Given the functions:

`f(x)= x^2 + 4x -3`

We need to find the point (x1,y1) on the graph of f(x) such that the slope is -3.

We know that the slope is the derivative at the point x1.

==> f'(x1)= -3

Now we need to find x1 such that f'(x1)= -3

`==gt f'(x)= 2x +4 =`

`=gt f'(x1)= 2x1+ 4 = -3 `

`==gt 2x1= -7 `

`==gt x1= -7/2`

Now we will find the image of x1 which is f(x1)= y1

`==gt f(x1)= f(-7/2) = (-7/2)^2 + 4(-7/2) -3 `

`= 49/4 - 28/2 -3 = (49-56 -12)/4 = -19/4`

**Then, the point is (-7/2, -19/4)**

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