Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 7.2
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Exercise 121 Page 348

Practice makes perfect
a

We want to write an exponential function for the graph that passes through the points (1,18) and (4,3888). Let's consider the general form for this type of function.

y=ab^x Since we want the points to lie on the graph, they must satisfy this equation. Let's substitute (1,18) into the above formula.

y=ab^x
18=ab^1
18=ab
ab=18
Next we will substitute the second given point, (4,3888).

y=ab^x
3888=ab^4
ab^4=3888

We received two equations, so to find the values of a and b we need to solve the system of equations. ab=18 & (I) ab^3=3888 & (II) The first equation says that the product of a and b is 18. Notice that the second equation contains the same expression multiplied by b^3. ab=18 ab^4=3888 ⇒ ab=18 ab* b^3=3888 This allows us to substitute the value of ab into the second equation and solve for b.

ab=18 & (I) ab* b^3=3888 & (II)
ab=18 18b^3=3888
(II):Solve for b
ab=18 b^3=216
ab=18 sqrt(b^3)=sqrt(216)
ab=18 b=sqrt(216)
ab=18 b=6

Now that we know the value of b, let's substitute it into the first equation to find a.

ab=18 b=6
a( 6)=18 b=6
a=3 b=6

Finally, we can write the full equation of the function. y= a b^x ⇒ y= 3* 6^x

b

Similarly, we can find the equation of the function passing through (-2,-8) and (3,-0.25). Let's substitute the first point into the general equation.

y=ab^x
-8=ab^(-2)
-8 = a*1/b^2
-8 =a/b^2
a/b^2=-8

Next, we will substitute the second given point, (3,-0.25).

y=ab^x
-0.25=ab^3
ab^3=-0.25
ab^3=-1/4
Now we need to solve the system of equations that we found. a/b^2=-8 & (I) ab^3=- 14 & (II) To solve it we can use the Substitution Method. Let's multiply both sides of the first equation by b^2 to isolate a. a/b^2=-8 ab^3=- 14 ⇒ a=-8 b^2 ab^3=- 14 This allows us to substitute the value of a into the second equation and solve for b.

a=-8 b^2 & (I) ab^3=- 14 & (II)
a=-8 b^2 ( -8 b^2)b^3=- 14
(II):Solve for b
a=-8 b^2 -8 b^5=- 14
a=-8 b^2 b^5=- 14÷(-8)
a=-8 b^2 b^5=- 1-32
a=-8 b^2 b^5= 132
a=-8 b^2 sqrt(b^5)=sqrt(132)
a=-8 b^2 b=sqrt(132)
a=-8b^2 b= 12

Once we know the value of b, let's substitute it to the first equation to find a.

a=-8b^2 b= 12
a=-8( 12)^2 b= 12
(I): Simplify right-hand side
a=-8( 14) b= 12
a= -84 b= 12
a=-2 b= 12

Finally, we can write the full equation of the function. y= -2( 1/2)^x ⇔ y= -2( 0.5)^x