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December 18, 2014

December 18, 2014

Posted by **Dana** on Wednesday, January 23, 2008 at 12:37am.

I tried

h(23) - h(7)

------------

23-7

= * in this step i subbed it into the equation.

= -3599 + 319

-------------

16

It seems odd...

2) What is the end behaviour of

f(x) = 2-3x+4x^2

-----------

8x^2+5x-11

I understand end behaviour..but can someone help rearrange the first equation?

3) 2^2x - 4 (2^x) + 3 = 0

Solve for x.

This one I don't get at all.m

- Math -
**Damon**, Wednesday, January 23, 2008 at 6:01amIt is odd all right. First of all if it is h(t), why is it written with x and no t on the right? The whole thing is a typo.

The -x^2 term means it will sink, not rise, as x gets large. That is why your average speed comes out negative.

2-3x+4x^2

-----------

8x^2+5x-11

4 x^2 -3 x + 2

------------------

8 x^2 + 5 x -11

as x gets big, the x^2 terms are much bigger than the terms in x and the constants so

4 x^2

------ = 1/2

8 x^2

2^2x - 4 (2^x) + 3 = 0

The trick is to see that:

2^2x = (2^x)^2

so say y = 2^x

and we have

y^2 - 4 y +3 = 0

(y-3)(y-1)=0

y = 3 or y = 1

so 2^x = 3 or 2^x = 1

x log 2 = log 3 or x log 2 = log 1

x =.477/.301 or x = 0

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