An algebra problem by Diego González

$2(x-2)^2-5=2x^2-8x+3$

Hence, $a=-8,b=3,ab=-24$

$f(x)=2x^2+ax+b\implies f’(x)=4x+a=0\implies 4(2)+a=0\implies a=-8$ we set the first derivative equal to zero to attain the minimum.

Now the minimum belongs to the graph, so the coordinates satisfy the graph. Thus, $2(2^2)-8(2)+b=-5\implies b=3\implies ab=-24$

We will find out what are $a$ & $b$ . We know that $f(2)=2(2)^2+2a+b=-5$ . We also know that the derivative at $(2,-5)$ must be $0$ , since it is a local minimum; so $f'(2)=4(2)+a=0$ . We solve the equation $8+a=0$ , and we obtain the value of $a$ , that is $-8$ . We can now substitute in our first equation: $8-16+b=-5$ and solve for $b$ and get $3$ . Now we finally multiply $ab$ in order to get -24.