Grade 12 math problems

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Lime
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Grade 12 math problems

Postby Lime » Mon Oct 18, 2010 12:08 am UTC

There was an issue where I was placed in the wrong class at first this year and it's only just been rectified. I'm now in my proper Math 30 Pure class. My teacher was kind enough to waive the first few cumulative reviews and tests, but I was just assigned one that is going to be weighted the same as all of those past ones combined. Fun. The problem is that I've literally been to 3 classes, so while I've got a grasp of 2 lessons (one day was assigned to work on this review), I'm at a loss for the other 20 or so that the class has covered in my absence. I'm not allowed to take the review home, but I DID write down a few problems that I was having trouble with. Any help you guys could offer would be fantastic. And feel free to laugh at me for not knowing this stuff.

Given the function y= x^2(x+3)-6, I need to make the following transformations.
-Stretch it horizontally by 1/3
-Reflect about the x axis
-Reflect about the y axis
-Move it up 6 units, left 4.

From just messing around with my TI-83, as well as prior knowledge, I THINK that this is the proper solution, but feel free to correct me if I'm wrong.
- y= 3x^2(3x+3)-6
- y= -1(3x^2(3x+3)-6)
- y= -1(-3x^2(-3x+3)-6)
- y= -1(-3x^2(-3x+7))

Like I said, feel free to eviscerate me if I'm wrong on any of these.

Next, there's this.
Given 1/2y= f(-3x-9)+2, I need to find out what the points
(x,y)
(-6,-3)
(0,7) would change to if the equation was changed to
y=(f(x))^-1
I believe that this is an inverse function (like I said, could be wrong. It might be a reciprocal), which means you switch x and y, right? So
(y,x)
(-3,-6)
(7,0)
Like I said though, not sure if it's inverse or reciprocal.

Then there's this.
Given f(x)=√(x)+7, determine
x=f(y)
X intercepts for y+f^-1(x)
Invariant points for y=f(x) and y=f^-1(x)

On this one I'm just lost.

Finally, I need to find the formula for coterminal angles for
25π/8 and
8.3

Both of those in radians, of course. I understand the coterminal formula is and angle + n360, nEI, but that's for degrees. How do I do it for radians?

Again, any help for any of this would be greatly appreciated. Math has never been my strongest suit, but it's even harder when I haven't had a chance to learn it.

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Yakk
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Re: Grade 12 math problems

Postby Yakk » Mon Oct 18, 2010 5:34 pm UTC

Lime wrote:There was an issue where I was placed in the wrong class at first this year and it's only just been rectified. I'm now in my proper Math 30 Pure class. My teacher was kind enough to waive the first few cumulative reviews and tests, but I was just assigned one that is going to be weighted the same as all of those past ones combined. Fun. The problem is that I've literally been to 3 classes, so while I've got a grasp of 2 lessons (one day was assigned to work on this review), I'm at a loss for the other 20 or so that the class has covered in my absence. I'm not allowed to take the review home, but I DID write down a few problems that I was having trouble with. Any help you guys could offer would be fantastic. And feel free to laugh at me for not knowing this stuff.

Given the function y= x^2(x+3)-6, I need to make the following transformations.
-Stretch it horizontally by 1/3
-Reflect about the x axis
-Reflect about the y axis
-Move it up 6 units, left 4.


First step -- solve for a simpler function.

Given the function y = x+1:
-Stretch it horizontally by 1/3
-Reflect about the x axis
-Reflect about the y axis
-Move it up 6 units, left 4.

To check your work, graph the new function you create (using graph paper). You should be able to do it for this simple a function, right?

After you have done that, try it with:
y = x^2 + 1

Then, try it on the function you where assigned.

Like I said, feel free to eviscerate me if I'm wrong on any of these.

I'm not going to mark your assignment/test/revue before you hand it in. ;)

Do you know how to check your answers? Take out some graph paper, and graph the resulting function. Does it do what you want it to do?

Next, there's this.
Given 1/2y= f(-3x-9)+2, I need to find out what the points
(x,y)
(-6,-3)
(0,7) would change to if the equation was changed to
y=(f(x))^-1
I believe that this is an inverse function (like I said, could be wrong. It might be a reciprocal), which means you switch x and y, right? So
(y,x)
(-3,-6)
(7,0)
Like I said though, not sure if it's inverse or reciprocal.

Yes, f-1 (or f^-1) is notation for the inverse function. It also is notation for reciprocal, but that is used more rarely in conjunction with functional notation.

The question as you described is ambiguous, either as a result of the original question being ambiguous, or your half-understanding of it and repeating it being a bit off. Your interpretation is consistent with at least one possible interpretation of what the teacher could have been asking.
Then there's this.
Given f(x)=√(x)+7, determine
x=f(y)
X intercepts for y+f^-1(x)
Invariant points for y=f(x) and y=f^-1(x)

Ok, once again, can you answer the question for f(x) = x+1? How about f(x) = x^2+1?

Solve it for a simpler function (but not too simple), and then work your way up to a more complex one.
Finally, I need to find the formula for coterminal angles for
25π/8 and
8.3

I'd have to google what a coterminal angle is, honestly.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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LaserGuy
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Re: Grade 12 math problems

Postby LaserGuy » Mon Oct 18, 2010 5:43 pm UTC

Lime wrote:Both of those in radians, of course. I understand the coterminal formula is and angle + n360, nEI, but that's for degrees. How do I do it for radians?

Again, any help for any of this would be greatly appreciated. Math has never been my strongest suit, but it's even harder when I haven't had a chance to learn it.


If you can solve the problem in degrees, you should be able to solve the problem in radians. Any angle in degrees can be converted to an angle in radians by a simple conversion factor, and vice versa.

Lime wrote:y=(f(x))^-1)
I believe that this is an inverse function (like I said, could be wrong. It might be a reciprocal), which means you switch x and y, right? So

...

y=f^-1(x)


For what it's worth, I would normally treat the first one [imath]f(x)^{-1}[/imath] as the reciprocal 1/f(x) and [imath]f^{-1}(x)[/imath] as the inverse function (switching y and x). I've seen a few texts that do this a bit differently, but this is by far the most common interpretation of these expressions in my experience.

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Lime
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Re: Grade 12 math problems

Postby Lime » Tue Oct 19, 2010 1:50 am UTC

Yakk wrote:First step -- solve for a simpler function.

Given the function y = x+1:
-Stretch it horizontally by 1/3
-Reflect about the x axis
-Reflect about the y axis
-Move it up 6 units, left 4.

To check your work, graph the new function you create (using graph paper). You should be able to do it for this simple a function, right?

After you have done that, try it with:
y = x^2 + 1

Then, try it on the function you where assigned.

The problem is that I don't know how the x's are affected when there are multiple instances of them.


I'm not going to mark your assignment/test/revue before you hand it in. ;)

Good to know :P

Do you know how to check your answers? Take out some graph paper, and graph the resulting function. Does it do what you want it to do?

Yeah, I mostly just graph em on my calculator, but I can use paper if you're convinced it'll be better.

Yes, f-1 (or f^-1) is notation for the inverse function. It also is notation for reciprocal, but that is used more rarely in conjunction with functional notation.

The question as you described is ambiguous, either as a result of the original question being ambiguous, or your half-understanding of it and repeating it being a bit off. Your interpretation is consistent with at least one possible interpretation of what the teacher could have been asking.

My teacher's been making a huge deal out of the difference between reciprocal and inverse. Honestly I can't tell the difference. I know that inverse is when x and y are switched, and reciprocal is when you do 1/x and 1/y, but I have no idea when it means which one. And the wording is a little different, but that's what the question says.

Ok, once again, can you answer the question for f(x) = x+1? How about f(x) = x^2+1?

The problem is that I'm not sure how to put that into my calculator, and I'm not sure what it means. As I said, the difference between inverse and reciprocal eludes me


I'd have to google what a coterminal angle is, honestly.

