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 2 years ago
Find the equation of the tangent to the curve at the given x value. y=sqrt(x^2+3),x=1
 2 years ago
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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1335241905273:dw That was my final answer.

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0The correct answer os x2y+3=0

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1find the derivative, replace x by 1 to get your slope, then use point slope formula

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0I tried to, but I reached a point where my denominator equals 0.

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0unless I can factor "h" out of my answer?

satellite73
 2 years ago
Best 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. \((1,2)\)

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0We're not using f′(x) yet.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1really? then what are you supposed to do ?

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0Well, we're either using substitution, factoring, or the conjugate method.

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1oh 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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1they still have calculus phobe videos? i thought they were gone

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0My teacher showed me in class.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1ok that was wrong, i forgot about the square start with \[\frac{\sqrt{(1+h)^2+3}2}{h}\]

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1well 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

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0They're saying its x2y+3=0

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1mutiply by 2 and put everything on one side of the equal sign and you get the same thing

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1no idea. i guess it says "standard form" or something? it matters not

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