Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 10.2
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Exercise 94 Page 503

Practice makes perfect
a We want to find an exponential function that passes through (2,2) and (4, 12). To do this, we will create a system of equations by substituting ( x, y) = ( 2, 2) into the general form of exponential function, y=ab^x, and then substituting again with ( x, y) = ( 4, 12 ).
2=a * b^2 & (I) 12=a * b^4 & (II) We want to solve the above system of equations. When solving a system of equations using substitution, there are three steps.
  1. Isolate a variable in one of the equations.
  2. Substitute the expression for that variable into the other equation and solve.
  3. Substitute this solution into one of the equations and solve for the value of the other variable.
Observing the given equations, it looks like it will be simplest to isolate a in the first equation.
2=a * b^2 & (I) 12=a * b^4 & (II)
â–Ľ
(I): Solve for a
2b^2=a 12=a * b^4
a= 2b^2 12=a * b^4
Now that we have isolated a we can solve the system by substitution.
a= 2b^2 12=a * b^4
a= 2b^2 12= 2b^2 * b^4
a= 2b^2 12= 2b^2 * b^2 * b^2
a= 2b^2 12=2 * b^2
â–Ľ
(II): Solve for b
a= 2b^2 14=b^2
a= 2b^2 sqrt(14)=sqrt(b^2)
a= 2b^2 12=b
a= 2b^2 b= 12
Great! Now, to find the value of a we need to substitute b= 12 into either one of the equations in the given system. Let's use the first equation.
a= 2b^2 b= 12
a= 2( 1/2)^2 b= 12
â–Ľ
(I): Evaluate
a= 21/4 b= 12
a= 81 b= 12
a=8 b= 12
The solution to this system of equations is a=8 and b=12. With this information, we can finally write an equation of the exponential function that passes through the given points. f(x)=ab^x ⇔ f(x)=8 (1/2)^x
b We want to sketch a graph of the given exponential function. To do so, we will use the equation found in Part A to make a table of values. This will help us obtain points on the curve. Let's now assign some values to the x-variable and calculate the corresponding values for the y-variable.
x 8(1/2)^x f(x)=8(1/2)^x
0 8(1/2)^0 8
1 8(1/2)^1 4
2 8(1/2)^2 2
3 8(1/2)^3 1
4 8(1/2)^4 1/2

We will now plot and connect the obtained points with a smooth curve.