# `f(x) = ksqrt(x), y = x + 4` Find k such that the line is tangent to the graph of the function.

The tangent line will touch one point of the original function.

Set the two equations equal to each other. Rewrite the square root as a fractional power.

`kx^(1/2)= x+4`

We need a relationship of x and k since we have 2 unknown variables.

Take the derivative of f(x) and set the derivative equal to the slope of the tangent line equation, which is 1, and solve for x.

`f(x)= kx^(1/2)`

`f'(x) = 1/2(k)x^(-1/2)`

`1=1/2(k)x^(-1/2)`

`2= kx^(-1/2)`

`2= k * (1/ x^(1/2))`

`k=2x^(1/2)`

Square both sides.

`k^2 = 4x`

`x= k^2 /4`

Substitute the x back into the first equation.

`k(k^2 /4)^(1/2)= (k^2/4)+4`

`k(k/2)= k^2/4 +4`

`k^2/2 = k^2/4+4`

Subtract `k^2/4` on both sides.

`k^2/4=4`

`k^2 = 16`

`k=+-4`

If we plugged both numbers back to recheck, only `k=4` will work.

The answer is: