SECTION 0.11: SOLVING EQUATIONS - · PDF file(Section 0.11: Solving Equations) 0.11.2 ......

(Section 0.11: Solving Equations) 0.11.1

SECTION 0.11: SOLVING EQUATIONS

LEARNING OBJECTIVES

• Know how to solve linear, quadratic, rational, radical, and absolute value equations.

PART A: DISCUSSION

• Much of precalculus is devoted to solving equations of various types. In this section, we will solve basic algebraic equations.

• We will solve polynomial equations more generally in Chapter 2, exponential and logarithmic equations in Chapter 3, and trigonometric equations in Chapter 4. We will solve systems of equations in Chapters 7 and 8.

PART B: SOLVING EQUATIONS

A solution to an equation in x is a number that makes the equation true when the number is substituted for x. For now, we only consider real solutions.

We solve an equation by finding its solution set, the set of all solutions.

When solving an equation, we often write a sequence of equivalent equations, which have the same solution set.

• Adding, subtracting, multiplying by, and dividing by the same nonzero number on both sides of an equation maintains equivalence.

Example Set 1 (Solving Equations)

• When we solve the linear equation 2x = 6 , we divide both sides by 2 and obtain the equivalent equation x = 3. The solution set of both equations is

3{ } . We can check a solution by verifying that it satisfies the original

equation: 2 3( ) = 6 , so 3 checks out.

• The equation x = x +1 has no real solutions. Its solution set is ∅ , the empty set (or null set). • The equation x +1= x +1 is solved by all real numbers. Its solution set is , so the equation is automatically called an identity.

• The equation 1x= 1

x is solved by all nonzero real numbers. Its solution set

is \ 0{ } , also written as −∞, 0( )∪ 0,∞( ) . The equation is an identity, in part because the only excluded real number (0) corresponds to a restriction. §

(Section 0.11: Solving Equations) 0.11.2 WARNING 1: There is a difference between simplifying an expression and solving an equation. For example, when we simplify the expression 2x + x , we write “ 2x + x = 3x ,” and 3x is our answer. On the other hand, when we solve the equation 2x + x = 3x , we state that the solution set is .

PART C: SOLVING QUADRATIC EQUATIONS

The general form of a quadratic equation in x is given by: ax2 + bx + c = 0 , where a ≠ 0 . Its solutions are given by the Quadratic Formula:

x =

−b ± b2 − 4ac2a

• Sometimes, the solutions are not real, but imaginary (see Chapter 2).

The discriminant of ax2 + bx + c is b2 − 4ac , the radicand in the formula.

• It may be denoted by D or Δ (uppercase delta). • It helps us classify (or “discriminate between”) types of solutions.

WARNING 2: Make sure the fraction bar in the formula goes all the way

across. The formula is not: x = −b ±

b2 − 4ac2a

.

WARNING 3: Here, the plus-minus sign ± indicates that we take both the result from the “+” case and the result from the “− ” case. (The results are equal ⇔ the discriminant b2 − 4ac is 0.) Sometimes in precalculus, the ± sign indicates that we do not yet know which sign to take.

Example 2 (Using the Quadratic Formula)

Solve the equation 2x2 − 7x = 15 using the Quadratic Formula.

§ Solution

WARNING 4: We must rewrite the equation in general form before we apply the Quadratic Formula. We must isolate 0 on one side of the equation. (Sometimes, it is easier to isolate 0 on the left-hand side.)

2x2 − 7x = 152x2 − 7x −15 = 0

TIP 1: If we had been asked to solve 4x2 −14x − 30 = 0 , we could have easily divided both sides by 2.

(Section 0.11: Solving Equations) 0.11.3 TIP 2: To avoid sign errors, we will identify a, b, and c before we apply the Quadratic Formula. Make sure to get them from the general form!

Here, a = 2 , b = −7 , and c = −15.

x = −b± b2 − 4ac2a

=− −7( ) ± −7( )2

− 4 2( ) −15( )2 2( ) = 7 ± 49+120

4= 7 ± 169

4

WARNING 5: We must simplify the radical.

(If we had obtained, say, 7 ± 2

4, there would be no need

to simplify.)

