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Chapter 2 Review and Tests

Review

Definitions, Concepts, and Formulas

Examples

2.1 The Coordinate Plane

Ordered pair. A pair of numbers in which the order is specified is called an ordered pair of numbers.
Distance formula. The distance between two points $P (x_{1}, y_{1})$ $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2}),$ $Q (x_{2}, y_{2}),$ denoted by d(P, Q), is given by

$d (P, Q) = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} .$ $d (P, Q) = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} .$
Midpoint formula. The coordinates of the midpoint M(x, y) on the line segment joining $P (x_{1}, y_{1})$ $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ $Q (x_{2}, y_{2})$ are given by

$M (x, y) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) .$ $M (x, y) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) .$

Find the distance between the two points P(2, 3) and $Q (- 2, 5) .$ $Q (- 2, 5) .$

Solution

d (P, Q) = \sqrt{{(- 2 - 2)}^{2} + {(5 - 3)}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5}

$d (P, Q) = \sqrt{{(- 2 - 2)}^{2} + {(5 - 3)}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5}$

Find the coordinates of the midpoint M(x, y) on the line segment joining $P (- 3, 8)$ $P (- 3, 8)$ and $Q (7, - 1) .$ $Q (7, - 1) .$

Solution

M = (\frac{- 3 + 7}{2}, \frac{8 + (- 1)}{2}) = (2, \frac{7}{2})

$M = (\frac{- 3 + 7}{2}, \frac{8 + (- 1)}{2}) = (2, \frac{7}{2})$

2.2 Graphs of Equations

The graph of an equation in two variables, such as x and y, is the set of all ordered pairs (a, b) in the coordinate plane that satisfies the equation. A graph of an equation, then, is a picture of its solution set.
Sketching the graph of an equation by plotting points
1. Step 1 Make a representative table of solutions of the equation.
2. Step 2 Plot the solutions as ordered pairs in the coordinate plane.
3. Step 3 Connect the solutions in Step 2 with a smooth curve.
Intercepts
1. The x-intercept is the x-coordinate of a point on the graph where the graph touches or crosses the x-axis. To find the x-intercepts, set $y = 0$ $y = 0$ in the equation and solve for x.
2. The y-intercept is the y-coordinate of a point on the graph where the graph touches or crosses the y-axis. To find the y-intercepts, set $x = 0$ $x = 0$ in the equation and solve for y.
Symmetry A graph is symmetric with respect to:
1. the y-axis if for every point (x, y) on the graph, the point $(- x, y)$ $(- x, y)$ is also on the graph. That is, replacing x with $- x$ $- x$ in the equation produces an equivalent equation.
2. the x-axis if for every point (x, y) on the graph, the point $(x, - y)$ $(x, - y)$ is also on the graph. That is, replacing y with $- y$ $- y$ in the equation produces an equivalent equation.
3. the origin if for every point (x, y) on the graph, the point $(- x, - y)$ $(- x, - y)$ is also on the graph. That is, replacing x with $- x$ $- x$ and y with $- y$ $- y$ in the equation produces an equivalent equation.
Circle A circle is the set of points in a Cartesian coordinate plane that are at a fixed distance r from a specified point (h, k). The fixed distance r is called the radius of the circle, and the fixed point (h, k) is called the center of the circle. The equation ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ is called the standard form of an equation of a circle.

The general form of an equation of a circle is

$x^{2} + y^{2} + a x + b y + c = 0 .$ $x^{2} + y^{2} + a x + b y + c = 0 .$

Refer to Example 1 on page 186 to review sketching the graph of an equation by plotting points.

Find the x- and y-intercepts and identify the symmetries for the graph of the equation $x^{2} + y^{2} = 25,$ $x^{2} + y^{2} = 25,$ shown below.

Solution

The graph is symmetric with respect to the x-axis, the y-axis, and the origin. Both the x- and y-intercepts are $- 5$ $- 5$ and 5.

Write the equation of a circle:

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

$x^{2} + y^{2} + 4 x - 6 y - 3 = 0$

in standard form and give the center and radius of the circle.

