Question

State the interval over which f(x)=-6x^2+12x-4 is decreasing.
A) (-infinity, -1)
B) (-1, infinity)
C) (1, infinity)
D) (-infinity, 1)

Answer 1

This question looks incomplete.

Answer 2

State the interval over which the function what?

Answer 3

Ah I see, fixed it!

Answer 4

Refresh

Answer 5

Ok.

Answer 6

Do you know calculus?

Answer 7

We can also do this with algebra.

Answer 8

The trick is finding the vertex.

Answer 9

I know pre-calculus.
Algebra is preferred.

Answer 10

Yess the vertex, I'm having issues with that. I tried to complete the square.

Answer 11

Graph this using this site https://www.desmos.com/calculator
It should be (1, +infinity)

Answer 12

No, I don't like depending on sites to graph things out for me. I want to know how to do it algebraically.

Answer 13

But I appreciate it.

Answer 14

well then use factoring, and then find the midpoint of the two zeros and plug the midpoint into the function and find the max.

Answer 15

\[f(x)=ax^2+bx+c \\ f(x)=ax^2+\frac{a}{a}bx+c \\ f(x)=a(x^2+\frac{1}{a}bx)+c \\ f(x)=a(x^2+\frac{b}{a}x)+c \\f(x)=a(x^2+\frac{b}{a}x+( \frac{b}{2a} )^2)+c-a ( \frac{b}{2a})^2 \\ f(x)=a(x+\frac{b}{2a})^2+c-a \frac{b^2}{4a^2} \\ f(x)=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ f(x)=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]

Answer 16

We know it will be max because the degree is even, and leading coeifficient is negative, so the parabola will look like this.

Answer 17

Ok ok hold on.

Answer 18

Therefore the vertex is \[(\frac{-b}{2a},\frac{4ac-b^2}{4a})\]
But we really only care about the x-coordinate of the vertex which is x=-b/(2a)

Answer 19

Before we move on.
Can I show you what I did?

Answer 20

Ok I will hold on. lol.

Answer 21

yes

Answer 22

Ok, would be it correct to complete the square?

Answer 23

Yes! You can follow that scheme I laid out above.

Answer 24

That involves completing the square for any quadratic

Answer 25

Freckles, yours is a bit different than mine.
Let me show you how I learned it.

Answer 26

Alright.

Answer 27

-6x^2+12x-4
-6(x^2-2x+____)=4+_____
(b/2)^2
(-2/2)^2=-1^2=1
-6(x^2-2x+1)=4+6
-6(x^2-2x+1)=10
-6(x-1)^2=10
-6(x-1)^2-10
Oh
LOL I did it wrong on paper then while typing it out I got it I think.
-10 is vertex?!

Answer 28

oh wait no

Answer 29

(1,-10) is vertex

Answer 30

Right?

Answer 31

Well I guess by you mean you did it different you mean put an equal sign and you didn't add and take away on the same side
but you added on both sides .
\[-6x^2+12x-4=f(x) \\-6(x^2-2x)-4=f(x) \\ -6(x^2-2x+1)-4=f(x)-6 \\ -6(x-1)^2-4=f(x)-6 \\ -6(x-1)^2-4+6=f(x)\]

Answer 32

so basically I'm saying you subtracted on one side but added on the other side

Answer 33

you added -6 on left hand side and added +6 on the other side

Answer 34

Ah I see
it would be -6(x-1)^2-2

Answer 35

So the vertex becomes (1,-2)

Answer 36

Well the vertex is (1,-4+6)

Answer 37

So I'm saying -6(x-1)^2+2=f(x)

Answer 38

and therefore the vertex is (1,2)

Answer 39

But anyways for your problem we only care about the x-coordinate of the vertex.

Answer 40

What other good info you can determine which interval it is decreasing on even increasing on is to determine if the parabola is open up or open down

Answer 41

OHH I SEEEEEEEEE WHY ITS 2 POSTIVE LOL

Answer 42

Right ok.

Answer 43

its open down

Answer 44

Since our a is negative, then our parabola is open down
which mean the function is increasing on (-inf,1)
and decreasing on (1,inf)

Answer 45

So it would be decreasing from 1 to the right? so 1 to infinity?

Answer 46

(1,infinity)?

Answer 47

High five!

Answer 48

High five!

Answer 49

Thank you very much for helping me! :)xxxxx

Answer 50

Np.

Answer 51

Have a great day wherever!

Answer 52

You know a easier method would be just to get the derivative of the function, and make it equal zero to get the value of x for the max. and just plug it into the function and you get the Max point. I dont know why i didnt think of that before lol.