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):
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)
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.,
