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joycelefebvre
find slope of tangent line y=f(x)=xsquared + 1 at (2,5) given f'(x)=2x
the slope = 2x at (2,5) x = 2 so slope of tangent = 2(2) = 4
So your question is to find the slope of the equation\(f(x)=x^2+1\) at (2,5) given f'(x)=2x
joycelefebvre are you new to OS?
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
Given any function, the derivative of the function is the equation to the slope of the given equation--------did you know that?
kinda I took calc in college in 93 and its not coming back
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)
So your given function is \(f(x)=x^2+1\) and \(f'(x)=2x\)
Now as I told you earlier, \(f'(x)\) should be the equation of the tangent at any point
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?
omg I was trying to do y2-y1/x2-x1
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 \(\frac {y2-y1}{x2-x1}\)
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)
Is that clear, any problem?
tu i couldn't find the praxis study guied for 0061 so i got the cliff notes and its all text no examples
Did you know you can chat with other users in the maths group?
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my lap top keeps telling me i cant run this script
OH, I see, what operating system are you using?