To Graph:
y=0⋅60x2⋅756
Given: points are given in (x,y) as (1,0⋅6) , (2,4⋅1) , (3,12.4) , (4,27) and (5,49⋅5).
Concept used:
If a function y=f(x) passes through a point (x0,y0) , then the point (x0,y0) satisfies the equation of the function, that is, y0=f(x0).
Calculation:
Table of data pairs (lnx,lny) is written as;
lnx00⋅691⋅091⋅381⋅60lny−0⋅511⋅412⋅513⋅293⋅90
Scatter graph of the (lnx,lny) is given as;
Now, the general exponential form is y=abx ;
Here, choose two points (0,−0⋅51) and (0⋅69,1⋅41) then the equation of the line;
lny−y1=y2−y1x2−x1(lnx−x1)
Now, substitute the value;
lny+0⋅51=3⋅90+0⋅511⋅60+0(lnx−0)lny+0⋅51=4⋅411⋅60lnxlny=2⋅756lnx−0⋅51
Now, exponentiation each side by e;
elny=e2⋅756lnx−0⋅51[elny=y]y=x2⋅756⋅e−0⋅51y=0⋅60x2⋅756[e−0⋅51=0⋅61]
Conclusion:
Hence, the exponential function that satisfies the given points is y=0⋅60x2⋅756.
Chapter 4 Solutions
Holt Mcdougal Larson Algebra 2: Student Edition 2012
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