=

7 ±134

“+” case : x = 7 +13

4= 20

4= 5

“− ” case: x = 7 −13

4= −6

4= − 3

2

The solution set is: − 3

2, 5

⎧⎨⎩

⎫⎬⎭

.

(Some instructors list solutions in increasing order, although solution sets are technically unordered, meaning order doesn’t matter.) §

We also use the Factoring Method, the Square Root Method, and the Completing the Square (CTS) Method to solve quadratic equations. The Factoring Method for solving equations relies on the following Zero Factor Property.

Zero Factor Property (or Zero Product Property)

If a and b represent real quantities, then:

ab = 0 ⇔ a = 0 or b = 0( ) .

• Essentially, a product is 0 ⇔ a factor is 0, provided all factors are defined.


Example 3 (Using the Factoring Method; Revisiting Example 2)

Solve 2x2 − 7x = 15 using the Factoring Method.

§ Solution

WARNING 6: Again, we must first isolate 0 on one side.

2x2 − 7x = 152x2 − 7x −15 = 02x + 3( ) x − 5( ) = 0

By the Zero Factor Property,

2x + 3= 0 or

x = − 32

x −5= 0x = 5

Again, the solution set is: −

32

, 5⎧⎨⎩

⎫⎬⎭

. §

In Example 2, the discriminant of 2x2 − 7x −15 was 169, a perfect square, which led to the elimination of the radical sign. As a result, 2x2 − 7x −15 = 0 had rational numbers as solutions.

Also, by the Test for Factorability from Section 0.7, 2x2 − 7x −15 can be factored over the integers, . In Example 3, we used the Factoring Method as a quicker alternative to the Quadratic Formula when solving 2x2 − 7x −15 = 0 , and we obtained the same rational numbers as solutions.

Test for Factorability and Types of Solutions

The Test for Factorability applies to ax2 + bx + c , where a, b, and c are nonzero integers. (Assume the GCF is 1 or −1; otherwise, factor it out.)

If the discriminant b2 − 4ac is … Then, ax2 + bx + c …

(Distinct) solutions to ax2 + bx + c = 0

… a perfect square … can be factored over two rational numbers

… in fact, 0

… and is a PST (Perfect Square Trinomial)

one rational number (a “double root”)

… not a perfect square

… is prime over

… and positive

two irrational numbers

… and negative

two imaginary numbers (see Chapter 2)


Square Root Method

If d > 0 , then x2 = d ⇔ x = ± d .

If d = 0 , then x2 = d ⇔ x2 = 0 ⇔ x = 0 .

If d < 0 , then x2 = d has no real solutions.

• For example, x2 = 3 ⇔ x = ± 3 , while x

2 = −3 has no real solutions. WARNING 7: Remember that squares of real numbers are never negative.

• This method can be extended to u2 = d , where u is an expression in x or some other variable.

• This is often the quickest method for solving ax2 + bx + c = 0 when b = 0 .

Example 4 (Using the Square Root Method)

Solve 3x2 − 4 = 0 using the Square Root Method.

§ Solution

We begin by isolating x2 on one side of the equation.

3x2 − 4 = 03x2 = 4

x2 =43

We now apply the Square Root Method.

x = ± 4

3

WARNING 8: Do not forget the “± ” sign.

WARNING 9: If we have a numerical fraction as a radicand, we usually have to simplify. Here, we will rationalize the denominator.

x = ± 23

x = ± 2 33

Technically, the solution set is: −

2 33

, 2 33

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪. §


Completing the Square (“CTS”) Method

This method creates a perfect square trinomial (PST), which can be factored as the square of a binomial. That square is then isolated, and the Square Root Method is applied.

• This method can be easy to apply if a = 1 and b is even. (Other cases will be discussed in Chapters 2 and 10.)

• The Quadratic Formula can be derived using this method.

• CTS will be used to set up standard forms for equations of conics in Sections 0.13 and Chapters 2 and 10.

Example 5 (Using the “CTS” Method)

Solve x2 + 8x + 5 = 0 using the “CTS” Method.

