graph of the nth power function !! :( help!

hopes

i hate this ..... i cant seem to understand what the hell to do. any help will be greatly appreciated.

How much do you know about differentiation? In each case you need to calculate the gradient of the desired line and then plug the relevant values into

y-y_1=m(x-x_1)

Reply 2

OP

BJack

How much do you know about differentiation? In each case you need to calculate the gradient of the desired line and then plug the relevant values into

y-y_1=m(x-x_1)

blah?

as far as i know its dy/dx right? so i got to 4x^-2 for the second one and dy/dx for that is -8? so the gradient of the normal is 1/8 bt when i plug the values in the answer doesnt match whats behind the book. i got 4y=x-3??

Reply 3

OP

im relying on u guys 2 help me out here!

Reply 4

Gowtham1991

Your differentiation is wrong. Do it correctly, and you ll get the right answer :smile:

Reply 5

OP

hmm i still cant seem to get the it right!?

Reply 6

hopes

hmm i still cant seem to get the it right!?

Let's look at the first one. Is it

\displaystyle y=\frac{x-9}{x}

or

\displaystyle y=x-\frac{9}{x}

?

Reply 7

OP

BJack

Let's look at the first one. Is it

\displaystyle y=\frac{x-9}{x}

or

\displaystyle y=x-\frac{9}{x}

?

its the second one but its x PLUS not minus 9/x

Reply 8

Sculler

5

hopes

its the second one but its x PLUS not minus 9/x

And what do you get when you differentiate it?

Reply 9

OP

Sculler

And what do you get when you differentiate it?

x+9x^-1= -9??

Reply 10

hopes

x+9x^-1= -9??

I think you must be going wrong with your differentiation.

\frac{d}{dx} ax^n = nax^{n-1}

hopes

its the second one but its x PLUS not minus 9/x

Oh yes, so it is. Any way, you should be able to work out that if

y=x+\frac{9}{x}

then

\frac{dy}{dx} = 1 - \displaystyle \frac{9}{x^2}

So the gradient when x=3 comes out as 0. Thus the equation for the tangent to the curve at x=3 is

y-6=0(x-3) \Rightarrow y=6

Is that correct? :s-smilie:

Reply 11

OP

BJack

Oh yes, so it is. Any way, you should be able to work out that if

y=x+\frac{9}{x}

then

\frac{dy}{dx} = 1 - \displaystyle \frac{9}{x^2}

So the gradient when x=3 comes out as 0. Thus the equation for the tangent to the curve at x=3 is

y-6=0(x-3) \Rightarrow y=6

Is that correct? :s-smilie:

my teacher didnt explain anything to us tbh, not using an excuse but how did u get from x+9/x to 1-9/x^2?

Reply 12

hopes

my teacher didnt explain anything to us tbh, not using an excuse but how did u get from x+9/x to 1-9/x^2?

If you don't understand this step, I think you need to go back and revise differentiation. This is fairly standard stuff so it's important that you understand it before you meet the trickier maths that's yet to come. :smile:

Reply 13

OP