The tangent line will touch a point on the original function f(x).

Set the two equations equal to each other since they intersect.

`kx^2=-2x+3`

We will need another relationship since we have two unknown variables.

Take the derivative of f(x) and set the derivative function equal to the slope of the tangent line, which is equal to negative 2. The k is a constant.

`f'(x)= 2kx`

`-2 = 2kx`

`-1 = kx`

`-1/k =x`

We can then substitute the value of x back into the first equation to solve for k.

`k(-1/k)^2=-2(-1/k)+3`

`k(1/k^2)=2/k+3`

`1/k= 2/k +3`

Multiply by k on both sides.

`1 = 2+3k`

`-1 = 3k`

`k=-1/3`

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