Linear and Quadratic Functions - skylit.com · 1 6 Linear and Quadratic Functions 6.1 Prologue...

1

6 Linear and Quadratic Functions

6.1 Prologue Definition 4 in Book I of Euclid’s Elements (3rd century BC) reads: “A straight line is a line that lies evenly with the points on itself.” No one knows for sure what exactly he meant by this — maybe the meaning got lost in translation. Perhaps he meant that if you aim from one point to another, all the points in between fall on that line and no points stick out. In any case, we know what a straight line is. Or do we? A stretched piece of string? A ray of light? This is probably the most deceptive intuitive notion. Einstein’s theory of relativity states that rays of light are actually bent by gravity. In general, our faith in Euclidean geometry has been shaken by more recent (18th century) models of non-Euclidean geometries, in which there exists more than one line through a given point that is parallel to a given line, or in which no parallel lines exist at all. What if our space is actually curved? Here we’ll avoid these mind-bending considerations and stick with the Euclidean point of view. We define a straight line on the Cartesian plane as a graph of a linear relation px qy C+ = . If 0q ≠ , this relation is a function of the form y mx b= + , where

pmq

= − and Cbq

= . A function y mx b= + is called a linear function. In

Sections <...>-<...>, we will discuss the properties of linear relations and functions. A function given by the formula 2( )f x ax bx c= + + , where a, b, and c are constants and 0a ≠ , is called a quadratic function. Its graph is a parabola, a curve with many wonderful properties. We will discuss the properties of quadratic functions and their graphs in Sections <...>-<...>.

6.2 Linear Relations

A relation from to of the form px qy C+ = , where p, q, and C are constants (and at least one of p and q is not equal to 0) is called a linear relation. The graph of a linear relation in the Cartesian plane is a straight line.

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2 CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS

It is not very hard to prove that this definition of a straight line is consistent with the postulates of Euclidean geometry. For example, Postulate 1 states that we can draw a straight line through any two points. Indeed, the points 1 1( , )x y and 2 2( , )x y lie on the graph of px qy C+ = , where 2 1( )p y y= − , 2 1( )q x x= − − , and 1 2 2 1C x y x y= − (see Question <...> in the exercises). It is also not very hard to prove that there is only one line that can be drawn through any two given points. A straight line is an abstraction: our intuition (or our education) tells us that we can align an imaginary ruler, called a straightedge, with any two points, and draw a straight line through them with an infinitely thin pencil.

Example 1 Find a linear relation that defines a straight line through the points (2, 5) and (1, 4).

Solution Using the above formulas, 2 1( )p y y= − , 2 1( )q x x= − − , and 1 2 2 1C x y x y= − , we get:

2 1 4 5 1p y y= − = − = − , 2 1( ) (1 2) 1q x x= − − = − − = , 1 2 2 1 2 4 1 5 3C x y x y= − = ⋅ − ⋅ = . Thus, the relation 3x y− + = or 3y x− = defines the line through (2, 5) and (1, 4).

By our definition, the x-axis is a straight line, because it is the graph of the relation 0 1 0 0x y y⋅ + ⋅ = ⇔ = . In general, if p = 0, the graph is the horizontal line

Cyq

= . Similarly, the y-axis is a straight line, because it is the graph of

1 0 0 0x y x⋅ + ⋅ = ⇔ = . In general, if q = 0, the graph is the vertical line Cxp

= .

Two straight lines that do not intersect (have no common points) are called parallel.

Clearly the graphs of 1px qy C+ = and 2px qy C+ = cannot share a point when 1C and 2C are different, so these lines are parallel. If p and q are fixed, and we vary C, we get a family of parallel lines (Figure 6-1). One of them, the graph of 0px qy+ = , passes through the origin.

