slope of a tangent

**ffezz** · October 9th 2007, 04:15 PM

hi guys

i need to find the equation of the tangent at a certain point in the curve

the point it (1,2)

the equation is y= $sqrt(x^2 + 3)$

now here is what i have so far.

m = $f(x) - f(a) / x-a$
m = $sqrt (x^2 + 3) - sqrt(1^2 + 3) / x-1$

from that point i get lost. im assuming i have to multiply by the conjugate but i think im screwing something up when i do that.
If someone could do that with the inbetween steps I would appreciate it.
Also, teach me how to put things under a square root in the math tags XD
thanks

dont answer llawll i was factoring out nonexistent termx XD

**Jhevon** · October 9th 2007, 04:17 PM

Originally Posted by ffezz

hi guys

i need to find the equation of the tangent at a certain point in the curve

the point it (1,2)

the equation is y= $sqrt(x^2 + 3)$

now here is what i have so far.

m = $f(x) - f(a) / x-a$
m = $sqrt (x^2 + 3) - sqrt(1^2 + 3) / x-1$

from that point i get lost. im assuming i have to multiply by the conjugate but i think im screwing something up when i do that.
If someone could do that with the inbetween steps I would appreciate it.
Also, teach me how to put things under a square root in the math tags XD
thanks

you are in calculus right? the slope is given by the derivative. find the derivative and plug in x = 1, that will give you the slope of the tangent line, then use the point-slope form ( $y - y_1 = m(x - x_1)$ ) to find the equation of the tangent line

**ffezz** · October 9th 2007, 04:19 PM

no im not in calculus. next semester. the think tank in ontario changed the cariculum for advanced functions so we dont actually get into any calculus untill I get to my calculus class. no idea what a derivative is. maybe i do but it doesnt ring a bell

**Jhevon** · October 9th 2007, 04:22 PM

Originally Posted by ffezz

no im not in calculus. next semester. the think tank in ontario changed the cariculum for advanced functions so we dont actually get into any calculus untill I get to my calculus class. no idea what a derivative is. maybe i do but it doesnt ring a bell

i'm sorry, but what you are asking here requires calculus. maybe you are using the definition of the derivative but they haven't told you yet what that is called. you are asking for the tangent at a point, so we need the instantaneous slope.

does $\lim_{x \to a}\frac {f(x) - f(a)}{x - a}$ ring a bell?

**ffezz** · October 9th 2007, 04:40 PM

I learnt something tonight XD
but Im messing up on my conjugate multiplication I think. Im getting a slope of -1/2 but should be getting 1/2.
heres what I did.

$m = \frac {sqrt(x^2 + 3)- 2 }{x-1}$ x $\frac {sqrt(x^2 + 3) + 2 }{sqrt(x^2 + 3) + 2}$

which gives me

$m = \frac {x^2 + 3 - 4}{ (x-1)(sqrt(x^2 +3) +2)}$

then i factored the top to (x-3)(x-1) and removed the factor of x-1 so im left with

$m = \frac {x - 3}{sqrt(x^2 + 3) + 2}$
I can put in my numbers from there on and get -2/4 which is -1/2

**Jhevon** · October 9th 2007, 04:58 PM

Originally Posted by ffezz

I learnt something tonight XD
but Im messing up on my conjugate multiplication I think. Im getting a slope of -1/2 but should be getting 1/2.
heres what I did.

$m = \frac {sqrt(x^2 + 3)- 2 }{x-1}$ x $\frac {sqrt(x^2 + 3) + 2 }{sqrt(x^2 + 3) + 2}$

which gives me

$m = \frac {x^2 + 3 - 4}{ (x-1)(sqrt(x^2 +3) +2)}$

then i factored the top to (x-3)(x-1) and removed the factor of x-1 so im left with

$m = \frac {x - 3}{sqrt(x^2 + 3) + 2}$
I can put in my numbers from there on and get -2/4 which is -1/2

you are indeed doing what i said. you are taking the limit. however, this is incorrect. you should end up with $\frac {x + 1}{\sqrt{x^2 + 3} + 2}$ and then plug in x = 1 (that is, take the limit as x goes to 1

**ffezz** · October 9th 2007, 04:59 PM

yus i figured it out! thanks so much

**Jhevon** · October 9th 2007, 05:07 PM

Originally Posted by ffezz

yus i figured it out! thanks so much

ok, so what is the equation of the tangent line?

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