equation of a tangent line

jayjay60 · #1 $Old$ September 23rd, 2009, 01:05 PM

hey guys. i'm new to the forum. i can tell i may end up here a lot. but for now, i cannot figure this problem out on my own! i'm trying hard, but i really need help...please kthx?

Find an equation of the tangent line to the curve at the point (7, 4).

i have another problem too. it just seems too easy.

Find f '(a). f(x) = 8 - 6x + 5x2
f '(a) =

thank you. i'm trying, but it seems i am not good at this kind of stuff. x_x

e^(i*pi) · #2 $Old$ September 23rd, 2009, 01:09 PM

Quote:

Originally Posted by jayjay60 $View Post$

hey guys. i'm new to the forum. i can tell i may end up here a lot. but for now, i cannot figure this problem out on my own! i'm trying hard, but i really need help...please kthx?

Find an equation of the tangent line to the curve at the point (7, 4).

i have another problem too. it just seems too easy.

Find f '(a). f(x) = 8 - 6x + 5x2
f '(a) =

thank you. i'm trying, but it seems i am not good at this kind of stuff. x_x

For number 1 differentiate using the quotient rule:

$y' = \frac{u'v - uv'}{v^2}$

where u is the numerator and v the denominator.

EDIT: once you've differentiated it find $f'(7)$ because that's the point of the x-coordinate. Now you have the gradient use the equation of a straight line to find the equation of the tangent: $y-y_1 = m(x-x_1)$ or $y=mx+c$

For 2 I get an answer of $f'(x) = 10x-6 = 2(5x-3)$ . Not sure what your answer is

Calculus26 · #3 $Old$ September 23rd, 2009, 01:10 PM

for y= f(x)

the general form of the tangent line at (a,f(a)) is :

y= f '(a)(x-a) +f(a)

in your eg

y = f ' (7)(x-7) + 4

your 2d pblm is as easy as you think

artvandalay11 · #4 $Old$ September 23rd, 2009, 01:11 PM

Quote:

Originally Posted by jayjay60 $View Post$

hey guys. i'm new to the forum. i can tell i may end up here a lot. but for now, i cannot figure this problem out on my own! i'm trying hard, but i really need help...please kthx?

Find an equation of the tangent line to the curve at the point (7, 4).

i have another problem too. it just seems too easy.

Find f '(a). f(x) = 8 - 6x + 5x2
f '(a) =

thank you. i'm trying, but it seems i am not good at this kind of stuff. x_x

Alright let's bang the second problem out first. It asks you to find out what the derivative of f(x) is when evaluated at the point "a".

So $f(x)=8-6x+5x^2$ and so $f'(x)=-6+10x$

Now $f'(a)=-6+10(a)=-6+10a$

Now for the second problem, we need the slope of the tangent line at the point (7,4) which is the slope of the curve at that point. So we need to find y' using the quotient rule

$y=\frac{x-3}{x-6}$

$y'=\frac{(x-6)(1)-(x-3)(1)}{(x-6)^2}$

Now we plug in (7,4) to get the slope of the tangent line. And using point slope form, the equation of the line is $y-4=y'(7)(x-7)$

Do you think you can fill in the details?

e^(i*pi) · #5 $Old$ September 23rd, 2009, 01:14 PM

Quote:

Originally Posted by artvandalay11 $View Post$

Alright let's bang the second problem out first. It asks you to find out what the derivative of f(x) is when evaluated at the point "a".

So $f(x)=8-6x-5x^2$ and so $f'(x)=-6-10x$

Now $f'(a)=-6-10(a)=-6-10a$

Where did you get those minus signs from? I thought the question said $8 - 6x+5x^2$

artvandalay11 · #6 $Old$ September 23rd, 2009, 01:15 PM

good call

jayjay60 · #7 $Old$ September 23rd, 2009, 01:27 PM

i thought it was f(x+a)-f(x)/h...? o_O,,
oh wait...that's how f'(x) though right?

idk...i have been having real probs with the first one. it's wrong, i can tell you that LOL.

LOL. it's ok. i was doing it right but it turns out i was plugging in "X" instead of "A"
man, sometimes they just trick you to no end.... :/ yikes

e^(i*pi) · #8 $Old$ September 23rd, 2009, 01:37 PM

Quote:

Originally Posted by jayjay60 $View Post$

idk...i have been having real probs with the first one. it's wrong, i can tell you that LOL.

$u = x-3 \: \rightarrow \: u' = 1$

$v = x-6 \: \rightarrow \: v' = 1$

$y' = \frac{1 \times (x-6) + 1 \times (x-3)}{(x-6)^2}$

$= \frac{x-6+x-3}{(x-6)^2} = \frac{2x-9}{(x-6)^2}$

$f'(7) = \frac{2(7)-9}{(7-6)^2} = 5$

$(x_1 ,y_1) = (7,4)$

$y-y_1 = m(x-x_1)$

$y-4 = 5(x-7) \: \rightarrow \: y = 5x - 31$

EDIT: $y=5x-31$ may also be expressed as $5x-y-31=0$

jayjay60 · #9 $Old$ September 23rd, 2009, 01:39 PM

alrighty guys...i got it all it was

(7,4)

y'=-3/(x-6)^2
substitute 7 for x and the slope of the tangent line=y'=-3
use point slope form of the equation of a line y-ysub1=m(x-xsub1) or y-4=-3(x-7)
simplify and y=-3x+25

whew! thank you all for all your help. x_x
and don't doubt for a second that: i'll be baaaaach X_X

guess · #10 $Old$ September 23rd, 2009, 07:21 PM

Quote:

Originally Posted by e^(i*pi) $View Post$

$u = x-3 \: \rightarrow \: u' = 1$

$v = x-6 \: \rightarrow \: v' = 1$

$y' = \frac{1 \times (x-6) + 1 \times (x-3)}{(x-6)^2}$

$= \frac{x-6+x-3}{(x-6)^2} = \frac{2x-9}{(x-6)^2}$

$f'(7) = \frac{2(7)-9}{(7-6)^2} = 5$

$(x_1 ,y_1) = (7,4)$

$y-y_1 = m(x-x_1)$

$y-4 = 5(x-7) \: \rightarrow \: y = 5x - 31$

EDIT: $y=5x-31$ may also be expressed as $5x-y-31=0$

Hi,

I think there might be some mistake here.

Shouldn't the quotient rule be:

(f(v)f'(u)-f(u)f'(v)) / (f(v))^2.

Hence the ans will be

y=-3x+25.

Or you can solve it in a easier way.

(x-3)/(x-6) = (x-6)/(x-6) + 3/(x-6) <-- When you decompose the numerator

f(x)=1+3(x-6)^-1

f'(x)=-3(x+6)^-2

f'(7)=-3

Equation of tangent:

y-y1=m(x-x1)

y-4=-3(x-7)

y-4=-3x+21

y=-3x+25

hope that you can understand