§ Solution

We begin by isolating the x2 and x terms on one side of the equation and isolating a constant term on the other side.

x2 + 8x + 5 = 0x2 + 8x = −5

The coefficient of x2 is 1, so we may now complete the square. We accomplish this by adding 16 to both sides of the equation. Why 16? We take the coefficient of x (here, 8), halve it (resulting in 4), and then square the result (the square of 4 is 16).

x2 +8x +16 = −5+16

WARNING 10: Remember to add 16 to the right-hand side, also. We now have a PST on the left-hand side.

x + 4( )2= 11 by factoring the PST( )

x + 4 = ± 11 by the Square Root Method( )x = −4 ± 11

Technically, the solution set is: −4− 11, −4+ 11{ } . §

(Section 0.11: Solving Equations) 0.11.7 PART D: SOLVING RATIONAL EQUATIONS

We often solve a rational equation by first multiplying both sides by the LCD.

WARNING 11: Indicate restrictions that are “hidden” by this step.

Example 6 (Solving a Rational Equation)

Solve

9x3

=1

4x.

§ Solution

We multiply both sides by the LCD, 4x3 .

9x3 =

14x

9x3 4x3( ) = 1

4x4x3( )

36 = x2 x ≠ 0( )x2 = 36 x ≠ 0( )x = ±6

The solution set is: −6, 6{ } .

WARNING 12: If we had obtained x = 0 , we would have had to reject it. § PART E: SOLVING RADICAL EQUATIONS

We often solve a radical equation by isolating radicals on one or both sides of the equation and then raising both sides to the appropriate positive integer power. WARNING 13: If we raise both sides of an equation to an even power at any step, we must check any tentative solutions at the end and reject extraneous solutions. (Raising to an odd power does not require such a check.)

• Observe that, if we square both sides of x = 2 , we obtain x2 = 4 , which has both 2 and −2 as solutions. However, −2 is an extraneous solution that must be rejected.

• Although x = 2 ⇒ x2 = 4 , the equations x = 2 and x2 = 4 are not equivalent. That is, x = 2 ⇔ x2 = 4 . We lose “reversibility” here.


Example 7 (Solving a Radical Equation)

Solve x + 4 = x2 + x − 21 .

§ Solution

We square both sides of the equation. Since it is cumbersome to write restrictions that are “hidden” by this step, namely x + 4 ≥ 0 and x2 + x − 21≥ 0 , we will instead check our tentative solutions at the end.

x + 4 = x2 + x − 21 ⇒

x + 4( )2= x2 + x − 21( )2

x + 4 = x2 + x − 2125= x2

x2 = 25x = ±5

We must check our tentative solutions.

x = 5 checks out: 5( ) + 4 = 5( )2+ 5( )− 21

TIP 3: Beyond mechanical errors, we need to check that radicands of even roots are nonnegative.

9 = 93= 3

x = −5 does not check out:

−5( ) + 4 = −5( )2+ −5( ) − 21

−1 = −1

WARNING 14: We must reject −5 , because it yields a

non-real expression, −1 , in the check.

The solution set is: 5{ } . §

(Section 0.11: Solving Equations) 0.11.9. PART F: SOLVING ABSOLUTE VALUE EQUATIONS

Solving Absolute Value Equations

If d > 0 , then

x = d ⇔ x = ± d .

If d = 0 , then x = d ⇔ x = 0 ⇔ x = 0 .

If d < 0 , then

x = d has no solutions.

• For example, x = 3 ⇔ x = ±3 , while x = −3 has no solutions.

WARNING 15: Remember that absolute values are never negative.

• This method can be extended to u = d , where u is an expression in x or some other variable.

Example 8 (Solving an Absolute Value Equation)

Solve x −1 = 2 .

§ Solution

x −1 = 2

x −1= ±2

“+” case : “− ” case:

x −1= 2x = 3

x −1= −2x = −1

The solution set is: −1, 3{ } .

• Observe that −1 and 3 lie at a distance of two units away from 1 on the real number line. This is consistent with our discussion of absolute value and distance in Section 0.4.

§

(Section 0.12: Solving Inequalities) 0.12.1

SECTION 0.12: SOLVING INEQUALITIES

LEARNING OBJECTIVES

• Know how to solve linear inequalities and absolute value inequalities. PART A: DISCUSSION

• We will solve inequalities when we perform sign analyses and find domains of radical functions (see Section 1.1 and Chapter 2).

