# `f(x) = k/x, y = (-3/4)x + 3` Find k such that the line is tangent to the graph of the function.

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### 1 Answer

The original equation will intersect a point on the tangent line.

Set both equations equal to each other.

`k/x = -3/4 x +3`

We have 2 unknowns and 1 equation.

Take the derivative of f(x) in order to find another relationship between k and x.

`f(x)=k/x`

`f(x)=kx^(-1)`

`f'(x) = -kx^-2`

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

Substitute the slope `-3/4` into `f'(x)` .

`-3/4=-k/x^2`

Cross multiply.

`3x^2=4k`

`k=(3x^2) /4`

Substitute k back to the original equation to solve for x.

`k/x = -3/4 x +3`

`k* 1/x = -3/4 x +3`

`((3x^2)) /4 * 1/x = -3/4 x +3`

` 3/4 x= -3/4 x+3`

`6/4 x = 3`

`3/2 x = 3`

`3x = 6`

`x=2`

Substitute the x value back to .

`k=(3(2)^2) /4 = (3*4) /4 `

The answer is: `k=3`