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Pearson Algebra 2 Common Core, 2011 View details

2. Standard Form of a Quadratic Function

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Exercise 48 Page 207

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a We are given the two functions below. One is linear, and the other is a quadratic function.

f (x) = 4 x + 3 g (x) = \frac{1}{2} x^{2} + 2

We have to find the rate of change of each function from

x = 0

x = 1 .

Before starting the computations, let's remember the expression that gives us the rate of change of a function

F (x)

between

x = a

and

x = b .

Rate of Change ⇕ Rate of Change = \frac{Change in y}{Change in x} = \frac{F ( b ) - F ( a )}{b - a}

Using this formula, we can find the rate of change of each of the given functions.

Rate of Change $= \frac{F ( b ) - F ( a )}{b - a}$
$\frac{f ( 1 ) - f ( 0 )}{1 - 0}$	$\frac{g ( 1 ) - g ( 0 )}{1 - 0}$
$\frac{4 ( 1 ) + 3 - ( 4 ( 0 ) + 3 )}{1 - 0}$	$\frac{\frac{1}{2} ( 1 ) ^{2} + 2 - ( \frac{1}{2} ( 0 ) ^{2} + 2 )}{1 - 0}$
$4$	$0.5$

From the table above, we conclude that the function $f (x)$ has a greater rate of change from $x = 0$ to $x = 1 .$ We can also check it graphically.

Looking at the graph, we can say that both functions have the same change in the $x -$ coordinates, but the change in the $y -$ coordinates is greater for the function $f (x) .$ Therefore, $f (x)$ has a greater rate of change.

b First, let's remember the formula to calculate the rate of change of a function

F (x)

from

x = a

x = b .

Rate of Change ⇕ Rate of Change = \frac{Change in y}{Change in x} = \frac{F ( b ) - F ( a )}{b - a}

As in Part A, let's find the rate of change of each of the given functions from

x = 2

and

x = 3 .

Rate of Change $= \frac{F ( b ) - F ( a )}{b - a}$
$\frac{f ( 3 ) - f ( 2 )}{3 - 2}$	$\frac{g ( 3 ) - g ( 2 )}{3 - 2}$
$\frac{4 ( 3 ) + 3 - ( 4 ( 2 ) + 3 )}{3 - 2}$	$\frac{\frac{1}{2} ( 3 ) ^{2} + 2 - ( \frac{1}{2} ( 2 ) ^{2} + 2 )}{3 - 2}$
$4$	$2.5$

From the table above, we conclude once more that the function $f (x)$ has a greater rate of change from $x = 2$ to $x = 3 .$ We can also check it graphically.

Looking at the graph, we can say that both functions have the same change in $x -$ coordinates, but the change in $y -$ coordinates is greater for the function $f (x) .$ Therefore, $f (x)$ has a greater rate of change.

c From the two previous results, and since the graph of

y = f (x)

is a line, we can affirm that the rate of change of

f (x)

is always constant and equal to

4 .

However, the rate of change of

g (x)

is not constant, and it increases as

x

increases.

Rate of Change = \frac{g ( b ) - g ( a )}{b - a}, a < b

We are interested in finding, if any, the values of

a

and

b

for which the expression above is greater than

4 .

\frac{g ( b ) - g ( a )}{b - a} > 4

SubstituteExpressions

Substitute expressions

\frac{( \frac{1}{2} b ^{2} + 2 ) - ( \frac{1}{2} a ^{2} + 2 )}{b - a} > 4

▼

Simplify

Distr

Distribute $- 1$

\frac{\frac{1}{2} b ^{2} + 2 - \frac{1}{2} a ^{2} - 2}{b - a} > 4

SubTerms

Subtract terms

\frac{\frac{1}{2} b ^{2} - \frac{1}{2} a ^{2}}{b - a} > 4

SubFrac

Subtract fractions

\frac{\frac{b ^{2} - a ^{2}}{2}}{b - a} > 4

DivFracD

$\frac{a / c}{b} = \frac{a}{b \cdot c}$

\frac{b ^{2} - a ^{2}}{2 ( b - a )} > 4

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

\frac{( b - a ) ( b + a )}{2 ( b - a )} > 4

CancelCommonFac

Cancel out common factors

\frac{b + a}{2} > 4

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

b + a > 8

Let's analyze the inequality

a + b > 8

depending on the value of

a .

If $a < 4,$ it will depend on the value of $b$ to achieve this. For example, for $a = 3$ and $b = 5.5,$ we will have $a + b > 8,$ but if $b = 4.5$ then $a + b < 8 .$
If $a \geq 4,$ then for all $b > 4$ we will have $a + b > 8 .$

We conclude that $a$ must be greater than or equal to $4 .$ Therefore, the rate of change of $g (x)$ is greater than $4$ for all $x \geq 4 .$

Subchapter links

Lesson Check p.206

Practice and Problem-Solving Exercises p.206-208

Standardized Test Prep p.208

Mixed Review p.208

Exercise 48 Page 207

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