• Absolute value inequalities allow us to write compound inequalities more efficiently. They also help us describe an interval on the real number line with respect to its center.

• We will solve nonlinear inequalities in Chapter 2.

PART B: SOLVING LINEAR INEQUALITIES Strict inequalities involve the “<” (is less than) or the “>” (is greater than) signs. Weak inequalities involve the “≤” (is less than or equal to) or the “≥” (is greater than or equal to) signs. Inequations involve the “≠ ” (is not equal to) sign. Solving linear inequalities is similar to solving linear equations, except that:

• An inequality typically has infinitely many solutions, and the solution set is often written in interval form.

• WARNING 1: We must reverse the direction of the inequality sign if we switch the sides of an inequality, or if we multiply or divide both sides by a negative number.

Solving inequations is also similar to solving equations, although “≠ ” never needs to be reversed.

(Section 0.12: Solving Inequalities) 0.12.2

Example 1 (Solving a Linear Inequality)

Solve −3x > x + 8 .

§ Solution Method 1

−3x > x + 8 Now subtract x from both sides.

−4x > 8 Now divide both sides by −4 .

We must reverse the direction of the inequality sign.

x < −2

The solution set …

… in set-builder form is: x ∈ x < −2{ } , or

x ∈ : x < −2{ }

… in graphical form is:

… in interval form is: −∞, −2( ) §

§ Solution Method 2

−3x > x + 8 Now add 3x to, and subtract 8 from, both sides.

−8 > 4x Now switch sides.

We must reverse the direction of the inequality sign.

4x < −8

x < −2

See Method 1 for the solution set. §

(Section 0.12: Solving Inequalities) 0.12.3 PART C: SOLVING ABSOLUTE VALUE INEQUALITIES

Solving Absolute Value Inequalities

If d > 0 , then:

x < d ⇔ − d < x < d , and

x ≤ d ⇔ − d ≤ x ≤ d .

Also,

x > d ⇔ x > d or x < − d( ) , and

x ≥ d ⇔ x ≥ d or x ≤ − d( ) .

• TIP 1: Think of

x as the distance between x and 0 on the real number line.

• For example, x <1 ⇔ −1< x <1. This is a compound inequality that means:

x > −1 and x <1. The solution set is the interval −1,1( ) . It is the set of numbers that lie strictly within one unit of 0 on the real number line.

• Also, x >1 ⇔ x >1 or x < −1( ) . This is a different kind of compound

inequality. The solution set is −∞, −1( )∪ 1, ∞( ) . It is the set of numbers that are further than one unit from 0 on the real number line.

WARNING 2: Students incorrectly write 1< x < −1 here. This actually means x >1 and x < −1( ) , which corresponds to ∅ , the empty (or null) set.

• These methods can be extended to u , where u is an expression in x or some other variable.

(Section 0.12: Solving Inequalities) 0.12.4.

Example 2 (Solving an Absolute Value Inequality; Related to Section 0.11, Example 8)

Solve x −1 < 2 .

§ Solution

x −1 < 2

−2 < x −1< 2

We can add 1 to all three parts of this compound inequality.

−1 < x < 3

The solution set …

… in set-builder form is: x ∈ −1< x < 3{ } , or

x ∈ : −1< x < 3{ }

… in graphical form is:

… in interval form is: −1, 3( )

• The solution set is the set of numbers that lie strictly within two units of 1 on the real number line. This is consistent with our discussion of absolute value and distance in Section 0.4.

§

(Section 0.13: The Cartesian Plane and Circles) 0.13.1

SECTION 0.13: THE CARTESIAN PLANE and CIRCLES

LEARNING OBJECTIVES

• Understand the Cartesian plane and associated terminology. • Know how to plot points and graph equations in the Cartesian plane. • Know the Distance and Midpoint Formulas. • Be able to recognize, write, standardize, and graph equations of circles.

PART A: DISCUSSION

• We typically graph an equation in x and y in the Cartesian (or rectangular) plane, named after René Descartes. We plot points using their Cartesian coordinates. • There are alternate coordinate systems. We will discuss polar coordinates in Chapter 6. In three dimensions, we use Cartesian, cylindrical, and spherical coordinates.

