anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
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anonymous
  • 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
  • anonymous
anonymous
  • anonymous

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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
@billj5 can you help me? :)
anonymous
  • anonymous
Ok, so the part I'm having trouble with is finding the maximum of f(x)
anonymous
  • 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
  • anonymous
this might help http://www.purplemath.com/modules/fcntrans.htm
anonymous
  • anonymous
Have you seen the -b/2a idea before, horse?
anonymous
  • anonymous
Ok thank you @billj5 ;)
anonymous
  • anonymous
Yes I have :) @Concentrationalizing
anonymous
  • anonymous
Okay, awesome. So for our problem, what is -b/2a?
anonymous
  • anonymous
I think it would be -3/2(-2)
anonymous
  • 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
  • anonymous
I meant those would be the values. So would the -3/2(-2) = 0?
anonymous
  • 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
  • anonymous
But f(x) only has two terms?`:|
anonymous
  • anonymous
So it's not a quadratic `:|
anonymous
  • 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
  • anonymous
Ok but I still don't get why b = 0.
anonymous
  • anonymous
|dw:1433381836170:dw|
anonymous
  • anonymous
|dw:1433381878344:dw|
anonymous
  • anonymous
Ooooh! That makes sense ;)
anonymous
  • anonymous
So it would all just equal 0. Got it :)
anonymous
  • 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
  • anonymous
Got it! That makes total sense now. I guess I must be a visual person lol :3 Thanks again @Concentrationalizing !
anonymous
  • anonymous
No problem :)

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