Question

What is the vertex of the following inequality: 2(2x^2 – 3x) > -9 ?i have to pick one of them for the answer (-3/4, 27/4) (3/4, -27/4) (-3/4, -27/4) (3/4, 27/4)

Answer 1

help

Answer 2

2(2x^2 – 3x) > -9
4x^2-6x>-9
4x^2-6x+9>0

f(x) = ax^2 + bx + c

The vertex = (-b/2a, f(-b/2a))

Find the vertex of f(x) = 4x^2 - 6x + 9

-b/2a=6/2*4
-b/2a=6/8
-b/2a=3/4

f(x) = 4x^2 - 6x + 9
f(3/4) = 4(3/4)^2 - 6(3/4) + 9
f(3/4)=4(9/16)-18/4+9
f(3/4)=9/4-18/4+9
f(3/4)=9-18+36/4
f(3/4)=27/4

So, the vertex is (3/4,27/4).

Answer 3

Welcome to Openstudy @juanaandrea

Answer 4

wow thank you.... you just explained everything ur great

Answer 5

My Pleasure.

Answer 6

ok i will thanks

Answer 7

Anytime.