To Graph: a
y=0.212⋅(4.88)x .
Given: points are given in (x,y) as (1,1.4) , (2,6.7) , (3,32.9) , (4,161.4) and (5,790.9) .
Concept used:
(1) 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) .
(2) A exponential model or function is defined as:
y=a⋅(b)x ; where a>0 and b≥0 .
(3) Equation of a line passing through the given two points as like (x1,y1) and (x2,y2) .
Then, (y−y1)=m(x−x1) .
Here, m is the slope. And m is as follows:
m=y2−y1x2−x1 .
Calculation:
The table of data pairs (x,lny) is written as;
x12345lny0.341.913.495.086.67
The Scatter graph of (x,lny) is given as;
Now, the general exponential form is y=abx ;
Here, choose two points (1,0.34) and (5,6.67) then the equation of the line;
lny−y1=y2−y1x2−x1(x−x1)
Now, substitute the value;
lny−0.34=6.67−0.345−1(x−1)lny−0.34=6.334(x−1)lny−0.34=1.585(x−1)lny−0.34=1.585x−1.585
Thus,
lny=1.585x−1.585+0.34lny=1.585x−1.551
Now, take the exponentiation of each side by e;
elny=e1.585x−1.551[elny=y]y=e−1.551⋅e1.585xy≈0.212⋅[(2.7182)1.585]x[e≈2.7182e1.585≈4.879e−1.551≈0.212]y≈0.212⋅(4.879)x
Therefore, the exponential function of the model is y=0.212⋅(4.88)x .
Conclusion:
Hence, the exponential function of the model that satisfies the given points is y=0.212⋅(4.88)x .
Chapter 4 Solutions
Holt Mcdougal Larson Algebra 2: Student Edition 2012
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