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

*print*Print*list*Cite

### 1 Answer

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: