C1 Question (stupid question)

OP

ive been waiting a whole hour and still no reply :frown:

Reply 2

Zezzy

What do you need help with? Do you need help with finding where the lines cross the axes in part c) of the question?

Reply 3

Hopple

18

1) You don't need to do any more working for the minimum because you know that since f(x) is always greater than or equal to -4, then f(stuff) is always greater than or equal to -4, whatever 'stuff' is.

2) I'm not sure what you're asking. They increase all y coordinates by 4 if y is now f(x)+4 as opposed to just f(x)

Reply 4

chrypton

5

QUESTION 2:

To find the y-intercept, put in x=0 in the function.

Not sure what your first question is, do you want to find the coordinates of the minimum point?

Reply 5

OP

Most of the replies seem confused by my quesiton. Sorry. Let me try and explain again.

I done part a and had no problems.

part bi is where im getting confused.

Basically the questions asking you to apply the transformation y=f(x+2) to the curve f(x)=x(x-4)

The curve is being moved by -2 along the x axis, this means the new x coordinates will be -2 and 2.

I get that bit.

So lets say i drew on my curve crossing the x-axis at -2 and 2.

I can see that the curve intercepts the y-axis. so i obviously have to label that on otherwise il loose marks and i dont know what it is. so i try and find it. I replaced x in the original equation f(x)=x(x-4) with (x+2) to get y= (x+2)(x-2).
i put x=0 and got 2x-2 = -4 . so the y coordinate -4

which i now label on the graph.

what i dont get is for their working on bi (take a look at the pic for Bi) you will see their is no working to find the y-coordinate. I understand why there isn't any working for the x coordinate , but i dont understand why there is no working for the y coordinate. this is were im thinking I am supposed to have calculated the y-intercept a different way. Because my way took some time and on their "how to do it" marking scheme. there is no calcultion for the y-intercept.

I hope that clears things

(edited 13 years ago)

Reply 6

ttoby

15

Original post by jayseanfan

Most of the replies seem confused by my quesiton. Sorry. Let me try and explain again.

I done part a and had no problems.

part bi is where im getting confused.

Basically the questions asking you to apply the transformation y=f(x+2) to the curve f(x)=x(x-4)

The curve is being moved by -2 along the x axis, this means the new x coordinates will be -2 and 2.

I get that bit.

So lets say i drew on my curve crossing the x-axis at -2 and 2.

I can see that the curve intercepts the y-axis. so i obviously have to label that on otherwise il loose marks and i dont know what it is. so i try and find it. I replaced x in the original equation f(x)=x(x-4) with (x+2) to get y= (x+2)(x-2).
i put x=0 and got 2x-2 = -4 . so the y coordinate -4

which i now label on the graph.

what i dont get is for their working on bi (take a look at the pic for Bi) you will see their is no working to find the y-coordinate. I understand why there isn't any working for the x coordinate , but i dont understand why there is no working for the y coordinate. this is were im thinking I am supposed to have calculated the y-intercept a different way. Because my way took some time and on their "how to do it" marking scheme. there is no calcultion for the y-intercept.

I hope that clears things

In the answers for part c, they do calculate the axis intersections (including the one you are referring to). So what they probably expect you to do is sketch the graphs in part b, then do part c and once you know the required axis intersections to go back and label them on your graphs.

Probably the reason why they split this bit into a separate part is to make sure that you do actually go to the trouble of calculating the intersections, rather than just leaving them unlabelled on your sketch.

Reply 7

Original post by jayseanfan

Most of the replies seem confused by my quesiton. Sorry. Let me try and explain again.

I done part a and had no problems.

part bi is where im getting confused.

Basically the questions asking you to apply the transformation y=f(x+2) to the curve f(x)=x(x-4)

The curve is being moved by -2 along the x axis, this means the new x coordinates will be -2 and 2.

I get that bit.

So lets say i drew on my curve crossing the x-axis at -2 and 2.

I can see that the curve intercepts the y-axis. so i obviously have to label that on otherwise il loose marks and i dont know what it is. so i try and find it. I replaced x in the original equation f(x)=x(x-4) with (x+2) to get y= (x+2)(x-2).
i put x=0 and got 2x-2 = -4 . so the y coordinate -4

which i now label on the graph.

what i dont get is for their working on bi (take a look at the pic for Bi) you will see their is no working to find the y-coordinate. I understand why there isn't any working for the x coordinate , but i dont understand why there is no working for the y coordinate. this is were im thinking I am supposed to have calculated the y-intercept a different way. Because my way took some time and on their "how to do it" marking scheme. there is no calcultion for the y-intercept.

I hope that clears things

It's probably because they didn't believe working was required, I mean you only have 2 terms to multiply which is reasonably easy to do. It's not like you would lose any marks for writing it out in an exam (in which i would recommend writing it out). What you have done is fine.

(edited 13 years ago)

Reply 8

chrypton

5

Original post by jayseanfan

Most of the replies seem confused by my quesiton. Sorry. Let me try and explain again.

