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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 91 Page 616

From the diagram we can determine the value of $a$ directly. To find $b,$ we have to identify a second point through which the graph of the function passes.

$f (x) = 3 (\frac{1}{5})^{x}$

Practice makes perfect

In an exponential function in $f (x) = a b^{x}$ form, the coefficient $a$ shows the function's $y -$ intercept, while $b$ is the function's multiplier. Examining the diagram, we can determine the $y -$ intercept directly.

Now we have half of what we need to write the equation.

f (x) = 3 b^{x}

Finally, to find

b

we have to substitute a point into the function that does not fall on either axes. From the diagram we can identify such a point.

When we know one more point on the graph we can substitute it into the function and solve for

b .

f (x) = 3 b^{x}

$x = - 1$ , $f (x) = 15$

15 = 3 b^{- 1}

▼

Solve for

b

NegOneExponentToFrac

$a^{- 1} = \frac{1}{a}$

15 = 3 \cdot \frac{1}{b}

$LHS \cdot b = RHS \cdot b$

15 b = 3

$LHS / 15 = RHS / 15$

b = \frac{3}{1 5}

$\frac{a}{b} = \frac{a / 3}{b / 3}$

b = \frac{1}{5}

Now we can complete the equation.

f (x) = 3 (\frac{1}{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