Question

How would you find the minimum value of f(x) = 3x^2 + 12x + 16

Answer 1

What have you tried?

Answer 2

I tried using this formula, because I remembered it, but it came out weird: (c - b^2/4a)

Answer 3

Rather c-(b^2/4a)

Answer 4

I got 4 by doing that, and I wasn't sure if I did that correctly or not

Answer 5

you can also do it by converting to vertex form . Complete the square.

Answer 6

That's the right answer. The simplest way is to differentiate: \(f'(x)=0\implies x=-2\implies f(-2)=4\). Your method only gives you the \(y\)-coordinate of the minimum, though: it does not tell you where it happens.

Answer 7

Okay, so then substitute in 4 for the y-value?

Answer 8

That's right: solve \(4=3x^2+12x+16\).

Answer 9

4 = 3x^2 + 12x + 16
-12=3x^2+12x
-4=x^2+3x
-4=(x+1.5)^2
-1.75=(x+1.5)^2

Did something go wrong, because I know that you can't square root a negative.

Answer 10

check third line

Answer 11

Wait, would it be 2 because x^2+4x (because I messed up and wrote 3 instead) +4 and if you use product sum, it would be 2 and 2 because 2*2 =4 and 2+2=4?

Answer 12

correct

Answer 13

But then you would negate that, correct

Answer 14

remember that the goal is solving for x.

Answer 15

Right. Would the function would turn into (x+2)(x+2) then?

Answer 16

correct. you get a repeated root at 2, which means that your min is at (2,4)

Answer 17

at -2*

Answer 18

(-2,4)*

Answer 19

Right, okay! Thank you very much!

Answer 20

simply take the equation and plug it into the y= screen on your calculator and then click 2nd - trace - click minimum, then go to the left and right of your minimum location and when you do that click enter. It will automatically give u it, no work. easier. just gotta remember the method to it.