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Core Connections Integrated I, 2013

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Core Connections Integrated I, 2013 View details

2. Section 11.2

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Exercise 86 Page 613

a By substituting the points into the general form of an exponential function, we get a system of equations.

{- 2 = a b^{- 1} - 162 = a b^{3}

We can solve this with the Substitution Method.

{- 2 = a b^{- 1} - 162 = a b^{3} (I) (II)

▼

(I):

Solve for

a

NegOneExponentToFrac

$(I):$ $a^{- 1} = \frac{1}{a}$

{- 2 = a \cdot \frac{1}{b} - 162 = a b^{3}

$(I):$ $LHS \cdot b = RHS \cdot b$

{- 2 b = a - 162 = a b^{3}

$(I):$ Rearrange equation

{a = - 2 b - 162 = a b^{3}

$(II):$ $a = - 2 b$

{a = - 2 b - 162 = (- 2 b) b^{3}

▼

(II):

Solve for

b

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

{a = - 2 b 81 = b \cdot b^{3}

$(II):$ $a \cdot a^{m} = a^{1 + m}$

{a = - 2 b 81 = b^{4}

$(II):$ Rearrange equation

{a = - 2 b b^{4} = 81

$(II):$ $4 LHS = 4 RHS$

{a = - 2 b b = \pm 3

$(II):$ $b > 0$

{a = - 2 b b = 3

$(I):$ $b = 3$

{a = - 2 (3) b = 3

$(I):$ Multiply

{a = - 6 b = 3

Now we can write the complete equation.

y = - 6 (3)^{x}

b Like in Part A, we begin by substituting the points into the general form of an exponential function and create a system of equations.

{1.75 = a b^{2} 28 = a b^{- 2}

We can solve this with the Substitution Method.

{1.75 = a b^{2} 28 = a b^{- 2} (I) (II)

▼

(II):

Solve for

a

NegExponentToFrac

$(II):$ $a^{- m} = \frac{1}{a ^{m}}$

{1.75 = a b^{2} 28 = a \cdot \frac{1}{b ^{2}}

$(II):$ $LHS \cdot b^{2} = RHS \cdot b^{2}$

{1.75 = a b^{2} 28 b^{2} = a

$(II):$ Rearrange equation

{1.75 = a b^{2} a = 28 b^{2}

$(I):$ $a = 28 b^{2}$

{1.75 = (28 b^{2}) b^{2} a = 28 b^{2}

▼

(I):

Solve for

b

$(I):$ $a^{m} \cdot a^{n} = a^{m + n}$

{1.75 = 28 b^{4} a = 28 b^{2}

$(I):$ $LHS / 28 = RHS / 28$

{\frac{1 . 7 5}{2 8} = b^{4} a = 28 b^{2}

$(I):$ Rearrange equation

{b^{4} = \frac{1 . 7 5}{2 8} a = 28 b^{2}

$(I):$ $4 LHS = 4 RHS$

{b = \pm 0.5 a = 28 b^{2}

$(I):$ $b > 0$

{b = 0.5 a = 28 b^{2}

$(II):$ $b = 0.5$

{b = 0.5 a = 28 (0.5)^{2}

$(II):$ Calculate power

{b = 0.5 a = 28 \cdot 0.25

$(II):$ Multiply

{b = 0.5 a = 7

Now we can write the complete equation.

y = 7 (0.5)^{x}

Subchapter links

11.2.1 Solving Work and Mixture Problems p.603-604

11.2.2 Solving Equations and Systems Graphically p.607-608

11.2.3 Using a Best-Fit Line to Make a Prediction p.610-614

11.2.4 Treasure Hunt p.616-617

11.2.5 Using Coordinate Geometry and Constructions to Explore Shapes p.620-621

11.2.6 Modeling with Exponential Functions and Linear Inequalities p.625-628