be on the right. So it has to be reflected first, like this, where f is in blue, the reflection is in green, and the final shifted function (except for the vertical stretch (or rescaling) is in red:

View attachment 34198​

The final step is to stretch it vertically to pass through (0,5), which they explain in more detail.

What you describe (making it symmetrical) sounds like what I did by using the absolute value, in effect making a reflected copy of the right half on the left. They didn't actually do that (since they made the wrong graph); and what they could have done instead would be to graph only over the stated domain (just the right half), taking it (as they originally said) not as a cross-section, but as showing the height as a function of distance from the center (which in reality is rotated all the way around).

So, no, I don't think your reasoning is quite correct.

Ohh ok I get it now. I really have to stop confusing and accidentally switching x and y values in coordinates, which is why the h=5 confused me to begin with! Again, thank you for the help!

 

[pope4]

Beer induced query follows.

What is the ISBN of your textbook?

not sure about the ISBN since I don't have the textbook with me at the moment (I use a digital copy), but it's the Pre-Calculus 12 textbook (McGraw-Hill Rayson)