Here's the question you clicked on:
IsTim
Find the equation of the tangent to the curve at the given x value. y=sqrt(x^2+3),x=1
|dw:1335241905273:dw| That was my final answer.
The correct answer os x-2y+3=0
find the derivative, replace x by 1 to get your slope, then use point slope formula
I tried to, but I reached a point where my denominator equals 0.
unless I can factor "h" out of my answer?
\[f(x)=\sqrt{x^2+3}\] \[f'(x)=\frac{x}{\sqrt{x^2+3}}\] \[f'(1)=\frac{1}{2}\] is your slope, and your point is \((1,f(1))\) i.e. \((1,2)\)
We're not using f′(x) yet.
really? then what are you supposed to do ?
Well, we're either using substitution, factoring, or the conjugate method.
Oops, I think I forgot to properly do the Conjugate Method (Working off the Calculus Phobe Videos at Calculis Help)
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
they still have calculus phobe videos? i thought they were gone
My teacher showed me in class.
ok that was wrong, i forgot about the square start with \[\frac{\sqrt{(1+h)^2+3}-2}{h}\]
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
@satellite73 Umm, my final answer is 1/2x+1.5, which isn't the correct answer...
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
They're saying its x-2y+3=0
mutiply by 2 and put everything on one side of the equal sign and you get the same thing
Just wondering, how would I know to multiply by 2 on a test?
@satellite73 Just wondering, how would I know to multiply by 2 on a test?
no idea. i guess it says "standard form" or something? it matters not
So I guess the teacher wouldn't deduct me for marks on a test for something like that right?