Solution

Complete the square by adding both 4 and 9 to both sides of the given equation.

\begin{array}{rcl} x^{2} + y^{2} + 4 x - 6 y - 3 & = & 0 \\ (x^{2} + 4 x + 4) + (y^{2} - 6 y + 9) - 3 & = & 4 + 9 \\ {(x + 2)}^{2} + {(y - 3)}^{2} & = & 16 = 4^{2} \end{array}

$\begin{array}{rcl} x^{2} + y^{2} + 4 x - 6 y - 3 & = & 0 \\ (x^{2} + 4 x + 4) + (y^{2} - 6 y + 9) - 3 & = & 4 + 9 \\ {(x + 2)}^{2} + {(y - 3)}^{2} & = & 16 = 4^{2} \end{array}$

The center is $(- 2, 3)$ $(- 2, 3)$ and the radius is 4.

2.3 Lines

The slope m of a nonvertical line through the points $P (x_{1}, y_{1})$ $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ $Q (x_{2}, y_{2})$ is

$\begin{array}{rcl} m & = & \frac{rise}{run} = \frac{(y -coordinate of Q) - (y -coordinate of P)}{(x -coordinate of Q) - (x -coordinate of P)} \\ = & \frac{Δ y}{Δ x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} . \end{array}$ $\begin{array}{rcl} m & = & \frac{rise}{run} = \frac{(y -coordinate of Q) - (y -coordinate of P)}{(x -coordinate of Q) - (x -coordinate of P)} \\ = & \frac{Δ y}{Δ x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} . \end{array}$

The slope of a vertical line is undefined. The slope of a horizontal line is zero.
Equation of a line

$\begin{array}{rcll} y - y_{1} & = & m (x - x_{1}) & Point-slope form \\ y & = & m x + b & Slope-intercept form \\ y & = & k & Horizontal line \\ x & = & h & Vertical line \\ a x + b y + c & = & 0 & General form \end{array}$ $\begin{array}{rcll} y - y_{1} & = & m (x - x_{1}) & Point-slope form \\ y & = & m x + b & Slope-intercept form \\ y & = & k & Horizontal line \\ x & = & h & Vertical line \\ a x + b y + c & = & 0 & General form \end{array}$
Parallel and perpendicular lines

Two distinct lines with respective slopes $m_{1}$ $m_{1}$ and $m_{2}$ $m_{2}$ are
1. parallel if $m_{1} = m_{2} .$ $m_{1} = m_{2} .$
2. perpendicular if $m_{1} m_{2} = - 1 .$ $m_{1} m_{2} = - 1 .$
Linear regression A method for modeling data using linear functions.

Find the slope, m, of the line through the points (4, 8) and $(1, - 1) .$ $(1, - 1) .$

Solution

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 1 - 8}{1 - 4} = \frac{- 9}{- 3} = 3

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 1 - 8}{1 - 4} = \frac{- 9}{- 3} = 3$

Convert the general form $2 x - 3 y - 5 = 0$ $2 x - 3 y - 5 = 0$ into the point–slope form of the line equation.

Solution

\begin{array}{rcll} 2 x - 3 y - 5 & = & 0 \\ 3 y & = & 2 x - 5 & Add 3 y to both sides; \\ interchange sides . \\ y & = & \frac{2}{3} x - \frac{5}{3} & Divide both sides by 3 . \end{array}

$\begin{array}{rcll} 2 x - 3 y - 5 & = & 0 \\ 3 y & = & 2 x - 5 & Add 3 y to both sides; \\ interchange sides . \\ y & = & \frac{2}{3} x - \frac{5}{3} & Divide both sides by 3 . \end{array}$

The slope is $m = \frac{2}{3} .$ $m = \frac{2}{3} .$ The y-intercept is $- \frac{5}{3} .$ $- \frac{5}{3} .$

The x-intercept is $\frac{5}{2} .$ $\frac{5}{2} .$ $Set y = 0 and solve for x .$ $Set y = 0 and solve for x .$

The slope of any line parallel to the line with equation

$2 x - 3 y - 5 = 0$ $2 x - 3 y - 5 = 0$ must also have slope $\frac{2}{3},$ $\frac{2}{3},$ and any perpendicular line must have slope $- \frac{3}{2} .$ $- \frac{3}{2} .$

Review linear regression on page 208.

2.4 Functions

Any set of ordered pairs is a relation. The set of all first components is the domain, and the set of all second components is the range. The graph of all of the ordered pairs is the graph of the relation.
A function is a relation in which each element of the domain corresponds to exactly one element in the range.
If a function f is defined by an equation, then its domain is the largest set of real numbers for which f(x) is a real number.
Vertical-line test. If each vertical line intersects a graph at no more than one point, the graph is the graph of a function.

