Find an equation for the tangent line to the curve Y=square root 3+x/4 at the point where x=-1. The square and x/4 in the square makes it difficult for me. My next step would be (3+1/4x)^1/2 (i think). Then 1/2(3+1/4x)^-1/2 and then after that I am lost. HELP!!!

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Find an equation for the tangent line to the curve Y=square root 3+x/4 at the point where x=-1. The square and x/4 in the square makes it difficult for me. My next step would be (3+1/4x)^1/2 (i think). Then 1/2(3+1/4x)^-1/2 and then after that I am lost. HELP!!!

Mathematics
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\[y=\sqrt{3+x/4}=(3+x/4)^{1/2}\] y'=\[0.5\times(3+x/4)^{-1/2}\times(1/4)\] plug -1 into the original equition we got x=-1 y=\[\sqrt{11}/2\] then plug -1 into y' got slope m by the function y=mx+b find the answer
I can get Y'=0.5(3+x/4)^-1/2 (1/4) but don't see where you are getting the answer. The square root is messing me up. HELP!!
m=y'(-1)=\[\sqrt{11}/44\] \[\sqrt{11}/2=\sqrt{11}/44\times(-1)+b\] \[b=23\sqrt{11}/44\] tangent line will be \[y=(\sqrt{11}/44) x+(23\sqrt{11}/44)\]

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I am sorry I am just not seeing it. What I see is: Y'=1/2(3+1/4x)^-1/2 (1/4). So if I replace X with -1 I get 1/2(3-1/4)^-1/2 (1/4). Then I subtract 3-1/4 and get 11/4. Am I doing this part wrong.
The derivative will give you the slope=m, plug in the x value into you original equation to obtain a y value at the point x=-1, then use the point slope formula to get the tangent line: y=y1+m(x-x1).... you have m form the derivative, you have x1 and y1 form the original equation.....
I think I got it now. I forgot to use the square root of 11/4 and forgot that I can take the square root of the 4 to get the square root 11/2. Thanks!
I know I am being a pain but how do you get the square root of 11/4? I have Y'= 1/2(3+x/4)^- 1/2 x (1/4). Then I replace X=-1 and get 1/2(3-1/4)^-1/2 (1/4) This is where I think I am going wrong with the problem, then I take (1/2)(1/4) (11/4)^-1/2.
This problem is driving me CRAZY!!!
your derivative is wrong
it should be \[y'=1/2(3+x/4)^{-1/2}(1/4)\] you have \[y'=1/2(3+x/4)^{-1/2}(1/4)*x\] the derivative of the inside function is \[1/4\] not \[(1/4)*x\]
Ok what do I do after 1/2(3+ x/4)^ -1/2(1/4)
I think I replace the X with -1 so I should have 1/2(3 - 1/4) ^ -1/2 (1/4)
plug in x=-1 to find the slope at -1 bc the derivative gives you the slope at any point of the fucntion
I think then I should do (1/2) (1/4) (3-1/4)^ -1/2
yes this will be the slope
so then I get 1/8 (11/4)^ -1/2
thats correct but this can be simplified
the ^ -1/2 is what is bothering me.
so you have\[(1/8)(3-1/4)^{-1/2}\] this turns into:\[(1/8)[(11/4)^{-1/2}] =(1/8)(1/\sqrt{11/4})\] \[=(1/8)(1/\sqrt{11}/\sqrt{4})\rightarrow (1/8)(\sqrt{4}/\sqrt{11})\] \[(1/8)(2/\sqrt{11})\rightarrow 2/8\sqrt{11}\rightarrow 1/4\sqrt{11}\] now rationalize the denominator: \[(1/4\sqrt{11})(\sqrt{11}/\sqrt{11})\rightarrow \sqrt{11}/(4\sqrt{11}\sqrt{11})\] you should finally get:\[\sqrt{11}/44\] as your slope
does it make sense?
the (1/8)(1/square of 11/square of 4) do you take the 11/4 and multiply 11/4 /1. I forgot what this is called. But that is how you got the 1 to go away. Hope this makes sense
Think of it this way how would you do this 1/1/2?
would it be 1/2 /1
no you flip the denominator so it would be 1/(1/2)....1*(2/1)=2
i know that. this problem just has me all confused!!!!
I'm sorry you were right earlier..... so you would do the same thing to 1/square11/square4
Thank you for spending this much time with me. I did not mean for that to sound mean if I did.
Now its fine..... I know how frustrating math can get and how tricky the algebra can be
may be this might help:\[a/(b/c)=ac/b \rightarrow a=1, b=\sqrt{11}, c=\sqrt{4}\]
I have trouble with fractions and square roots together
so with the identity i gave you above you would get:\[1/(\sqrt{11}/\sqrt{4})=(1*\sqrt{4})/\sqrt{11}\]

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