quadratic function help (1 Viewer)

[kaha167]

find the least value of 3(1 - 4x)^2 + 5 = 0
I have gotten this down to:
6x^2 - 3x + 1 = 0
but am unable to get past this point,
any help is appreciated,
cheers.

 

[ohexploitable]

You mean minimum value? Just use x=-b/2a and sub it in.

 

[omgmynameislong]

that or differentiate and find minimum tp.

 

[hscishard]

You chose to expand. In this case, -b/2a is pretty slow.

Should've just differentiated from there.
First use the chain rule. Then differentiate 5. (= 0)

Expand derivative and find stationary point. Sub back in and thats your min value.

x=1/4. y=5

 

[Drongoski]

find the least value of 3(1 - 4x)^2 + 5 = 0
I have gotten this down to:
6x^2 - 3x + 1 = 0
but am unable to get past this point,
any help is appreciated,
cheers.

You have the expression in a nice form already;

3(1-4x)^2 >= 0

.: 3(1-4x)^2 + 5 >= 5

i.e. min value = 5

 

[ohexploitable]

You have the expression in a nice form already;

3(1-4x)^2 >= 0

.: 3(1-4x)^2 + 5 >= 5

i.e. min value = 5

Oh lol, I forgot about that form!

 

[hscishard]

I was thinking about the turning point form.
Nice.