Find the values of x that give critical points of y=ax^2 + bx + c, where a,b,c are constants. Under what conditions on a,b,c is the critical value a maximum? A minimum? x=__________ The critical value is a ***min/max(??)**** if a0. thank you! :D

Question

Find the values of x that give critical points of y=ax^2 + bx + c, where a,b,c are constants. Under what conditions on a,b,c is the critical value a maximum? A minimum?

x=__________

The critical value is a ***min/max(??)**** if a<0 and a ****min/max(??)**** if a>0.

thank you! :D

terenzreignz · Answer

critical value is the value of x that would make the derivative zero.

You probably already learned this some time ago, even if you didn't know that it WAS the critical value.

Anyway, differentiate the function :)

anonymous · Answer

like this?

y ' = 2ax + b + 0 

? :/

terenzreignz · Answer

Yup.
Now set y' = 0
and solve for x.

anonymous · Answer

2ax + b =0

2ax = -b
$$x=-\frac{ b }{ 2a }$$ ?

terenzreignz · Answer

Yup.

Look familiar? :D
Vertex formula of a parabola ^_^

Remember, the x-value of the vertex takes THAT exact value... right?

anonymous · Answer

ahh i see:)

yes:)

so that's the x value for my problem?

anonymous · Answer

:O

anonymous · Answer

and would it be like this then? 

The critical value is a ***min**** if a<0 and a ****max**** if a>0.

^^not too sure about this part :/

terenzreignz · Answer

That's a value (and from the looks of it, THE ONLY value) that would make the derivative equal to zero.

anonymous · Answer

okie:)

terenzreignz · Answer

And hold on.  Minima and maxima depend on the second derivative.
Find y''

anonymous · Answer

okay:)

so y '' = 2a + 0 ?

terenzreignz · Answer

Okay, good.
Now, as you can see, the second derivative is constant...

For a critical value to be a minimum, the second derivative has to be POSITIVE.

So, if a > 0, then the critical value would be... what?

terenzreignz · Answer

Okay, stuck?
Hint:
y'' = 2a

If a is positive, then what is 2a?

anonymous · Answer

ermm positive?

anonymous · Answer

2(5) = 10
2(100)=200

so pos?

terenzreignz · Answer

Yup. That means the second derivative y'' is positive.

Which makes the critical value... a min or a max?

Look back :)

anonymous · Answer

a max?

anonymous · Answer

or sorry a min when a>0 ?

anonymous · Answer

and a max the other way?

terenzreignz · Answer

There you go :)

terenzreignz · Answer

Does it not make sense?
When a > 0
then your parabola opens up, means it gets bigger and bigger; there CAN'T be a maximum, so your critical value must be a min.

When a< 0
your parabola opens down, means it gets smaller and smaller, and there can't be a minimum, so your critical value must be a max.

anonymous · Answer

yes:) 

hehe yayy!! thank you!! :D