Shown below is the graph of the relation $y = x^{3},$ $y = x^{3},$ which has the set of all real numbers as both domain and range. The relation $y = x^{3}$ $y = x^{3}$ defines a function because each element of the domain corresponds to exactly one element in the range.

Notice that each vertical line intersects the graph of $y = x^{3}$ $y = x^{3}$ at no more than one point.

The graph of $y^{2} = - 8 x$ $y^{2} = - 8 x$ shown below does not pass the vertical-line test, so it is not the graph of a function.

2.5 Properties of Functions

Let $x_{1}$ $x_{1}$ and $x_{2}$ $x_{2}$ be any two numbers in the interval (a, b).
1. f is increasing on (a, b) if $x_{1} < x_{2}$ $x_{1} < x_{2}$ implies $f (x_{1}) < f (x_{2}) .$ $f (x_{1}) < f (x_{2}) .$
2. f is decreasing on (a, b) if $x_{1} > x_{2}$ $x_{1} > x_{2}$ implies $f (x_{1}) > f (x_{2}) .$ $f (x_{1}) > f (x_{2}) .$
3. f is constant on (a, b) if $x_{1} < x_{2}$ $x_{1} < x_{2}$ implies $f (x_{1}) = f (x_{2}) .$ $f (x_{1}) = f (x_{2}) .$
Assume $(x_{1}, x_{2})$ $(x_{1}, x_{2})$ is an interval in the domain of f containing a number a.
1. f(a) is relative minimum of f if $f (a) \leq f (x)$ $f (a) \leq f (x)$ for every x in $(x_{1}, x_{2}) .$ $(x_{1}, x_{2}) .$
2. f(a) is relative maximum of f if $f (a) \geq f (x)$ $f (a) \geq f (x)$ for every x in $(x_{1}, x_{2}) .$ $(x_{1}, x_{2}) .$
3. A point (a, b) on the graph of f is a turning point if the graph of f changes direction at (a, b) from increasing to decreasing or from decreasing to increasing.
Suppose that for each x in the domain of f, $- x$ $- x$ is also in the domain of f.
1. f is called an even function if $f (- x) = f (x) .$ $f (- x) = f (x) .$ The graph of an even function is symmetric about the y-axis.
2. f is called an odd function if $f (- x) = - f (x) .$ $f (- x) = - f (x) .$ The graph of an odd function is symmetric about the origin.
The average rate of change of f(x) as x changes from a to b is defined by $\frac{f (b) - f (a)}{b - a}, b \neq a .$ $\frac{f (b) - f (a)}{b - a}, b \neq a .$
The expression $\frac{f (x) - f (a)}{x - a}, x \neq a$ $\frac{f (x) - f (a)}{x - a}, x \neq a$ , is called a difference quotient at a. The expression $\frac{f (x + h) - f (h)}{h}, h \neq 0,$ $\frac{f (x + h) - f (h)}{h}, h \neq 0,$ is also called a difference quotient.

The graph of $f (x) = x^{2}$ $f (x) = x^{2}$ is decreasing on $(- \infty, 0)$ $(- \infty, 0)$ and increasing on $(0, \infty), f (0) = 0$ $(0, \infty), f (0) = 0$ is a relative minimum of f, and (0, 0) is turning point. Also, because $f (- x) = {(- x)}^{2} = x^{2}, f (x) = f (- x)$ $f (- x) = {(- x)}^{2} = x^{2}, f (x) = f (- x)$ and f is an even function.

Find the average rate of change of $f (x) = x^{2}$ $f (x) = x^{2}$ as x changes from $- 3$ $- 3$ to 1.

Solution

The average rate of change is:

\frac{f (1) - f (- 3)}{1 - (- 3)} = \frac{1 - 9}{4} = - 2 .

$\frac{f (1) - f (- 3)}{1 - (- 3)} = \frac{1 - 9}{4} = - 2 .$

2.6 A Library of Functions

Basic Functions

Identity function	$f (x) = x$ $f (x) = x$
Constant function	$f (x) = c$ $f (x) = c$
Squaring function	$f (x) = x^{2}$ $f (x) = x^{2}$
Cubing function	$f (x) = x^{3}$ $f (x) = x^{3}$
Absolute value function	$f (x) = \| x \|$ $f (x) = \| x \|$
Square root function	$f (x) = \sqrt{x}$ $f (x) = \sqrt{x}$
Cube root function	$f (x) = \sqrt[3]{x}$ $f (x) = \sqrt[3]{x}$
Reciprocal function	$f (x) = \frac{1}{x}$ $f (x) = \frac{1}{x}$
Reciprocal square function	$f (x) = \frac{1}{x^{2}}$ $f (x) = \frac{1}{x^{2}}$
Greatest integer function	$f (x) = 〚 x 〛$ $f (x) = 〚 x 〛$

For piecewise functions, different rules are used in different parts of the domain. See page 253.

