# Find the y-intercept of the tangent line to: -.4/ sqrt(3 + x) at

by Lo.Lee.Ta.
Tags: 4 or, line, sqrt3, tangent, yintercept
 P: 209 1.Find the y-intercept of the tangent line to: y= -.4/√(3 + x) at [2.5, -.170560573084488] 2. So I thought the first step would be to find the slope of the tangent line. I think we find the slope of the tangent line by taking the derivative. So I am going to use the Quotient Rule to take the derivative. [(√3 + x)(0) - (-.4)(.5(x + 3)^-1/2) * 1] / √(x + 3)^2 .2√(x + 3) / (x + 3) <----- This is the slope of the tan line to y= -.4/√(3 + x) If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x? So, [.2√(3 + 2.5)]/(2.5 + 3) = .085<----- This is the slope Now, I thought we could substitute everything into y = mx + b in order to find out what b is. -.1705606 = .085(2.5) + b b = -.3831 3. But this is NOT the right answer! Please tell me what I am doing wrong! Thank you! :)
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P: 25,235
 Quote by Lo.Lee.Ta. 1.Find the y-intercept of the tangent line to: y= -.4/√(3 + x) at [2.5, -.170560573084488] 2. So I thought the first step would be to find the slope of the tangent line. I think we find the slope of the tangent line by taking the derivative. So I am going to use the Quotient Rule to take the derivative. [(√3 + x)(0) - (-.4)(.5(x + 3)^-1/2) * 1] / √(x + 3)^2 .2√(x + 3) / (x + 3) <----- This is the slope of the tan line to y= -.4/√(3 + x) If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x? So, [.2√(3 + 2.5)]/(2.5 + 3) = .085<----- This is the slope Now, I thought we could substitute everything into y = mx + b in order to find out what b is. -.1705606 = .085(2.5) + b b = -.3831 3. But this is NOT the right answer! Please tell me what I am doing wrong! Thank you! :)
Your derivative is wrong. What is u^(-1/2)/u? You aren't combining the exponents correctly. u^(-1/2)=1/sqrt(u). Not sqrt(u).
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,722 For calculating the derivative, instead of using the quotient rule, rewrite the equation: y = -0.4/SQRT(x+3) as y = -0.4 * (x+3)^(-1/2) Now, you have an eqn. of the form y = a u^n and you can use the power rule to get the derivative. The Quotient Rule is useful when you are dealing with the ratio of two functions.
Mentor
P: 21,397
Find the y-intercept of the tangent line to: -.4/ sqrt(3 + x) at

 Quote by Lo.Lee.Ta. 1.Find the y-intercept of the tangent line to: y= -.4/√(3 + x) at [2.5, -.170560573084488] 2. So I thought the first step would be to find the slope of the tangent line. I think we find the slope of the tangent line by taking the derivative. So I am going to use the Quotient Rule to take the derivative.
A better choice (because it's simpler and less prone to errors) would be write your equation as y = -.4 * (x + 3)-1/2
 Quote by Lo.Lee.Ta. [(√3 + x)(0) - (-.4)(.5(x + 3)^-1/2) * 1] / √(x + 3)^2
It's a good idea to keep track of what you're doing. For the above, you're working with the derivative, so this line should start with y' = or dy/dx = .

I see that you are including brackets around your numerator, so you've been listening. Good!

You have a mistake above that might just be a transcription error. At the beginning you have (√3 + x)(0) ...
That should be [(√(3 + x)(0)
 Quote by Lo.Lee.Ta. .2√(x + 3) / (x + 3) <----- This is the slope of the tan line to y= -.4/√(3 + x)
You have a mistake above. You should have gotten
dy/dx = .2(x + 3)-1/2/(x + 3)

 Quote by Lo.Lee.Ta. If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x? So, [.2√(3 + 2.5)]/(2.5 + 3) = .085<----- This is the slope Now, I thought we could substitute everything into y = mx + b in order to find out what b is. -.1705606 = .085(2.5) + b b = -.3831 3. But this is NOT the right answer! Please tell me what I am doing wrong! Thank you! :)
 Mentor P: 21,397 To expand on what SteamKing said, you should NEVER use the quotient rule if either the numerator or denominator is a constant. It's not incorrect to use it in this case, but it's more complicated, and the chances are greater of making a mistake. For example, if f(x) = 2/(x2 + 1), write this as f(x) = 2(x2 + 1)-1 and use the chain rule to get f'(x) = 2(-1)(x2 + 1)-2(2x).
 P: 209 When I was working it out on paper, I left out the (-) on the -1/2 exponent! Thanks for pointing that out! So the Quotient Rule method is: [.4(1/2(x + 3)^-1/2)]/(x + 3) .2/(x + 3)^1/2 * (x + 3) = .2/ (x + 3)^3/2 If I substitute the x = 2.5, then I get: .0155 as the slope. I tried the Product Rule as well to see if I'd get the same thing. So, -.4*-1/2(x + 3)^-3/2 + [(x + 3)^-1/2](0) = .2(x +3)^-3/2 When substituting x = 2.5, I get the same slope of .0155 When I substitute everything into y = mx + b, I get: -.17056 = .0155(2.5) + b b = -.209 ...But this is not the right answer... Do you know what could be wrong? Is my derivative still wrong? Thanks! :)