Coterminal angles are any angles that share the same terminal arm. Like 10 degrees, 370 degrees, -350 degrees, ect.

LaserGuy wrote:If you can solve the problem in degrees, you should be able to solve the problem in radians. Any angle in degrees can be converted to an angle in radians by a simple conversion factor, and vice versa.

I THINK I have the first one. You need to get down to the angle in standard position. 25pi/8 - 2pi would be 25pi/8 - 16pi/8, which leaves me with 9pi/8 as the angle in standard position. So the formula would be 9pi/8 + n2pi, nEi. The other one I don't really get. We haven't been spending a lot of time on radians without pi being included.

For what it's worth, I would normally treat the first one [imath]f(x)^{-1}[/imath] as the reciprocal 1/f(x) and [imath]f^{-1}(x)[/imath] as the inverse function (switching y and x). I've seen a few texts that do this a bit differently, but this is by far the most common interpretation of these expressions in my experience.

So when the ^-1 is on the right side of the brackets it's reciprocal, and left side is inverse. I think.

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Yakk
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Re: Grade 12 math problems

Postby Yakk » Tue Oct 19, 2010 2:28 am UTC

Lime wrote:The problem is that I don't know how the x's are affected when there are multiple instances of them.

Can you do those problems?

Then, try f(x) = x^2 + x. That is a really simple problem with two different x's.
Do you know how to check your answers? Take out some graph paper, and graph the resulting function. Does it do what you want it to do?

Yeah, I mostly just graph em on my calculator, but I can use paper if you're convinced it'll be better.

But you don't know how to use your calculator to solve every problem. With graph paper, you can just do each operation in order. It is much slower...
My teacher's been making a huge deal out of the difference between reciprocal and inverse. Honestly I can't tell the difference. I know that inverse is when x and y are switched, and reciprocal is when you do 1/x and 1/y, but I have no idea when it means which one. And the wording is a little different, but that's what the question says.

Ok. Do you know how to manually graph the inverse of a function? Turn the paper ninty degrees, and relabel axis.

Do you know how to manually take 1/f(x)? Graph each point. Take the result of each point, and calculate 1/y of it.

Try this with a bunch of simple functions. You'll see how they differ.
Ok, once again, can you answer the question for f(x) = x+1? How about f(x) = x^2+1?

The problem is that I'm not sure how to put that into my calculator, and I'm not sure what it means. As I said, the difference between inverse and reciprocal eludes me

The reciprocal of a function is 1/f(x). You take the function, evaluate it, then take the reciprocal (1/y) of it at each x point.

The inverse of a function is f^-1(x). For each point y, you map it to the point x that under the operation f(x) would reach that point y.

One of them "turns the graph on its side", the other ... doesn't.
Coterminal angles are any angles that share the same terminal arm. Like 10 degrees, 370 degrees, -350 degrees, ect.

Ok.

In radians, what angles are coterminal with 0? Give at least 5, at least 2 positive and 2 negative.

In radians, what angles are coterminal with pi? Give at least 5, at least 2 positive and 2 negative.

(If you need to, convert the radians to degrees. Then find coterminal angles there. Then convert back to radians. There are easier ways, but that might give you a pattern. Avoid using decimals for any of this -- if you use your calculator and just give decimal approximations, the result won't be all that enlightening.)
I THINK I have the first one. You need to get down to the angle in standard position. 25pi/8 - 2pi would be 25pi/8 - 16pi/8, which leaves me with 9pi/8 as the angle in standard position. So the formula would be 9pi/8 + n2pi, nEi. The other one I don't really get. We haven't been spending a lot of time on radians without pi being included.

You don't have to solve it to standard position.

Ie, give one angle that is coterminal with 110000000000 degrees? 110000000360!

For what it's worth, I would normally treat the first one [imath]f(x)^{-1}[/imath] as the reciprocal 1/f(x) and [imath]f^{-1}(x)[/imath] as the inverse function (switching y and x). I've seen a few texts that do this a bit differently, but this is by far the most common interpretation of these expressions in my experience.

