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

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IsTimBest ResponseYou've already chosen the best response.0
dw:1335241905273:dw That was my final answer.
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
The correct answer os x2y+3=0
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
find the derivative, replace x by 1 to get your slope, then use point slope formula
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
I tried to, but I reached a point where my denominator equals 0.
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
unless I can factor "h" out of my answer?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\[f(x)=\sqrt{x^2+3}\] \[f'(x)=\frac{x}{\sqrt{x^2+3}}\] \[f'(1)=\frac{1}{2}\] is your slope, and your point is \((1,f(1))\) i.e. \((1,2)\)
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
We're not using f′(x) yet.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
really? then what are you supposed to do ?
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Well, we're either using substitution, factoring, or the conjugate method.
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Oops, I think I forgot to properly do the Conjugate Method (Working off the Calculus Phobe Videos at Calculis Help)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
they still have calculus phobe videos? i thought they were gone
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
My teacher showed me in class.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
ok that was wrong, i forgot about the square start with \[\frac{\sqrt{(1+h)^2+3}2}{h}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
@satellite73 Umm, my final answer is 1/2x+1.5, which isn't the correct answer...
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
well slope is \[\frac{1}{2}\] and the point is \((1,2)\) so lets see \[y2=\frac{1}{2}(x1)\] \[y=\frac{1}{2}x+\frac{3}{2}\] looks good to me
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
They're saying its x2y+3=0
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
mutiply by 2 and put everything on one side of the equal sign and you get the same thing
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Just wondering, how would I know to multiply by 2 on a test?
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
@satellite73 Just wondering, how would I know to multiply by 2 on a test?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
no idea. i guess it says "standard form" or something? it matters not
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
So I guess the teacher wouldn't deduct me for marks on a test for something like that right?
 one year ago
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