Help? C3 Problem

[Ayesha B]

cant understamd properly ... can u guys solve it? aniekan and Ayesha B

Okay its pretty simple...
First you need to understand the basic graph, which is y = x + 1
The graph in the question is y = x + 1 with many transformations done to it.
I recommend that you sketch all the stages and so it will be easier to understand.

First transformation is the modulus, so | x + 1 |. This means all the y coordinates that lie below y = 0 will be reflected in the x- axis.
Then this modulus is multiplied by negative 1, which is (-1). This means all the y- coords will be multiplied by (-1)
And then finally you're adding 2 to the y-coords. Writing "2 - |x+1|" is the same as writing "-|x+1| + 2"

Now, for P, if you look at your initial sketch, it was (-1, 0). The modulus transformation didnt change it because it was on the x axis anyway. The (-1) trans didnt change it either because it was on the x- axis. The last trans was +2 to the y coords, so the answer is (-1, 2)

As for Q and R, you can use the fact that they are on the axes to determine their other coords.

Q --> lies on y-axis so that means x = 0. Sub it into the formula y = 2 - ( [0] +1 ) therefore y = 1... so Q(0, 1)
R --> lies on x- axis so that means y = o. Sub it into the formula [0] = 2 - (x +1) therefore x = 1... so R(1, 0)


I know this looks long and confusing but thats because im explaining it... if you try it yourself its really easy and the main thing is grasping the concept. Just rmmr with the transformation questions to go back to the basic graph and work upwards

 

[urwahboy]

Okay its pretty simple...
First you need to understand the basic graph, which is y = x + 1
The graph in the question is y = x + 1 with many transformations done to it.
I recommend that you sketch all the stages and so it will be easier to understand.

First transformation is the modulus, so | x + 1 |. This means all the y coordinates that lie below y = 0 will be reflected in the x- axis.
Then this modulus is multiplied by negative 1, which is (-1). This means all the y- coords will be multiplied by (-1)
And then finally you're adding 2 to the y-coords. Writing "2 - |x+1|" is the same as writing "-|x+1| + 2"

Now, for P, if you look at your initial sketch, it was (-1, 0). The modulus transformation didnt change it because it was on the x axis anyway. The (-1) trans didnt change it either because it was on the x- axis. The last trans was +2 to the y coords, so the answer is (-1, 2)

As for Q and R, you can use the fact that they are on the axes to determine their other coords.

Q --> lies on y-axis so that means x = 0. Sub it into the formula y = 2 - ( [0] +1 ) therefore y = 1... so Q(0, 1)
R --> lies on x- axis so that means y = o. Sub it into the formula [0] = 2 - (x +1) therefore x = 1... so R(1, 0)


I know this looks long and confusing but thats because im explaining it... if you try it yourself its really easy and the main thing is grasping the concept. Just rmmr with the transformation questions to go back to the basic graph and work upwards

Thanks I did understood about Q and R only P was a problem... btw we can find the x coordinate of P using symmetry (-3+1)/2 =-1 then we can find y coordinate using the equation but this way is kind of risky...

 

[Ayesha B]

Thanks I did understood about Q and R only P was a problem... btw we can find the x coordinate of P using symmetry (-3+1)/2 =-1 then we can find y coordinate using the equation but this way is kind of risky...

Hmmm... good point. I'm just wondering if these kinda graphs would always be symmetrical or not... if they're not then obviously symmetry wouldnt work... But in this case yes i think that's right. For a calculation that is exactly what you would show. I actually dk how to show the transformations thing as working... so i guess thats right. but going back to the basics actually just helps me understand it better...

 

[Ayesha B]

In this ques tho they dont look for working, they just want the coords, so any way would be acceptable

 

[urwahboy]

Hmmm... good point. I'm just wondering if these kinda graphs would always be symmetrical or not... if they're not then obviously symmetry wouldnt work... But in this case yes i think that's right. For a calculation that is exactly what you would show. I actually dk how to show the transformations thing as working... so i guess thats right. but going back to the basics actually just helps me understand it better...

symmetry will work if there is only one graph.. if two line with differemt equations are joined then symmetry wont work but they gave only one equation and the graph was linear so it worked .. there is another way if u wanna show it with working .. but its complicated.. first find the eq of the line at the right hand side from P and Q.. the equation comes -x+1 then find the equation of the line at left hand side we can esitmate it to be x+3(from the basics u explained) then equate the two equation x+3=-x+1 this also gives x=-1 ... if we cant estimate then we can use the method given in the book at pg no. 70 example 9. and example 10.
first we use original then equate it to the reflected.. 2+(-|x+1|)=2-(-|x+1) ... equating give x=1 but this is not possible as seen from the graph so we will take x=-1

 

[Ayesha B]

symmetry will work if there is only one graph.. if two line with differemt equations are joined then symmetry wont work but they gave only one equation and the graph was linear so it worked .. there is another way if u wanna show it with working .. but its complicated.. first find the eq of the line at the right hand side from P and Q.. the equation comes -x+1 then find the equation of the line at left hand side we can esitmate it to be x+3(from the basics u explained) then equate the two equation x+3=-x+1 this also gives x=-1 ... if we cant estimate then we can use the method given in the book at pg no. 70 example 9. and example 10.
first we use original then equate it to the reflected.. 2+(-|x+1|)=2-(-|x+1) ... equating give x=1 but this is not possible as seen from the graph so we will take x=-1

Mhmm yeah original and reflected lines can be equated... good

So basically there's many ways to do it, whatever you're most comfortable with

 

[Ayesha B]

I need help...

June 2009
7(c)

Find the exact values of x for which e^x/(e^x - 2)^2 = 1

Can someone do this?

 

[Hadi Murtaza]

eˣ / (eˣ - 2)² = 1
eˣ = (eˣ - 2)²
eˣ = e²ˣ - 4eˣ + 4
e²ˣ - 5eˣ + 4 = 0

Let eˣ = a
a² - 5a + 4 = 0
a² - 4a - a + 4 = 0
a(a - 4) - 1(a - 4) = 0
(a - 4)(a - 1) = 0
a = 1 , a = 4

Therefore
eˣ = 1 , eˣ = 4
x = ln(1) , x = ln(4)
[ x = 0 , ln(4) ]

 

[Ayesha B]

eˣ / (eˣ - 2)² = 1
eˣ = (eˣ - 2)²
eˣ = e²ˣ - 4eˣ + 4
e²ˣ - 5eˣ + 4 = 0

Let eˣ = a
a² - 5a + 4 = 0
a² - 4a - a + 4 = 0
a(a - 4) - 1(a - 4) = 0
(a - 4)(a - 1) = 0
a = 1 , a = 4

Therefore
eˣ = 1 , eˣ = 4
x = ln(1) , x = ln(4)
[ x = 0 , ln(4) ]

Thank youuu!

 

[Hadi Murtaza]

Thank youuu!

Ur most welkum