IsTim
  • IsTim
Find the equation of the tangent to the curve at the given x value. y=sqrt(x^2+3),x=1
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

IsTim
  • IsTim
|dw:1335241905273:dw| That was my final answer.
IsTim
  • IsTim
The correct answer os x-2y+3=0
anonymous
  • anonymous
find the derivative, replace x by 1 to get your slope, then use point slope formula

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

IsTim
  • IsTim
I tried to, but I reached a point where my denominator equals 0.
IsTim
  • IsTim
unless I can factor "h" out of my answer?
anonymous
  • 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
  • IsTim
We're not using f′(x) yet.
anonymous
  • anonymous
really? then what are you supposed to do ?
IsTim
  • IsTim
Well, we're either using substitution, factoring, or the conjugate method.
IsTim
  • IsTim
Oops, I think I forgot to properly do the Conjugate Method (Working off the Calculus Phobe Videos at Calculis Help)
anonymous
  • 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
  • anonymous
they still have calculus phobe videos? i thought they were gone
IsTim
  • IsTim
My teacher showed me in class.
anonymous
  • anonymous
ok that was wrong, i forgot about the square start with \[\frac{\sqrt{(1+h)^2+3}-2}{h}\]
anonymous
  • 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
  • IsTim
@satellite73 Umm, my final answer is 1/2x+1.5, which isn't the correct answer...
anonymous
  • 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
  • IsTim
They're saying its x-2y+3=0
anonymous
  • anonymous
mutiply by 2 and put everything on one side of the equal sign and you get the same thing
IsTim
  • IsTim
Just wondering, how would I know to multiply by 2 on a test?
IsTim
  • IsTim
@satellite73 Just wondering, how would I know to multiply by 2 on a test?
anonymous
  • anonymous
no idea. i guess it says "standard form" or something? it matters not
IsTim
  • IsTim
So I guess the teacher wouldn't deduct me for marks on a test for something like that right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.