Download section 0.11: solving equations

(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.
1 1
= is solved by all nonzero real numbers. Its solution set
x x
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. §
• The equation
(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:
ax 2 + bx + c = 0 , where a ≠ 0 .
Its solutions are given by the Quadratic Formula:
− b ± b2 − 4ac
x=
2a
• Sometimes, the solutions are not real, but imaginary (see Chapter 2).
The discriminant of ax 2 + 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
b2 − 4ac
across. The formula is not: x = − b ±
.
2a
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 2x 2 − 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.)
2x 2 − 7x = 15
2x 2 − 7x − 15 = 0
TIP 1: If we had been asked to solve 4x 2 − 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 .
− b ± b2 − 4ac
x=
2a
=
− (− 7) ±
( − 7 ) − 4 ( 2)( −15) = 7 ±
2 ( 2)
2
49 + 120 7 ± 169
=
4
4
WARNING 5: We must simplify the radical.
(If we had obtained, say,
to simplify.)
=
7± 2
, there would be no need
4
7 ± 13
4
“+” case : x =
7 + 13 20
=
=5
4
4
“ − ” case: x =
7 − 13 − 6
3
=
=−
4
4
2
⎧ 3 ⎫
The solution set is: ⎨− , 5⎬ .
⎩ 2 ⎭
(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.
(Section 0.11: Solving Equations) 0.11.4
Example 3 (Using the Factoring Method; Revisiting Example 2)
Solve 2x 2 − 7x = 15 using the Factoring Method.
§ Solution
WARNING 6: Again, we must first isolate 0 on one side.
2x 2 − 7x = 15
2x 2 − 7x − 15 = 0
( 2x + 3) ( x − 5) = 0
By the Zero Factor Property,
2x + 3 = 0
x=−
or
3
2
x −5= 0
x=5
⎧ 3 ⎫
Again, the solution set is: ⎨− , 5⎬ . §
⎩ 2 ⎭
In Example 2, the discriminant of 2x 2 − 7x − 15 was 169, a perfect square,
which led to the elimination of the radical sign. As a result, 2x 2 − 7x − 15 = 0 had
rational numbers as solutions.
Also, by the Test for Factorability from Section 0.7, 2x 2 − 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 2x 2 − 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 ax 2 + 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 …
… a perfect square
… in fact, 0
… not a perfect square
Then, ax 2 + bx + c …
(Distinct) solutions to
ax 2 + bx + c = 0
… can be factored over 
… and is a PST
two rational numbers
one rational number
(Perfect Square Trinomial)
(a “double root”)
… is prime over 
… and positive
two irrational numbers
… and negative
two imaginary numbers
(see Chapter 2)
(Section 0.11: Solving Equations) 0.11.5
Square Root Method
If d > 0 , then x 2 = d ⇔ x = ± d .
If d = 0 , then x 2 = d ⇔ x 2 = 0 ⇔ x = 0 .
If d < 0 , then x 2 = d has no real solutions.
• For example, x 2 = 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 u 2 = d , where u is an expression in x or some
other variable.
• This is often the quickest method for solving ax 2 + bx + c = 0 when b = 0 .
Example 4 (Using the Square Root Method)
Solve 3x 2 − 4 = 0 using the Square Root Method.
§ Solution
We begin by isolating x 2 on one side of the equation.
3x 2 − 4 = 0
3x 2 = 4
4
x2 =
3
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=±
x=±
2
3
2 3
3
⎧⎪ 2 3 2 3 ⎫⎪
,
Technically, the solution set is: ⎨−
⎬. §
3
3
⎪⎩
⎪⎭
(Section 0.11: Solving Equations) 0.11.6
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 x 2 + 8x + 5 = 0 using the “CTS” Method.
§ Solution
We begin by isolating the x 2 and x terms on one side of the equation and
isolating a constant term on the other side.
x 2 + 8x + 5 = 0
x 2 + 8x = − 5
The coefficient of x 2 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).
x 2 + 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
x + 4 = ± 11
( by factoring the PST )
( 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
9
1
=
.
x 3 4x
§ Solution
We multiply both sides by the LCD, 4x 3 .
9
1
=
3
4x
x
9
1
4x 3 =
4x 3
3
4x
x
36 = x 2
( x ≠ 0)
( )
( )
x 2 = 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 x 2 = 4 , which
has both 2 and − 2 as solutions. However, − 2 is an extraneous solution
that must be rejected.
