What is the smallest possible real value of x^2+8x?

Explanation:

Let's complete the square

y = x^2 + 8x

y + 16 = x^2 + 8x + 16 adding 16 to both sides

y + 16 = (x+4)^2

y = 1(x+4)^2 - 16

We added 16 to both sides to allow the right hand side to factor into a perfect square.

The last line shown above is in the form y = a(x - h)^2 + k where

a = 1

h = -4

k = -16

Note the positive 'a' value indicates we have a minimum.

The vertex is (h,k) = (-4, -16)

The lowest y value possible is at the vertex.

So the smallest y value is y = -16.

This means the smallest possible value of x^2+8x is -16.

abde or 1,2,4, 5 is the answer

step-by-step explanation:

squares, rhombus, rectangles