Find the equation of the tangent to the curve at the given x value. y=sqrt(x^2+3),x=1

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- IsTim

Find the equation of the tangent to the curve at the given x value. y=sqrt(x^2+3),x=1

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- IsTim

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That was my final answer.

- IsTim

The correct answer os x-2y+3=0

- anonymous

find the derivative, replace x by 1 to get your slope, then use point slope formula

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## More answers

- IsTim

I tried to, but I reached a point where my denominator equals 0.

- IsTim

unless I can factor "h" out of my answer?

- anonymous

\[f(x)=\sqrt{x^2+3}\]
\[f'(x)=\frac{x}{\sqrt{x^2+3}}\]
\[f'(1)=\frac{1}{2}\] is your slope, and your point is \((1,f(1))\) i.e. \((1,2)\)

- IsTim

We're not using f′(x) yet.

- anonymous

really? then what are you supposed to do ?

- IsTim

Well, we're either using substitution, factoring, or the conjugate method.

- IsTim

Oops, I think I forgot to properly do the Conjugate Method
(Working off the Calculus Phobe Videos at Calculis Help)

- anonymous

oh ok, in that case start with
\[\frac{\sqrt{1+h+3}-2}{h}\] or rather
\[\frac{\sqrt{4+h}-2}{h}\] and multiply top and bottom by conjugate

- anonymous

they still have calculus phobe videos? i thought they were gone

- IsTim

My teacher showed me in class.

- anonymous

ok that was wrong, i forgot about the square
start with
\[\frac{\sqrt{(1+h)^2+3}-2}{h}\]

- anonymous

or with
\[\frac{\sqrt{1+2h+h^2+3}-2}{h}=\frac{\sqrt{4+2h+h^2}-2}{h}\] and then multiply by the conjugate

- IsTim

@satellite73 Umm, my final answer is 1/2x+1.5, which isn't the correct answer...

- anonymous

well slope is \[\frac{1}{2}\] and the point is \((1,2)\) so lets see
\[y-2=\frac{1}{2}(x-1)\]
\[y=\frac{1}{2}x+\frac{3}{2}\] looks good to me

- IsTim

They're saying its x-2y+3=0

- anonymous

mutiply by 2 and put everything on one side of the equal sign and you get the same thing

- IsTim

Just wondering, how would I know to multiply by 2 on a test?

- IsTim

@satellite73 Just wondering, how would I know to multiply by 2 on a test?

- anonymous

no idea. i guess it says "standard form" or something? it matters not

- IsTim

So I guess the teacher wouldn't deduct me for marks on a test for something like that right?

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