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...
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= 2^{2}+ 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= x^{2}+ bx+ c at x= 2?
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.