Determine the value of k if g(x)=4x-k is a tanget to f(x)=-x²+8x+20​

This seems to be incomplete because we do not know where g(x) must be tangent to f(x). So i will assume that we want g(x) to be a tangent to f(x) in the point x0.

We know that for a function f(x), f'(x) is the tangent function in at the point x.

Where f'(x) = df(x)/dx

So we can find a value of k that is equal (for some value of x) to the first derivative of f(x).

g(x) = 4*x - k

f(x) = -x^2 + 8*x + 20

f'(x) = -2*x + 8

Now we want to find a value of k such that:

4*x0 - k = -2*x0 + 8

4*x0 - k = -2*x0 + 8

4*x0 + 2*x0 - 8 = k

6*x0 - 8 = k

Then if we want g(x) to be a tangent to f(x) = -x²+8x+20​ in the point x0, we must have k = 6*x0 - 8

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.