19thomasar
26.06.2019 •
Mathematics
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
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Ответ:
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.
Ответ:
answer:
c is the correct answer