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find slope of tangent line y=f(x)=xsquared + 1 at (2,5) given f'(x)=2x
 2 years ago
 2 years ago
find slope of tangent line y=f(x)=xsquared + 1 at (2,5) given f'(x)=2x
 2 years ago
 2 years ago

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cwrw238Best ResponseYou've already chosen the best response.0
the slope = 2x at (2,5) x = 2 so slope of tangent = 2(2) = 4
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
So your question is to find the slope of the equation\(f(x)=x^2+1\) at (2,5) given f'(x)=2x
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
joycelefebvre are you new to OS?
 2 years ago

joycelefebvreBest ResponseYou've already chosen the best response.0
yes could you show me your work on that please i'm studying for praxis 0061 my study guied doesnt show work its horrible tu
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Given any function, the derivative of the function is the equation to the slope of the given equationdid you know that?
 2 years ago

joycelefebvreBest ResponseYou've already chosen the best response.0
kinda I took calc in college in 93 and its not coming back
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
OK, then I am trying to refresh that. Suppose you have a function y=f(x), so the function y=f'(x) is the equation of the tangent of the function y=f(x)
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
So your given function is \(f(x)=x^2+1\) and \(f'(x)=2x\)
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Now as I told you earlier, \(f'(x)\) should be the equation of the tangent at any point
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Your question asks you to find the equation of the tangent at (2,5) so just put x=2 in the equation to tangent i.e. 2x2=4. So the answer is 4 Got it?
 2 years ago

joycelefebvreBest ResponseYou've already chosen the best response.0
omg I was trying to do y2y1/x2x1
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
This is actually a theorem.. If you just remember it you don't have to find the tangent that way.......... I mean the way you are trying using the formula \(\frac {y2y1}{x2x1}\)
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
The theorem is, Let y=f(x) be a given equation, to find the equation to the tangent at any point is y=f'(x) where f'(x) is the derivative of the function f(x)
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Is that clear, any problem?
 2 years ago

joycelefebvreBest ResponseYou've already chosen the best response.0
tu i couldn't find the praxis study guied for 0061 so i got the cliff notes and its all text no examples
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Did you know you can chat with other users in the maths group?
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
In that way you can get help from experts?
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Come join me in the chat room (do you know how?)
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Do you see those green boxes below
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
See there I have posted you a chat message
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
Yes, I am 31, thats my level.
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
But chatting is a different thing
 2 years ago

joycelefebvreBest ResponseYou've already chosen the best response.0
my lap top keeps telling me i cant run this script
 2 years ago

2bornot2bBest ResponseYou've already chosen the best response.2
OH, I see, what operating system are you using?
 2 years ago
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