F(x)=4(2)^x what would a graph of this look like?
1 answer:
9514 1404 393
Answer:
see attached
Step-by-step explanation:
A graphing calculator can do a nice job of showing you what the graph looks like.
The initial factor of 4 is the value when x=0, the y-intercept. The base of 2 tells you the function value is multiplied by 2 for each unit of x to the right, and divided by 2 for each unit of x to the left. (The curve quickly goes off the top of the graph.)
The horizontal asymptote is y=0, as it is for all exponential functions (that have not been translated).
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The level of the curves are all the eclipses with the major axis.
100 50 1
a2 = ------ - 1 , b2 = ------- - -----
C C 2
Given that:
100
T(x, y) = ------------
1 + x2 + y2
This implies that the level curves of a function(f) of two variables relate to the curves with equation f(x, y) = c
where c is the constant.
100
C = ------------ ----- (1)
1 + x2 + y2
By cross multiply
C (1 + x2 + y2) = 100
1 + x2 + y2 = 100/C
100
x2 + y2= --------- - 1 -----(2)
C
From (2); let's assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.
Now,
(x)2 (y)2
------ + ---------------
100 50 1
------ - 1 ------- - -----
C C 2
This is the equation for the family of the eclipses centered at (0,0) is :
x2/a2 + y2/b2 = 1
100 50 1
a2 = ------ - 1 b2 = ------- - -----
C C 2
Therefore; the level of the curves are all the eclipses with the major axis:
100 50 1
a2 = ------ - 1 b2 = ------- - -----
C C 2
and a minor axis which satisfies the values for which 0< c < 100.
The sketch of the level curves can be see in the attached image below.
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