How do I determine the value of k such that the graph of the function is tangent to the line?

Question

Determine the value of k such that the graph of the function is tangent to the line y = 4.

y = -x^2 + 2+ k

y = x^2 - 4kx + 8

Thanks for your help. :)

Answer 1

y = 4 is a horizontal line.  It's slope is 0.  So, you're looking for values of k such that the slope is 0 and somewhere the function touches the line.

For the first function, that's easy.  dy/dx = -2x = 0 at x = 0.  So, we need k such that the value of the function at x = 0 is 4.  -(0^2) + 2 + k = 4, obviously k = 2.  So, y = -x^2 + 4.  At x = 0, the tangent is horizontal and the function touches y = 4.

For the second:

y = x^2 - 4kx + 8

We need a value of k such that at some point A the derivative is 0 and the function value is 4.  So,

y' = 2x - 4k

y'(A) = 2A - 4k = 0

k = A/2

y(A) = A^2 - 4kA + 8 = 4

We have two unknowns:  A and k.  We have two equations.  This can be solved.