How Do U Find The Tangent Of This??

**calchelp** · June 22nd 2007, 07:39 PM

i really don't understand what to do...

it says to determine the equation of the tangen to the curve

y= 1/x at the point on the curve where x=2 ?????

and then the following question is like to find the equation of the tangent to y=x^2 -2y+5=0 which is perpendicular to the line x-2y+5=0
????? how do u do it wen u hav to hav find it perpendicular to that

**Jhevon** · June 22nd 2007, 08:21 PM

Originally Posted by calchelp

i really don't understand what to do...

it says to determine the equation of the tangen to the curve

y= 1/x at the point on the curve where x=2 ?????

The derivative gives the slope of the tangent line at any point. simply find the derivative and plug in x = 2 to find the slope at the point on the curve where x is 2. then after finding the corresponding y-coordinate, you can use the point-slope form of a line to find the tangent line. see my response below for directions to where you can get more information

and then the following question is like to find the equation of the tangent to y=x^2 -2y+5=0 which is perpendicular to the line x-2y+5=0
????? how do u do it wen u hav to hav find it perpendicular to that

I did a question very similar to this just a while ago. see here. if you don't get it say so...but please try to get it.

**calchelp** · June 22nd 2007, 09:09 PM

so u said to take the derivative and plug the number in...and then use the slope formula....please tell me im doing it the right way

y= x^-1
y'= -x^-2

f'(2)= -(2)^-2
=-1/4

m=(y2-y1)/(x2-x1)
m=(-1/4 - y)/(2-x)
m*(2-1)=(-1/4 - y)
what do i do from here?

**Jhevon** · June 22nd 2007, 09:11 PM

Originally Posted by calchelp

so u said to take the derivative and plug the number in...and then use the slope formula....please tell me im doing it the right way

y= x^-1
y'= -x^-2

f'(2)= -(2)^-2
=-1/4

m=(y2-y1)/(x2-x1)
m=(-1/4 - y)/(2-x)
m*(2-1)=(-1/4 - y)
what do i do from here?

m is the slope, so m should be -1/4. x = 2. you have to find what y is by plugging in 2 into the original equation, then use the formula $y - y_1 = m(x - x_1)$

EDIT: I have to leave now, but I see Krizalid online, he should be able to help you with any questions you have

**earboth** · June 23rd 2007, 01:31 AM

Originally Posted by calchelp

i really don't understand what to do...

it says to determine the equation of the tangen to the curve

y= 1/x at the point on the curve where x=2 ?????

...

Hello,

I'm going to take over all the results and put them together:

1. You know: $f(x) = \frac{1}{x}$ and $x = 2$

Now plug in the value 2 instead of x and you get: $f(2) = \frac{1}{2}$ That means: The point $T \left( 2, \frac{1}{2} \right)$ lies on the curve (the graph of f). It is the point where the tangent touches the curve.

2. The slope of the tangent has the same value as the derivative of the given function. You already did this part: $f'(x) = -\frac{1}{x^2}$ Plug in again the value 2 instead of x: $f'(2) = -\frac{1}{4}$

3. Now you know a point of the tangent (it's T) and the slope of the tangent. The point-slope-formula is:
$\frac{y - y_1}{x - x_1} = m$ where $(x_1 , y_1)$ are the coordinates of the point and m is the value of the slope. Plug in all values you know:

$\frac{y - \frac{1}{2}}{x - 2} = - \frac{1}{4}$ Now solve for y:

$y - \frac{1}{2} = - \frac{1}{4} \cdot (x-2)$

$y - \frac{1}{2} = - \frac{1}{4} \cdot x + \frac{1}{2}$

$y = - \frac{1}{4} \cdot x + 1$ And that's the equation of the tangent line.

**calchelp** · June 23rd 2007, 06:10 AM

hey thanks a lot i get it now

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