Finding a tangent point to a curve given the x-intercept of the tangent line?

Question

The line tangent to the graph of f at the point (k,f(x)) intercepts the x-axis at x=4. What is the value of K if f(x)=12-x^2 for x is greater than or equal to 0 and f(x) is greater than or equal to 0 for this interval?

Answer 1

f(x) = 12-x²

 

Slope of tangent line at (k,f(k)) = f'(k) = -2k

 

Point on tangent line = (4, 0)

 

Equation of tangent line:  y-0 = (-2k)(x-4)

 

                                       y = (-2k)(x-4)

 

The tangent line and the graph of f(x) intersect when x = k.

 

So, 12-k² = (-2k)(k-4)

 

      12-k² = -2k²+8k

 

          k²-8k+12=0

 

          (k-6)(k-2)=0

 

          k = 2 or k = 6

 

Since f(x) is assumed to be positive but f(6)<0, we can disregard k=6.

 

k = 2 is the only value of k that satisfies the given conditions.