Respuesta :
So we are given a function, [tex]y = 2x^2-32x + 56[/tex].
Proceed like this:
[tex]y = 2(x^2-16x + 28)\\ =2(((x-8)^2-64)+28)\\=2(x-8)^2-128+28\\=2(x-8)^2-100[/tex]
The vertex, which is also the maximum, is the following point:
[tex](8,100)[/tex]
Proceed like this:
[tex]y = 2(x^2-16x + 28)\\ =2(((x-8)^2-64)+28)\\=2(x-8)^2-128+28\\=2(x-8)^2-100[/tex]
The vertex, which is also the maximum, is the following point:
[tex](8,100)[/tex]
Answer:
y=2(x-8)^2 +(-72)
x-coordinate of minimum is 8
Step-by-step explanation:
