Question

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

Answer 1

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).

Answer 2

@geerky42 @pooja195

Answer 3

@Concentrationalizing @sammixboo

Answer 4

@amistre64 @TheSmartOne

Answer 5

@wio @IrishBoy123

Answer 6

@Vocaloid @hartnn

Answer 7

@Hero @perl

Answer 8

@billj5 can you help me? :)

Answer 9

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

Answer 10

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 \(-b/2a\).

Answer 11

this might help

http://www.purplemath.com/modules/fcntrans.htm

Answer 12

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

Answer 13

Ok thank you @billj5 ;)

Answer 14

Yes I have :) @Concentrationalizing

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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?

Answer 20

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

Answer 21

So it's not a quadratic `:|

Answer 22

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.

Answer 23

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