So when the ^-1 is on the right side of the brackets it's reciprocal, and left side is inverse. I think.

Ah -- your teacher is trying to confuse you. I thought the ^-1 position was a typo on your or the teachers part.

Ya, (f(x))^-1 is the function such that ((f(x))^-1 ) * f(x) = 1.
f^-1(x) is the function such that f(f^-1(x)) = x.

One is function composition, the other is function multiplication.

So if you have f(x) = x,

f(x)^-1 = 1/x
but f^-1(x) = x

If you have g(x) = x^2,
f(x)^-1 = 1/x^2
f^-1(x) = sqrt(x) (on the positive real numbers)
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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Qaanol
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Re: Grade 12 math problems

Postby Qaanol » Tue Oct 19, 2010 2:44 am UTC

To reflect about the x-axis, replace each occurrence of ‘y’ with ‘-y’.
To reflect about the y-axis, replace each occurrence of ‘x’ with ‘-x’.

In particular, if you have a function of x, replace the x’s before performing any operations. As an example, if [imath]f(x) = x^2+x[/imath], then [imath]g(x) = (-x)^2+(-x)[/imath] means the graph of y = g(x) is the mirror image of the graph of y = f(x) across the y-axis. Does that make sense?
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Re: Grade 12 math problems

Postby keeperofdakeys » Tue Oct 19, 2010 11:15 am UTC

Lime wrote:
Yakk wrote:First step -- solve for a simpler function.

Given the function y = x+1:
-Stretch it horizontally by 1/3
-Reflect about the x axis
-Reflect about the y axis
-Move it up 6 units, left 4.

To check your work, graph the new function you create (using graph paper). You should be able to do it for this simple a function, right?

After you have done that, try it with:
y = x^2 + 1

Then, try it on the function you where assigned.

The problem is that I don't know how the x's are affected when there are multiple instances of them.

It might help if you try to think exactly what you are doing when you perform these operations, then why these processes work with multiple x's.

If I have a function f(x)=x2+x, and I want to stretch it vertically by 2, then I would simply do 2*f(x) = 2*(x2+x). If I wanted to stretch it horizontally by 2, then I would probably want f(0.5*x) = (0.5*x)2 + (0,5*x); do you see how this would work?

The other three are similar, in that in order to perform the manipulations, you only need to play around with them in the form f(x). You will then find that the function will come out easily, after you do the substitutions.

Edit: fixed the horizontal stretch, thanks Yakk
Last edited by keeperofdakeys on Tue Oct 19, 2010 8:53 pm UTC, edited 1 time in total.

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Yakk
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Re: Grade 12 math problems

Postby Yakk » Tue Oct 19, 2010 1:13 pm UTC

Keeper, your horizontal "stretch" is less of a stretch than you'd imagine.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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Re: Grade 12 math problems

Postby Something Awesome » Tue Oct 19, 2010 1:25 pm UTC

Yakk wrote:
Lime wrote:My teacher's been making a huge deal out of the difference between reciprocal and inverse. Honestly I can't tell the difference. I know that inverse is when x and y are switched, and reciprocal is when you do 1/x and 1/y, but I have no idea when it means which one. And the wording is a little different, but that's what the question says.

Ok. Do you know how to manually graph the inverse of a function? Turn the paper ninty degrees, and relabel axis.

That's not quite right. Try that with f(x) = x^3 and you get that the inverse f^-1(x) is negative if x is positive, while the function we want is the cube root, so f^-1(x) should still be positive.

The inverse is actually a reflection across the line y = x, which you can obtain by flipping the paper over diagonally... and then I guess holding it up to the light so you can see the graph, since it's on the other side of the paper now.

Though I suppose you're right if you meant to relabel the axes by making "down" positive y.

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Re: Grade 12 math problems

Postby Yakk » Tue Oct 19, 2010 1:29 pm UTC

*nod*, you do have to do that as well. I should have said "rotate 90 degrees, then flip horizontally".

My bad.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.


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