WILL FAN AND MEDAL :) Plz help me walk through this!
Functions f(x) and g(x) are described as follows:
f(x) = -2x^2 + 3
x g(x)
0 0
1 2
2 3
3 2
4 0

- anonymous

- katieb

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- anonymous

Which statement best compares the maximum value of the two functions?
It is equal for both functions.
It is 2 units higher for f(x) than g(x).
It is 2 units lower for f(x) than g(x).
It is 1 unit higher for f(x) than g(x).

- anonymous

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- anonymous

- anonymous

- anonymous

@billj5 can you help me? :)

- anonymous

Ok, so the part I'm having trouble with is finding the maximum of f(x)

- anonymous

f(x) is a quadratic. The graph of a quadratic is a parabola, which would either have a finite minimum value or a finite maximum value. In this case you have a finite maximum value (because the question implies this and because the leading coefficient is negative).
Given a quadratic with a general form of \(ax^{2} + bx + c = 0\), the x-coordinate of the maximum or minimum value is equal to \(-b/2a\).

- anonymous

this might help
http://www.purplemath.com/modules/fcntrans.htm

- anonymous

Have you seen the -b/2a idea before, horse?

- anonymous

Ok thank you @billj5 ;)

- anonymous

Yes I have :) @Concentrationalizing

- anonymous

Okay, awesome. So for our problem, what is -b/2a?

- anonymous

I think it would be -3/2(-2)

- anonymous

Actually, it would be 0. Note that the b in the general form is the coefficient of the x to the first power variable. Yet our equation has no x to the first power, we only have an x^2 and a constant term. Thus b is 0, which makes -b/2a = 0

- anonymous

I meant those would be the values. So would the -3/2(-2) = 0?

- anonymous

b is not -3, though, b is 0.
Since a quadratic is \(ax^{2} + bx + c\), a is what goes with the x^2 term. So a for us is -2. b is what goes with the x term. But our equation is -2x^2 + 3, we dont have an x term. Thus b has to be 0. Then our constant, c, is 3. So a = -2, b = 0, c = 3. Which means
-b/2a
= 0/2(-2)
= 0/-4
= 0
Does that make sense?

- anonymous

But f(x) only has two terms?`:|

- anonymous

So it's not a quadratic `:|

- anonymous

A quadratic is any function in which all the powers are integers and the highest power is a 2. So all of these would be quadratics:
\(x^{2} + 5x + 6\)
\(\frac{-3x^{2}}{5} - 9\)
\(x^{2}\)
The first has 3 terms, the 2nd has 2, the last one has 1 term. But they're all quadratics because all of the exponents are integers and the highest one is a 2.

- anonymous

Ok but I still don't get why b = 0.

- anonymous

|dw:1433381836170:dw|

- anonymous

|dw:1433381878344:dw|

- anonymous

Ooooh! That makes sense ;)

- anonymous

So it would all just equal 0. Got it :)

- anonymous

Lol, glad the visual helped. But yes, b is 0 because that part is missing :3
So that means the x-coordinate of the maximum value is 0. So since we need x to be 0, we can just plug in 0. If we plug in 0, we get
\(-2(0)^{2} + 3 = 0 + 3 = 3\)
Thus the maximum value is 3 for f(x)

- anonymous

Got it! That makes total sense now. I guess I must be a visual person lol :3 Thanks again @Concentrationalizing !

- anonymous

No problem :)

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