It 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
1) A helium balloon is rising according to the function H(t) = -7x^2 + 5x - 11 where h(t) is the height in meters after t seconds. Determine the average rates of increase of height from t=7 to t=23 seconds.
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
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