Which equation when graphed has a minimum value at x = - 5/2?

A) y = x^2 + 8x + 16

B) y = 2x^2 + 8x + 16

C) y = x^2 + 10x + 12

D) y = 2x^2 + 10x + 12

D. y = 2x^2 + 10x + 12.

Step-by-step explanation:

All these have a minimum value beacuse the coefficient of x^2 is positive.

OK, so we have to convert the equations to vertex form. Then we can read off the minimum value.

We do this by completing the square:-

A y = x^2 + 8x + 16

We divide the + 8 by 2 which gives us + 4 in the parentheses:-

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

Here the minimum value is when x + 4 = 0 giving x = - 4.

B. This gives x = - 2 so its not B.

C. y = (x + 5) ^2 - 25 + 12 so x = - 5 so its not C.

D. y = 2x^2 + 10x + 12

y = 2 (x^2 + 5x) + 12

Dividing the + 5 by 2:-

y = 2 (x + 5/2) ^2 - (5/2) 2 + 12

x = - 5/2 at minimum.