PART B: THE CARTESIAN (OR RECTANGULAR) PLANE The Cartesian (or rectangular) plane is a plane with the Cartesian coordinate system imposed on it. We usually graph in the Cartesian xy-plane, though other variables could be used. We locate a horizontal line called the x-axis and a vertical line called the y-axis. These coordinate axes are real number lines; at least one nonzero tick mark should be placed on each axis to indicate scale. The axes intersect at the origin, O. A point in the plane corresponds to an ordered pair of the form x-coordinate, y-coordinate( ) . For example, the origin O corresponds to the ordered

pair 0, 0( ) , and the red point below corresponds to 2, 3( ) . If we name the red point

P, we can write P 2, 3( ) .


PART C: DISTANCE AND MIDPOINT FORMULAS

Distance Formula

The distance between points P x1, y1( ) and Q x2 , y2( ) in the Cartesian plane is given by:

d = x2 � x1( )2+ y2 � y1( )

2 or, equivalently, x1 � x2( )

2+ y1 � y2( )

2 • This is proven using the Pythagorean Theorem, which we will discuss in Chapter 4 on trigonometry.

Midpoint Formula

The midpoint of PQ , the line segment with endpoints P and Q, is given by:

x1 + x22

,y1 + y22

�

��

��

• Observe that the x-coordinate is the average of the x-coordinates of the endpoints, and the y-coordinate is the average of the y-coordinates.

• For example, in the figure below, the distance between the points �2,1( )

and 3, 3( ) is: d = 3� �2( )( )2

+ 3�1( )2

= 29 . If the coordinate axes are

scaled in (say) meters, then the distance is 29 meters.

• The midpoint M of the red line segment is:

�2 + 32

, 1+ 32

��

��=

12

, 2��

��


PART D: THE GRAPH OF AN EQUATION and CIRCLES

The Graph of an Equation; the “Basic Principle of Graphing”

The graph of an equation consists of all points whose coordinates satisfy the equation. The points correspond to solutions of the equation.

Circles A circle is the set of all points in a plane that are a fixed distance (r, the radius) away from a fixed point (the center). The diameter d is twice the radius.

Distance and Squared Distance from the Origin x2 + y2 is the distance between a point x, y( ) and the origin 0, 0( ) .

x2+ y2 is the squared distance between them.

Equation of a Circle with Center 0, 0( ) The standard form of the equation of a circle (in the xy-plane) with center 0, 0( ) and radius r, where r > 0 , is given by:

x2+ y

2= r

2

• This is because such a circle consists of all points x, y( ) whose squared

distance from 0, 0( ) is r 2 .


Example 1 (The Graph of an Equation; A Circle Centered at the Origin)

The graph of x2 + y2 = 9 is the circle below with center 0, 0( ) and radius 3.

The ordered pairs 3, 0( ) , �3, 0( ) , 0, 3( ) , and 0, �3( ) are all solutions of

the equation x2 + y2 = 9 , and their corresponding points lie on the circle.

Other points such as 7, 2( ) also lie on the circle.

§

Equation of a Circle with Center h, k( )

More generally, if the circle has center h, k( ) , the standard form is given by:

x � h( )2

+ y � k( )2

= r2

• The left-hand side is the squared distance between the points x, y( ) and

h, k( ) . The circle consists of all points x, y( ) whose squared distance from

h, k( ) is r 2 .


Example 2 (Finding the Equation of a Circle)

Find an equation of the circle in the xy-plane with center �2,1( ) and radius 3.

§ Solution

We obtain the equation

x � �2( )( )2+ y �1( )

2= 3( )

2, which we usually

rewrite as: x + 2( )2+ y �1( )

2= 9 .

WARNING 1: The center of the circle is �2,1( ) , not 2, �1( ) . If we are given the equation of a circle in standard form, we can find the center by asking, “What makes the left-hand side equal to 0?” (The center has a squared distance of 0 from itself.) §

The circles from Examples 1 and 2 are graphed below:

The circles are translations of one another. (See Section 1.4.)


