It's a quadratic with a positive x² coefficient, so a CUP, with a minimum
f(x) = 2x² -12 x + 13
= 2(x² - 6x) + 13
= 2(x - 3)² - 2(9) + 13
f(x) = 2(x - 3)² - 5
Since x-3 is always zero or positive the least value is
f(3) = -5
Answer: -5