I done part a and had no problems.

part bi is where im getting confused.

Basically the questions asking you to apply the transformation y=f(x+2) to the curve f(x)=x(x-4)

The curve is being moved by -2 along the x axis, this means the new x coordinates will be -2 and 2.

I get that bit.

So lets say i drew on my curve crossing the x-axis at -2 and 2.

I can see that the curve intercepts the y-axis. so i obviously have to label that on otherwise il loose marks and i dont know what it is. so i try and find it. I replaced x in the original equation f(x)=x(x-4) with (x+2) to get y= (x+2)(x-2).
i put x=0 and got 2x-2 = -4 . so the y coordinate -4

which i now label on the graph.

what i dont get is for their working on bi (take a look at the pic for Bi) you will see their is no working to find the y-coordinate. I understand why there isn't any working for the x coordinate , but i dont understand why there is no working for the y coordinate. this is were im thinking I am supposed to have calculated the y-intercept a different way. Because my way took some time and on their "how to do it" marking scheme. there is no calcultion for the y-intercept.

I hope that clears things

I think the reason there is no working for the y-coordinate is because it is very clear from the graphs.

Reply 9

Original post by chrypton

I think the reason there is no working for the y-coordinate is because it is very clear from the graphs.

ah yes that's true, in this sort of equation, distance between the two points of intersection=distance above or below x axis, ie 2-(-2)=4, as the graph has a U shape, the point of intersection =-4

Reply 10

OP

Original post by ttoby

In the answers for part c, they do calculate the axis intersections (including the one you are referring to). So what they probably expect you to do is sketch the graphs in part b, then do part c and once you know the required axis intersections to go back and label them on your graphs.

Probably the reason why they split this bit into a separate part is to make sure that you do actually go to the trouble of calculating the intersections, rather than just leaving them unlabelled on your sketch.

This is what I though too. But then i dont get how you are supposed to sketch BII.

Because yes you know the graph is translated by +4 in the y axis. and since the original y interception is at 0 this means the curve will cut 4. but how are you supposed to know that the graph just touches the x-axis at 2? I could have draw it like this

i wouldn't have know the curve touches the x-axis at 2 till i get to part c and calculated the x intercept (which is not a intercept in this case but you know what i mean). say there wasn't a part c ?

Im so confused. I understand the general concept but im getting confused by when your supposed to do what.

(edited 13 years ago)

Reply 11

chrypton

5

Original post by jayseanfan

This is what I though too. But then i dont get how you are supposed to sketch BII.

Because yes you know the graph is translated by +4 in the y axis. and since the original y interception is at 0 this means the curve will cut 4. but how are you supposed to know that the graph just touches the x-axis at 2? I could have draw it like this

i wouldn't have know the curve touches the x-axis at 2 its till i get to part c and calculated the x intercept (which is not a intercept in this case but you know what i mean). say there wasn't a part c ?

Im so confused. I understand the general concept but im getting confused by when your supposed to do what.

Since you were able to sketch b)i) and found the y-intercept to be (-4,0) you know that when you are translating the graph +4, the y-coordinate of the minimum point will be -4+4=0.

Reply 12

Original post by jayseanfan

This is what I though too. But then i dont get how you are supposed to sketch BII.

Because yes you know the graph is translated by +4 in the y axis. and since the original y interception is at 0 this means the curve will cut 4. but how are you supposed to know that the graph just touches the x-axis at 2? I could have draw it like this

i wouldn't have know the curve touches the x-axis at 2 till i get to part c and calculated the x intercept (which is not a intercept in this case but you know what i mean). say there wasn't a part c ?

Im so confused. I understand the general concept but im getting confused by when your supposed to do what.

again refer to my last post, use that on the first graph and you'll see that the greatest depth that the curve may be below the x axis is 4, raising the curve by 4 will therefore mean that the curve only touches the x axis.

(edited 13 years ago)

Reply 13

OP

Original post by roar558

ah yes that's true, in this sort of equation, distance between the two points of intersection=distance above or below x axis, ie 2-(-2)=4, as the graph has a U shape, the point of intersection =-4

I didn't know that rule.

Can i even use it or C1? and when can i use it ? Or am i better of translating the graph accoding to the rule and then put x=0 or y=0 to see where the curve intercepts the x or y axis?

Reply 14

OP

so what would u guys say to conclude

is it okay if i just apply the transformaiton first

and then just calculate the x intercept by y=0 OR the y intercept by x=0

Reply 15

Original post by jayseanfan

I didn't know that rule.

Can i even use it or C1? and when can i use it ? Or am i better of translating the graph accoding to the rule and then put x=0 or y=0 to see where the curve intercepts the x or y axis?

I'm not sure it is in a text book, it's caused by the shape of the graph and so is an easy way to find certain points of intersection. However I would check with your teacher/lecturer whether or not it would be accepted.

Reply 16