# What is intervals for y=3x^4 +16x^3 +24x^2 - 6 and what way u use? completing the square?

1

## Jawapan

• Pengguna Brainly
2016-02-28T18:17:46+08:00

### Ini Jawapan Diperakui

Jawapan diperakui mengandungi maklumat yang boleh dipercayai dan diharapkan yang dijamin dipilih dengan teliti oleh sepasukan pakar. Brainly mempunyai berjuta-juta jawapan berkualiti tinggi, semuanya disederhanakan dengan teliti oleh ahli komuniti kami yang paling dipercayai, tetapi jawapan diperakui adalah terbaik di kalangan terbaik.
Hello.. mmm im not sure for that question..but this is same thing
Best Answer:  Concavity is found via the 2nd derivative:

y = 3x^4 - 16x^3 + 24x^2 + 48
y' = 12x^3 - 48x^2 + 48x
y" = 36x^2 - 96x + 48

Find the inflection points (where y" = 0), and you'll have the concavity inetervals

36x^2 - 96x + 48 = 0
12(3x^2 - 8x + 4) = 0
3x^2 - 8x + 4 = 0

(3x - 2)(x - 2) = 0
Either
x = (2/3)
or
x = 2

So your intervals are (-infinity,(2/3)), ((2/3),2), (2,+infinity)

A graph is concave down when y" < 0, so test a value from each interval

(-infinity,(2/3))
Try x = 0
y"(0) = 36(0)^2 - 96(0) + 48 = 48, thus the graph is concave up.

((2/3),2)
Try x = 1
y"(1) = 36(1)^2 - 96(1) + 48 = -12, thus the graph is concave down.

(2,+infinity)
Try x = 3
y"(3) = 36(3)^2 - 96(3) + 48 = 84, thus the graph is concave up.