i really need help please.. The focal length of a diverging lens is negative. If f = −23 cm for a particular divergin

Question

The focal length of a diverging lens is negative. If f = −23 cm

 for a particular diverging lens, where will the image be formed of an object located 59 cm

 to the left of the lens on the optical axis?

16.5487804878 cm to the left of the lens

What is the magnification of the image?

i tried to solve this but the answer kept said i am wrong

Answer 1

1/d + 1/f = 1/F; 1/59 + 1/f = - 1/23; f = - (59·23/(59 + 23) = - 16.54878049 cm

This meeans that image also left from lens

k = 59/(-16.549) = - 3.6