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.  

the answer is a

step-by-step explanation:

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: