Question

Identify the minimum y=x^2-4x-32

Answer 1

y=x^2-4x-32

if you solve for x

you get 8 and -4 the max should be in the middle of those 2 points right?

Answer 2

2

Answer 3

or just differentiate, set dy/dx=0
y=x^2-4x-32
dy/dx=2x-4=0
x=2
then find y

Answer 4

assuming he knows calculus

Answer 5

but yes that is easier

Answer 6

we could use completing the square
x^2-4x-32
(x-2)^2-4-32
(x-2)^2-36
symmetry at x=2

Answer 7

Yup he should definately know CTS