• Although x = 2 ⇒ x 2 = 4 , the equations x = 2 and x 2 = 4 are not
equivalent. That is, x = 2 ⇔ x 2 = 4 . We lose “reversibility” here.
(Section 0.11: Solving Equations) 0.11.8
Example 7 (Solving a Radical Equation)
Solve
x + 4 = x 2 + 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
x 2 + x − 21 ≥ 0 , we will instead check our tentative solutions at the end.
x + 4 = x 2 + x − 21 ⇒
(
x+4
) =(
2
x + x − 21
2
)
2
x + 4 = x 2 + x − 21
25 = x 2
x 2 = 25
x = ±5
We must check our tentative solutions.
(5) + 4 = (5) + (5) − 21
2
x = 5 checks out:
TIP 3: Beyond mechanical errors,
we need to check that radicands of
even roots are nonnegative.
9= 9
3= 3
x = − 5 does not check out:
( − 5) + 4 = ( − 5) + ( − 5) − 21
2
−1 = −1
WARNING 14: We must reject − 5 , because it yields a
non-real expression,
{}
The solution set is: 5 . §
−1 , in the check.
(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= 2
x=3
{
x −1= −2
x = −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
− 8 > 4x
Now add 3x to, and subtract 8 from, both sides.
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
)
or x ≤ − d ) .
x > d ⇔ x > d or x < − d , and
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 .
(Section 0.13: The Cartesian Plane and Circles) 0.13.2
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 ) + ( y2
2
y1 )
2
or, equivalently,
( x1
x2 ) + ( y1
y2 )
2
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 + x2 y1 + y2
,
2
2
• 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
( )
and 3, 3 is: d =
( 3 ( 2 )) + ( 3 1)
2
2
(
)
2, 1
= 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:
1
2 + 3 1+ 3
=
,
,2
2
2
2
(Section 0.13: The Cartesian Plane and Circles) 0.13.3
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
(
)
( )
x 2 + y 2 is the distance between a point x, y and the origin 0, 0 .
x 2 + y 2 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:
( )
x 2 + y2 = r 2
(
)
• This is because such a circle consists of all points x, y whose squared
( )
distance from 0, 0 is r 2 .
(Section 0.13: The Cartesian Plane and Circles) 0.13.4
Example 1 (The Graph of an Equation; A Circle Centered at the Origin)
( )
The graph of x 2 + y 2 = 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 x 2 + y 2 = 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) + ( y k ) = r 2
2
2
(
)
• 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
(Section 0.13: The Cartesian Plane and Circles) 0.13.5
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
( ( )) + ( y 1) = (3) , which we usually
We obtain the equation x
(
) (
2
2
)
2
2
2
2
rewrite as: x + 2 + y 1 = 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.)
(Section 0.13: The Cartesian Plane and Circles) 0.13.6
Example 3 (Finding the Standard Form of the Equation of a Circle)
A circle has as its equation:
4x 2 + 4 y 2 16x + 4 y 11 = 0
Find the standard form of this equation, and identify the center and the
radius of the circle.
§ Solution
The common coefficient of x 2 and y 2 is 4, so we will divide both sides of
the equation by 4.
4x 2 + 4 y 2 16x + 4 y 11 = 0
x2 + y2
4x + y
11
=0
4
Now, we group together the x 2 and x terms, and we group together the
y 2 and y terms. We isolate constant terms on the right-hand side.
(x
) (
)
4x + y 2 + y =
2
11
4
We now Complete the Square (CTS) in both groups.
(x
2
)
4x + 4 + y 2 + y +
11
1
1
= +4+
4
4
4
WARNING 2: Do not forget to add 4 and
the right-hand side, also.
1
to
4
We now factor both of the resulting Perfect Square Trinomials (PSTs).
(
1
x 2 + y+
2
)
2
2
=7
We now have the desired form, although the equation could be rewritten as:
( x 2)
The circle has center 2,
2
1
2
+ y
1
2
and radius
2
=7
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.
(Section 0.14: Lines) 0.14.2
PART B: NOTATION
Assume that m, a, b, c, x1 , x2 , y1 , y2 , A, B, and C represent real numbers.
