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

1. IsTim

|dw:1335241905273:dw| That was my final answer.

2. IsTim

3. satellite73

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

4. IsTim

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

5. IsTim

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

6. satellite73

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

7. IsTim

We're not using f′(x) yet.

8. satellite73

really? then what are you supposed to do ?

9. IsTim

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

10. IsTim

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

11. satellite73

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

12. satellite73

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

13. IsTim

My teacher showed me in class.

14. satellite73

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

15. satellite73

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

16. IsTim

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

17. satellite73

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

18. IsTim

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

19. satellite73

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

20. IsTim

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

21. IsTim

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

22. satellite73

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

23. IsTim

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