Answer :
[tex]y=f\left( x \right) \\ \\ f\left( x \right) ={ x }^{ 2 }-3x+1\\ \\ \therefore \quad f'\left( x \right) =2x-3[/tex]
Now, when f'(x)=-5,
[tex]2x-3=-5\\ \\ 2x=-2\\ \\ x=-1[/tex]
When x=-1,
[tex]f\left( -1 \right) ={ \left( -1 \right) }^{ 2 }-3\left( -1 \right) +1\\ \\ =1+3+1\\ \\ =5[/tex]
Therefore, the co-ordinates you're looking for are:
(-1, 5)
Now, when f'(x)=-5,
[tex]2x-3=-5\\ \\ 2x=-2\\ \\ x=-1[/tex]
When x=-1,
[tex]f\left( -1 \right) ={ \left( -1 \right) }^{ 2 }-3\left( -1 \right) +1\\ \\ =1+3+1\\ \\ =5[/tex]
Therefore, the co-ordinates you're looking for are:
(-1, 5)