Let ( x1 , y1 ) and ( x2 , y2 ) be two distinct points on a line.
m is the slope of the line (if defined).
( )
b, or the point ( 0, b) , is the y-intercept of the line (if there is exactly one).
a, or the point a, 0 , is the x-intercept of the line (if there is exactly one), and
(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
y y2
=
=
run
x x2
y1
x1
or, equivalently,
y1
x1
y2
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.
(Section 0.14: Lines) 0.14.4
Interpreting the Sign of Slope m
m>0
m=0
m<0
The line slopes upward (from left to right).
The line is horizontal (“flat”).
The line slopes downward.
m measures the steepness of the line.
Think: /
Think:
Think: \
(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 x1 ) ,
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 y
+ = 1,
a b
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
y
+ = 1 . Observe that 2,0 and 0,4 satisfy this equation.
2 4
(
)
( )
(Section 0.14: Lines) 0.14.7
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=
6 5
y
11
=
=
x 2 ( 4)
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,
y
y1 = m ( x
y 5=
(
4, 5 ) .
x1 )
( Point-Slope Form )
( ( 4 ))
11
x
6
We now solve for y and write the equation in Slope-Intercept Form,
y = mx + b .
(Section 0.14: Lines) 0.14.8
y=
y=
y=
y=
11
(x + 4) + 5
6
11
44
x
+5
6
6
11
22 15
x
+
6
3
3
11
7
x
6
3
Method 2 (Substitute Directly into Slope-Intercept Form)
In Slope-Intercept Form, y = mx + b , x and y are variables that must
11
. We need to find b.
stay in the final equation. We know that m =
6
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
22
+b
3
7
b=
3
5=
Replace m and b in y = mx + b with their values.
Again, x and y must stay in the final equation.
y=
§
11
x
6
7
3
(Section 0.14: Lines) 0.14.9
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.
their slopes are
Otherwise, two lines are perpendicular (“ ”)
opposite reciprocals of each other.
• For example, any line of slope 3 is parallel to any line of slope 3 in the
1
same plane, and it is perpendicular to any line of slope
in the same
3
plane.
• Also, any line of slope
same plane.
7
4
is perpendicular to any line of slope
in the
4
7
WARNING 2: A pair of perpendicular lines may not “appear”
perpendicular if the x- and y-axes are scaled differently.
(Section 0.14: Lines) 0.14.10
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)
y = 4.
Find the x- and y-intercepts of the line 2x
§ Solution
Find the x-intercept.
We substitute y = 0 and solve for x.
2x
2x
y= 4
(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
()
y= 4
2 0
y=4
( )
The y-intercept is 4, or the point 0,4 .
The equation 2x
y = 4 has as its Two-Intercept Form:
x
y
+ = 1.
2 4
(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 = s 2
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 =
Trapezoid
with bases b1 and b2 and
height h
Circle
with radius r
Area =
1
bh
2
b1 + b2
2
h
(the average of the bases
times the height)
Area = r 2
Circumference = 2 r
(the distance around)
(Section 0.15: Plane and Solid Geometry) 0.15.2
PART C: SOLID GEOMETRY
Description
Rectangular Box
with dimensions l, w, and h
Right Circular Cylinder
with base radius r and
height h
Right Circular Cone
with base radius r and
height h
Sphere
with radius r
Solid
Formulas
Volume = lwh
Surface Area = 2lw + 2wh + 2lh
(See Note 1.)
Volume = r 2 h
Lateral Surface Area = 2 rh
Total Surface Area = 2 rh + 2 r 2
(See Note 2.)
1 2
Volume =
r h
3
Lateral Surface Area = rl , with
slant height l = r 2 + h2
Total Surface Area = rl + r 2
(See Note 3.)
4 3
r
3
Surface Area = 4 r 2
Volume =
• 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
4 3
r does, in fact, yield a
measured in meters, then the volume formula V =
3
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=
k
for some nonzero k.
x
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=
kV
,
r2
where k is the (nonzero) constant of proportionality.
To find k, we use the fact that h = 2 when V = 18
2=
and r = 3.
( )
(3)
k 18
2
18 k
9
2=2 k
2
=k
2
1
k=
2=
To obtain our particular model, we must substitute this value of k into our
general model.
1
V
h=
r2
V
h= 2
r
Note: If we solve for V, we get V = r 2 h . What formula is this? §