The circle touches the sides of AABC, AB, BC and
CA at points P, Q and R respectively. Prove that:
AP+BQ + CR = PB + QC + RA= (AB + BC + AC.)

It is the top one. the answer is -7. that is smallest

Idon't see any graph, however the graph of the function : f(x) = x²-2x+3 is a parabola: 1) open upward (since a> 0)2) it intersects y-axis in 3 (y=3)3) its axis of symmetry is x = +1 ( -b/2a)4) it's minimum is y = -2, so its vertex is (1,-2)5) it doesn't intersect the x-axis since discriminant < 0