Refer to the Library of Functions in Section 2.6. The graphs of the basic functions can also be found on the final pages just inside the back cover of the text.

2.7 Transformations of Functions

Vertical and horizontal shifts. Let f be a function and c and d be positive numbers.

To graph	Shift the graph of `f`
$f (x) + d$ $f (x) + d$	up `d` units
$f (x) - d$ $f (x) - d$	down `d` units
$f (x - c)$ $f (x - c)$	right `c` units
$f (x + c)$ $f (x + c)$	left `c` units

Reflections:

To graph	Reflect the graph of `f` about the
$- f (x)$ $- f (x)$	`x`-axis
$f (- x)$ $f (- x)$	`y`-axis

Stretching and compressing. The graph of $g (x) = a f (x)$ $g (x) = a f (x)$ for $a > 0$ $a > 0$ has the same shape as the graph of f(x) and is stretched vertically away from the x-axis if $a > 1$ $a > 1$ and compressed vertically toward the x-axis if $0 < a < 1 .$ $0 < a < 1 .$ If a is negative, the graph of $| a | f (x)$ $| a | f (x)$ is reflected about the x-axis.

The graph of $g (x) = f (b x)$ $g (x) = f (b x)$ for $b > 0$ $b > 0$ is obtained from the graph of f by stretching it away from the y-axis if $0 < b < 1$ $0 < b < 1$ and compressing it horizontally toward the y-axis if $b > 1 .$ $b > 1 .$ If b is negative, the graph of $f (| b | x)$ $f (| b | x)$ is reflected about the y-axis.

Stretch the graph of $f (x) = - 3 \sqrt{- x} - 1 .$ $f (x) = - 3 \sqrt{- x} - 1 .$

Solution

The graph of $f (x) = - 3 \sqrt{- x} - 1$ $f (x) = - 3 \sqrt{- x} - 1$ is the graph of $f (x) = \sqrt{x}$ $f (x) = \sqrt{x}$ reflected about the y-axis, then reflected about the x-axis, stretched vertically by a factor of 3, and then shifted 1 unit down.

2.8 Combining Functions; Composite Functions

Given two functions f and g, for all values of x for which both f(x) and g(x) are defined, the functions can be combined to form sum, difference, product, and quotient functions.
The composition of f and g is defined by $(f \circ g) (x) = f (g (x)) .$ $(f \circ g) (x) = f (g (x)) .$

The input of f is the output of g. The domain of $f \circ g$ $f \circ g$ is the set of all x in the domain of g such that g(x) is in the domain of f.

In general, $f \circ g \neq g \circ f .$ $f \circ g \neq g \circ f .$

Let $f (x) = x^{2} - 5 x + 4$ $f (x) = x^{2} - 5 x + 4$ and $g (x) = x - 4 .$ $g (x) = x - 4 .$

Find $f + g, f - g, f \cdot g,$ $f + g, f - g, f \cdot g,$ and $\frac{f}{g}$ $\frac{f}{g}$ and the domain of each function.

Solution

\begin{array}{rcl} (f + g) (x) & = & x^{2} - 4 x \\ (f - g) (x) & = & x^{2} - 6 x + 8 \\ (f \cdot g) (x) & = & x^{3} - 9 x^{2} + 24 x - 16 \\ (\frac{f}{g}) (x) & = & \frac{x^{2} - 5 x + 4}{x - 4} = \frac{(x - 4) (x - 1)}{x - 4} = x - 1, x \neq 4 \end{array}

$\begin{array}{rcl} (f + g) (x) & = & x^{2} - 4 x \\ (f - g) (x) & = & x^{2} - 6 x + 8 \\ (f \cdot g) (x) & = & x^{3} - 9 x^{2} + 24 x - 16 \\ (\frac{f}{g}) (x) & = & \frac{x^{2} - 5 x + 4}{x - 4} = \frac{(x - 4) (x - 1)}{x - 4} = x - 1, x \neq 4 \end{array}$