Example 3 (Finding the Standard Form of the Equation of a Circle)

A circle has as its equation:

4x2+ 4y2

�16x + 4y �11= 0

Find the standard form of this equation, and identify the center and the radius of the circle.

§ Solution

The common coefficient of x2 and y2 is 4, so we will divide both sides of

the equation by 4.

4x2 + 4y2 �16x + 4y �11= 0

x2 + y2 � 4x + y � 114= 0

Now, we group together the x2 and x terms, and we group together the

y2 and y terms. We isolate constant terms on the right-hand side.

x2� 4x( ) + y2

+ y( ) =114

We now Complete the Square (CTS) in both groups.

x2 � 4x + 4( ) + y2

+ y + 14

��

��=

114+ 4 +

14

WARNING 2: Do not forget to add 4 and 14

to

the right-hand side, also.

We now factor both of the resulting Perfect Square Trinomials (PSTs).

x � 2( )

2+ y + 1

2��

��

2

= 7

We now have the desired form, although the equation could be rewritten as:

x � 2( )

2+ y � �

12

��

��

�

��

��

2

= 7

The circle has center 2, � 12

��

��

and radius 7 . §

(Section 0.14: Lines) 0.14.1

SECTION 0.14: LINES

LEARNING OBJECTIVES

• Understand and compute the slope of a line. • Distinguish between equations of horizontal lines and those of vertical lines. • Know how to write equations of lines in various forms, including Point-Slope Form, Slope-Intercept Form, and Two-Intercept Form. • Understand parallel and perpendicular lines and relate their slopes. • Know how to find the intercepts of a line.

PART A: DISCUSSION

• We frequently graph lines in precalculus and calculus.

• In this section, we will graph lines in the xy-plane, though we can work with different variables. • There are many ways to write an equation for a line. The form we select may depend on the information we have about the line, or on the information we want to find or display.

• In Section 1.11, we will discuss linear approximations of functions and graphs by tangent lines, a crucial idea in calculus.


PART B: NOTATION

Assume that m, a, b, c, x1, x

2, y

1, y

2, A, B, and C represent real numbers.

Let x1, y1( ) and x

2, y

2( ) be two distinct points on a line.

m is the slope of the line (if defined).

a, or the point a, 0( ) , is the x-intercept of the line (if there is exactly one), and

b, or the point 0, b( ) , is the y-intercept of the line (if there is exactly one).

(Section 0.14: Lines) 0.14.3 PART C: m, THE SLOPE OF A LINE

Formulas for m, the Slope of a Line

m =rise

run=�y�x

=y2 � y1

x2 � x1

or, equivalently, y1 � y2

x1 � x2

��

��

• If the line is vertical, then m is undefined. • A negative rise can be interpreted as a drop.

• � (uppercase delta) denotes “change in.”

Interpretation of Slope m as Marginal Change

For every unit increase in x along the line, y changes by m.

• This idea will be developed in Section 1.11.


Interpreting the Sign of Slope m

m > 0 � The line slopes upward (from left to right). Think: / m = 0 � The line is horizontal (“flat”). Think: � m < 0 � The line slopes downward. Think: \

m measures the steepness of the line.

(Section 0.14: Lines) 0.14.5 PART D: HORIZONTAL AND VERTICAL LINES

Equations of Horizontal and Vertical Lines

The graph of y = c is a horizontal line.

• For instance, the graph of y = 0 is the x-axis.

The graph of x = c is a vertical line.

• For instance, the graph of x = 0 is the y-axis.

WARNING 1: The two equation forms are frequently confused. TIP 1: Remember the Basic Principle of Graphing. For example, the graph of x = 2 consists of all points with x-coordinate 2. The graph is a vertical line. The graph of y = 2 is a horizontal line.

Slopes of Horizontal and Vertical Lines

m = 0 for a horizontal line.

m is undefined for a vertical line.

(Section 0.14: Lines) 0.14.6 PART E: FORMS FOR THE EQUATION OF A LINE

The following forms are used to write the equation of a line in the xy-plane.

General Form

Ax + By = C , where A and B are not both 0

Point-Slope Form

y � y1= m x � x

1( ) , where m is defined

• This is derived from the idea that m =�y

�x.

