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Quick C1 question

This has preceding questions so I'll just fill you in:

Part I)

Describe fully the transformation that transforms the curve

y = \frac{1}{x}

to the curve

y = \frac{1}{x + 2}

.

I know that's a translation of 2 units in the negative x direction parallel to the x axis.

Then it asks to differentiate the first curve which gives

-x^{-2}

Now this last part I can't understand how I'm doing it wrong:

Part iii)

Use the preceding parts of the question to find the gradient of the curve

y = \frac{1}{x + 2}

at the point where it crosses the y axis.

Well surely when it crosses the y axis, x = 0 - if I differentiate the curve and put in x as zero you get infinity..?

Reply 1

Dededex

Hmm I understand but you get the same result dont you?

Call me an idiot but can it not be written as:

y = \frac{1}{2} + x^{-1}

Reply 2

Dededex

Differentiating that gives exactly the same as the first curve - anyone help? :smile:

Reply 3

GabsyWabsyWoo

Original post by Dededex

This has preceding questions so I'll just fill you in:

-x^{-2}

find the gradient of the curve

y = \frac{1}{x + 2}

at the point where it crosses the y axis.

Well surely when it crosses the y axis, x = 0 - if I differentiate the curve and put in x as zero you get infinity..?

I think it's asking for the gradient value of the (x+2)^-1 not the x^-1 graph,

for y = (x+2)^-1
when x = 0, y = 1/2

The y = (x+2)^-1 graph has only been translated to the left by 2 relative to y=x^-1
So the gradient values of the new curve could be thought of as translated to the left by 2 as well.

Sketch the graphs.

(edited 13 years ago)

Reply 4

Dededex

Original post by GabsyWabsyWoo

The y = (x+2)^-1 graph has only been translated to the left by 2 relative to y=x^-1
So the gradient values of the new curve could be thought of as translated to the left by 2 as well.

Sketch the graphs.

Kind of - I've drawn them but still don't understand why you cant get the correct answer with x = 0

Reply 5

GabsyWabsyWoo

Original post by Dededex

Kind of - I've drawn them but still don't understand why you cant get the correct answer with x = 0

on y=(x+2)^-1 the gradient at x=0 is what we want

on y=x^-1, the gradient at x=2 is -1/4 as found by the gradient function y'=-(x^-2)

if tangents are equal but translated we can assume what the y=(x+2)^-1 gradient is at x=0