The domain of $f + g, f - g,$ $f + g, f - g,$ and $f \cdot g$ $f \cdot g$ is $(- \infty, \infty) .$ $(- \infty, \infty) .$

The domain of $\frac{f}{g}$ $\frac{f}{g}$ is $(- \infty, 4) \cup (4, \infty) .$ $(- \infty, 4) \cup (4, \infty) .$

Let $f (x) = x^{2} - 3$ $f (x) = x^{2} - 3$ and $g (x) = \sqrt{x} .$ $g (x) = \sqrt{x} .$ Find $f \circ g$ $f \circ g$ and its domain.

Solution

\begin{array}{rcl} (f \circ g) (x) & = & f (g (x)) = {(\sqrt{x})}^{2} - 3 \\ = & x - 3, x \geq 0. \end{array}

$\begin{array}{rcl} (f \circ g) (x) & = & f (g (x)) = {(\sqrt{x})}^{2} - 3 \\ = & x - 3, x \geq 0. \end{array}$

The domain of $f \circ g$ $f \circ g$ is $[0, \infty) .$ $[0, \infty) .$

2.9 Inverse Functions

A function f is one-to-one if for any $x_{1}$ $x_{1}$ and $x_{2}$ $x_{2}$ in the domain of f, $f (x_{1}) = f (x_{2})$ $f (x_{1}) = f (x_{2})$ implies $x_{1} = x_{2} .$ $x_{1} = x_{2} .$
Horizontal-line test. If each horizontal line intersects the graph of a function f in at most one point, then f is a one-to-one function.
Inverse function. Let f be a one-to-one function. Then g is the inverse of f, and we write $g = f^{- 1}$ $g = f^{- 1}$ if $(f \circ g) (x) = x$ $(f \circ g) (x) = x$ for every x in the domain of g and $(g \circ f) (x) = x$ $(g \circ f) (x) = x$ for every x in the domain of f. The graph of $f^{- 1}$ $f^{- 1}$ is the reflection of the graph of f about the line $y = x .$ $y = x .$
If f is a one-to-one function, we find $f^{- 1}$ $f^{- 1}$ by the four-step procedure on page 299.

The function $f (x) = x^{2}$ $f (x) = x^{2}$ is not a one-to-one function because, for example, $y = 4$ $y = 4$ corresponds to both $x = 2$ $x = 2$ and $x = - 2 .$ $x = - 2 .$ That is, $f (2) = f (- 2)$ $f (2) = f (- 2)$ but $2 \neq - 2 .$ $2 \neq - 2 .$

Determine that $f (x) = 3 x - 6$ $f (x) = 3 x - 6$ is a one-to-one function. Find.

Solution

The graph of $f (x) = 3 x - 6$ $f (x) = 3 x - 6$ is a straight line with slope 3. By the horizontal-line test f is a one-to-one function.

To find the inverse of f:

\begin{array}{l} f (x) = 3 x - 6 & Given function \end{array}

$\begin{array}{l} f (x) = 3 x - 6 & Given function \end{array}$

$\begin{array}{l} y = 3 x - 6 & Replace f (x) with y . \end{array}$ $\begin{array}{l} y = 3 x - 6 & Replace f (x) with y . \end{array}$
$\begin{array}{l} x = 3 y - 6 & Interchange x and y . \end{array}$ $\begin{array}{l} x = 3 y - 6 & Interchange x and y . \end{array}$
$\begin{array}{l} y = \frac{x + 6}{3} & Solve for y . \end{array}$ $\begin{array}{l} y = \frac{x + 6}{3} & Solve for y . \end{array}$
$\begin{aligned} f^{- 1} (x) = \frac{1}{3} x + 2 & \frac{x + 6}{3} = \frac{x}{3} + \frac{6}{3} = \frac{x}{3} + 2 \\ Replace y with f^{- 1} (x) . \end{aligned}$ $\begin{aligned} f^{- 1} (x) = \frac{1}{3} x + 2 & \frac{x + 6}{3} = \frac{x}{3} + \frac{6}{3} = \frac{x}{3} + 2 \\ Replace y with f^{- 1} (x) . \end{aligned}$

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Table of Contents for Chapter 2 Review and Tests

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Table of Contents for
Chapter 2 Review and Tests