Question 1183695
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Find the value of K for which y=x+k is a tangent to the curve y=x²+5x+2
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<pre>
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 = {{{(1-5)/2}}} = {{{-4/2}}} = -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.


<U>ANSWER</U>.  The value of "k" is -2.



                         <U>VISUAL CHECK</U>



    {{{graph( 400, 400, -5, 5, -10, 10,        
              x^2 + 5x + 2, x - 2 
)}}}


     Plot y = x^2 + 5x + 2 (red),  y = x - 2 (green)
</pre>

Solved.