Calculus, again

  • Thread starter danne89
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  • #1
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Consider the pababola y=x^2+bx+c. Find the values of b and c such that the line y=2x is tangent to the point (2,4).

I've no clue at all...
 

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  • #2
HallsofIvy
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Really, no clue at all? Did it not occur to you that if the line is tangent to the parabola at the point (2,4), then the parabola must go through (2,4)- that is, that
4= 22+ b(2)+ c.

Has no one told you that the derivative at a point IS the slope of the tangent line at that point? What is the slope of the line y= 2x? Can you find the derivative of
y= x2+ bx+ c at x= 2?
 
  • #3
danne89 said:
Consider the pababola y=x^2+bx+c. Find the values of b and c such that the line y=2x is tangent to the point (2,4).

I've no clue at all...
You have y = 2x as the tangent line.. and you know that y' = 2x + b and thus you can substitute x = 2 into y' to get y' = 4 + b and from the question you know y' = 2... and so 2 = 4 + b; b = -2
Next step is finding c, just plug in:
4 = 4 - 4 + c
c = 4
And the parabola is y = x^2 - 2x + 4
And to test it..
y' = 2x - 2
And at the point (2,4); y - 4 = 2(x-2); y = 2x - 4 + 4 = 2x
And that's the answer...
Or I could be completely wrong. :approve:
 

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