Slope-Intercept Form

y = mx + b , where m is defined

Two-Intercept Form

x

a+y

b= 1,

where the line has a unique x-intercept a, or a, 0( ) , and

a unique y-intercept b, or 0, b( ) , and neither a nor b is 0.

For example, the Two-Intercept Form of the equation of the line below is: x

�2+y

4= 1 . Observe that �2,0( ) and 0,4( ) satisfy this equation.


Two distinct points determine a line. • In other words, given two different points, exactly one line can pass through them.

Example 1 (Finding Forms of the Equation of a Line) Find the Slope-Intercept Form of the equation of the line passing through the points � 4, 5( ) and 2, �6( ) .

§ Solution

Find m, the slope of the line:

m =�y

�x=

�6 � 5

2 � � 4( )= �

11

6

Method 1 (First Find a Point-Slope Form)

Either of the two given points may be used as our “point.” TIP 2: There are infinitely many Point-Slope Forms for the equation of this line, since any point on the line can be used as our “point.” However, the Slope-Intercept Form is unique, because a nonvertical line has only one slope and one y-intercept. Let’s use the first given point, � 4,5( ) .

y � y1 = m x � x1( ) Point-Slope Form( )

y � 5 = �11

6x � � 4( )( )

We now solve for y and write the equation in Slope-Intercept Form, y = mx + b .


y = �11

6x + 4( ) + 5

y = �11

6x �

44

6+ 5

y = �11

6x �

22

3+15

3

y = �11

6x �

7

3

Method 2 (Substitute Directly into Slope-Intercept Form)

In Slope-Intercept Form, y = mx + b , x and y are variables that must

stay in the final equation. We know that m = �11

6. We need to find b.

According to the Basic Principle of Graphing, x = � 4, y = 5( ) must satisfy the equation of the line. We now solve for b:

y = mx + b �

5( ) = �11

6

��

�� 4( ) + b

5 =22

3+ b

b = �7

3

Replace m and b in y = mx + b with their values. Again, x and y must stay in the final equation.

y = �11

6x �

7

3

§


PART F: PARALLEL AND PERPENDICULAR LINES

Parallel and Perpendicular Lines

Vertical lines are parallel. Otherwise, two lines are parallel (“||”) � their slopes are equal.

A horizontal line and a vertical line are perpendicular. Otherwise, two lines are perpendicular (“� ”) � their slopes are opposite reciprocals of each other.

• For example, any line of slope 3 is parallel to any line of slope 3 in the

same plane, and it is perpendicular to any line of slope �1

3 in the same

plane.

• Also, any line of slope �4

7 is perpendicular to any line of slope

7

4 in the

same plane.

WARNING 2: A pair of perpendicular lines may not “appear” perpendicular if the x- and y-axes are scaled differently.


PART G: FINDING INTERCEPTS

Finding Intercepts

Consider the graph of an equation in x and y. To find its x-intercept(s), set y equal to 0 (the x-axis has equation y = 0 ), and solve for x. To find its y-intercept(s), set x equal to 0 (the y-axis has equation x = 0 ), and solve for y. WARNING 3: Remember which variable to set equal to 0 and which variable to solve for.

(Section 0.14: Lines) 0.14.11 Example 2 (Finding Intercepts)

Find the x- and y-intercepts of the line 2x � y = �4 .

§ Solution Find the x-intercept.

We substitute y = 0 and solve for x.

2x � y = �4 �

2x � 0( ) = �4

x = �2

The x-intercept is �2 , or the point �2,0( ) . Find the y-intercept.

We substitute x = 0 and solve for y.

2x � y = �4 �

2 0( ) � y = �4

y = 4

The y-intercept is 4, or the point 0,4( ) .

The equation 2x � y = �4 has as its Two-Intercept Form: x

�2+y

4= 1 .

(To see this, divide both sides of the first equation by the right-hand side, �4 .) These are equivalent equations, and they have the same graph.

§

(Section 0.15: Plane and Solid Geometry) 0.15.1

SECTION 0.15: PLANE AND SOLID GEOMETRY

LEARNING OBJECTIVES

• Know formulas for area, perimeter, and circumference in plane geometry. • Know formulas for volume and surface area in solid geometry. • Be able to use dimensional analysis to check the validity of formulas.

