SOLUTION: Find the value of K for which y=x+k is a tangent to the curve y=x²+5x+2

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Question 1183695: Find the value of K for which y=x+k is a tangent to the curve y=x²+5x+2
Found 2 solutions by ikleyn, robertb:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
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Find the value of K for which y=x+k is a tangent to the curve y=x²+5x+2
~~~~~~~~~~~~~~~ 

Notice that the slope of the line y = x+k  is equal to  1  (one).


Hence, to solve the problem, we should find a point on the parabola, where the slope is 1,
and then to pass the line y = x+k through this point.


The derivative of the quadratic function is  y' = 2x + 5.

It is equal to  1  when  2x+5 = 1,  or  x =  =  = -2.


Next, the value of the quadratic function at x= -2 is  (-2)^2 + 5*(-2) + 2 = 4 - 10 + 2 = -4.


Now we find the value of "k" from this equation  x + k = -4  at  x= -2, i.e.

       -2 + k = -4   ---->    k = -4 + 2 = -2.


ANSWER.  The value of "k" is -2.



                         VISUAL CHECK



    


     Plot y = x^2 + 5x + 2 (red),  y = x - 2 (green)

Solved.



Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
Without using calculus, a simpler solution would be as follows:

Equate and .
===> , since at their intersection point(s) they're supposed to have the same y-values.
<===> .

Since the line is supposed to be tangent to the curve, it means that there are two identical roots for the preceding equation.
This only happens when the discriminant is equal to 0, i.e.,

===> ===> .

Solved.
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