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Core Connections Integrated I, 2013

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Core Connections Integrated I, 2013 View details

2. Section 8.2

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Exercise 121 Page 483

a If we substitute the points

(2, 2)

and

(4, \frac{1}{2})

into the functions we can write two equations that describe the same function.

{2 = a b^{2} \frac{1}{2} = a b^{4}

Let's solve this system of equations.

{2 = a b^{2} \frac{1}{2} = a b^{4}

$(II):$ Calculate quotient

{2 = a b^{2} 0.5 = a b^{4}

$(I):$ $LHS / b^{2} = RHS / b^{2}$

{\frac{2}{b ^{2}} = a 0.5 = a b^{4}

$(I):$ Rearrange equation

{a = \frac{2}{b ^{2}} 0.5 = a b^{4}

$(II):$ $a = \frac{2}{b ^{2}}$

{a = \frac{2}{b ^{2}} 0.5 = \frac{2}{b ^{2}} \cdot b^{4}

▼

(II):

Solve for

b

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$(II):$ $\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

{a = \frac{2}{b ^{2}} 0.5 = \frac{2 b ^{4}}{b ^{2}}

$(II):$ Calculate quotient

{a = \frac{2}{b ^{2}} 0.5 = 2 b^{2}

$(II):$ Rearrange equation

{a = \frac{2}{b ^{2}} 2 b^{2} = 0.5

$(II):$ $LHS / 2 = RHS / 2$

{a = \frac{2}{b ^{2}} b^{2} = 0.25

$(II):$ $LHS = RHS$

{a = \frac{2}{b ^{2}} b = \pm 0.5

$(II):$ $b > 0$

{a = \frac{2}{b ^{2}} b = 0.5

Having solved for

b,

we can substitute this value into the first equation to solve for

a .

{a = \frac{2}{b ^{2}} b = 0.5

$(I):$ $b = 0.5$

{a = \frac{2}{0 . 5 ^{2}} b = 0.5

$(I):$ Calculate power

{a = \frac{2}{0 . 2 5} b = 0.5

$(I):$ Calculate quotient

{a = 8 b = 0.5

Now we can write the equation.

y = 8 (0.5)^{x}

b From Part A, we know that the function's equation is

y = 8 (0.5)^{x} .

In addition to the two point we have been given in the exercise, we also know that it intercepts the

y -

axis at

(0, 8) .

Let's plot all of these points and draw our graph.

Subchapter links

8.2.1 Curve Fitting p.471-472

8.2.2 Curved Best-Fit Models p.478-480

8.2.3 Solving a System of Exponential Functions Graphically p.483-484