PART A: DISCUSSION

• A number of geometric formulas must be memorized in preparation for calculus. We use them in the study of rates of change, related rates, optimization, and mass.

PART B: PLANE GEOMETRY

Description Plane Figure Formulas Square with side length s

Area = s2 Perimeter = 4s (the distance around)

Rectangle with base b and height h (covers Square)

Area = bh Perimeter = 2b + 2h

Parallelogram with base b and height h (covers Rectangle, Square)

Area = bh

Triangle with base b and height h (think: half a Parallelogram)

Area =

12

bh

Trapezoid with bases b1 and b2 and height h

Area =

b1 + b2

2�

��

��h

(the average of the bases times the height)

Circle with radius r

Area = �r 2 Circumference = 2�r (the distance around)

(Section 0.15: Plane and Solid Geometry) 0.15.2 PART C: SOLID GEOMETRY

Description Solid Formulas Rectangular Box with dimensions l, w, and h

Volume = lwh Surface Area = 2lw + 2wh + 2lh (See Note 1.)

Right Circular Cylinder with base radius r and height h

Volume = �r 2h Lateral Surface Area = 2�rh Total Surface Area = 2�rh + 2�r 2

(See Note 2.)

Right Circular Cone with base radius r and height h

Volume = 1

3�r 2h

Lateral Surface Area = �rl , with slant height l = r 2 + h2

Total Surface Area = �rl + �r 2 (See Note 3.)

Sphere with radius r

Volume = 4

3�r3

Surface Area = 4�r 2

• In calculus, you may verify some of these formulas. • We can use dimensional analysis to help check our formulas. If lengths are measured in meters, say, then surface areas are measured in square meters, and volumes are measured in cubic meters. For example, if the radius r of a sphere is

measured in meters, then the volume formula V =

43�r3 does, in fact, yield a

volume in cubic meters. This analysis prevents us from accidentally switching this formula with the formula for surface area.

Note 1 (Box)

The volume equals the rectangular base area times the height. The surface area is the sum of the areas of the six sides. Think of the walls, floor, and ceiling of a room.

(Section 0.15: Plane and Solid Geometry) 0.15.3

Note 2 (Cylinder)

The volume equals the circular base area times the height. The total surface area equals the sum of the lateral surface area and the two circular base areas.

The lateral surface area equals the base circumference times the height. • Consider the area of a soup can label. Imagine slitting the label along the red dashed line segment below and spreading it out as the rectangle on the right.

Note 3 (Cone)

The volume equals one-third of the volume of the right circular cylinder with the same base radius and height. (The cone “fits snugly” within this cylinder.) The total surface area equals the sum of the lateral surface area and the circular base area.

(Section 0.16: Variation) 0.16.1

SECTION 0.16: VARIATION

LEARNING OBJECTIVES

• Know how to model direct, inverse, and joint variation. • Be able to find constants of proportionality (or variation).

PART A: DISCUSSION

• The terminology and modeling techniques of this section are used in physics and calculus, particularly in applications involving mass and force.

PART B: VARIATION

y is directly proportional to x (or y varies directly as x) �

y = kx for some nonzero constant of proportionality (or constant of variation) k.

WARNING 1: k could be negative, so y does not necessarily increase as x increases.

y is inversely proportional to x (or y varies inversely as x) �

y =

kx

for some nonzero k.

z is jointly proportional to x and y (or z varies jointly as x and y) �

z = kxy for some nonzero k.

(Section 0.16: Variation) 0.16.2

Example 1 (Modeling Using Variation)

h is directly proportional to V and is inversely proportional to the square of r.

Find the particular mathematical model related to this statement if h is 2 when V is 18� and r is 3.

§ Solution

Our general model is:

h =

kVr 2

,

where k is the (nonzero) constant of proportionality. To find k, we use the fact that h = 2 when V = 18� and r = 3.

2 =k 18�( )

3( )2

2 =18�k

92 = 2�k

22�

= k

k =1�

To obtain our particular model, we must substitute this value of k into our general model.

h =

1�

��

��

V

r 2

h =V�r 2

Note: If we solve for V, we get V = �r 2h . What formula is this? §

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