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CHAPTER 6 ~ LINEAR AND QUADRATIC FUNCTIONS 3

x

y

1px qy C+ =

2px qy C+ =

3px qy C+ =

0px qy+ =

O

Figure 6-1. The graphs of px + qy = C for different values of C

produce a family of parallel lines

This model of Euclidean geometry is also consistent with Euclid’s fifth postulate, which is equivalent to the statement that given a straight line l and a point P outside it, we can draw precisely one line through P parallel to l:

P

l

If a straight line is neither vertical nor horizontal, that is, 0p ≠ and 0q ≠ , and it doesn’t pass through the origin, that is 0C ≠ , then it has one x-intercept, 0a ≠ and one y-intercept, 0b ≠ :

O x

y

a

b

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An equation of such a line can be written as 1x ya b+ = , where 1p

a= , 1q

b= , and

1C = ).

This form of the equation of the line is called the intercept form.

Example 2 Sketch a straight line through (2, 0) and (0, 1) and determine its equation.

Solution

2 -1

1

-2 -1 O 1 x

y

The x-intercept is 2 and the y-intercept is 1. Using the intercept form, we

immediately find the equation: 12 1x y+ = or 2 2x y+ = .

Example 3 Sketch the graph of x + 2y = 3.

Solution We know from Example 2 what the graph of x + 2y = 2 looks like. The graph x + 2y = 3 is a line parallel to the line x + 2y = 2, but its x-intercept is 3, not 2:

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2 -1

1

3 -1 O 1 x

y

To solve this problem “from scratch,” set y to 0 to get the x-intercept, 3; then set x to

0 to get the y-intercept, 32

, then connect the two intercept points with the straight

line.

Any line 1px qy C+ = is perpendicular to any line 2( )q x py C− + = .

Recall from Chapter <....> that two vectors are perpendicular to each other if and only if their dot product is equal to 0. Also recall that the vector v , which represents the difference of the vectors 2 2( , )x y and 1 1( , )x y , 2 1 2 1( , )v x x y y= − − , is parallel to the line that passes through the points 1 1( , )x y and 2 2( , )x y :

2 2( , )x y

O x

y

1 1( , )x y v

Suppose an equation of this line is px qy C+ = . If ( , )u p q= , then the dot product

2 1 2 1( ) ( )u v p x x q y y⋅ = − + − = 2 2 1 1( ) ( ) 0px qy px qy C C+ − + = − = . This means u is perpendicular to v . Thus a line px qy C+ = is perpendicular to the vector

( , )u p q= . We can see it graphically, too. For example:

2 -1

1

3 -1 O 1 x

y

x + 2y = 3

2 (1, 2)

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If we rotate u by 90 degrees, we get the vector ( , )u q p⊥ = − : it has the same length as u and is perpendicular to it. Indeed, their dot product, ( ) 0u u p q qp⊥⋅ = − + = .

Example 4 Find an equation of the line that contains the point ( 1, 3)− and is perpendicular to the line 3 5 1x y+ = .

Solution The equation has the form ( 5) 3x y C− + = . For the point ( 1, 3)− to be on that line, we must have ( 5)( 1) 3 3 14C C− − + ⋅ = ⇒ = . The answer is 5 3 14x y− + = .

Exercises 1. When and where did Euclid live? How many books make up Euclid’s

Elements? In Questions 2-5, write an equation px qy C+ = for the straight line that passes through the two given points. 2. (-2, -1) and (3, 5) 3. (2, 4) and (-2, -2) 4. (2, 2) and (2, 5) 5. (1, -2) and (-3, -2) In Questions 6-9, write an equation for the straight line that passes through the given point and is parallel to the given line. 6. Point (-3, 1) and line 0x y+ = . 7. Point (2, 5) and line 0y = . 8. Point (4, -1) and line 3 7x y+ = .

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9. Point (-1, 2) and line 2 2x y− = . In Questions 10-13, write an equation for the straight line through the two given

points in the intercept form, 1x ya b+ = .

10. (0, 3) and (4, 0). 11. (0, -2) and (3, 0). 12. (4, 0) and (0, -1). 13. (-3, 2) and (2, -3). Hint: First write px qy C+ = , then find the

intercepts. In Questions 14-17, write an equation for the straight line that passes through the given point and is perpendicular to the given line. 14. Point (0, 0) and line 1x y+ = . 15. Point (6, -3) and line 2x = . 16. Point (3, -3) and line 0x y− = . 17. Point (-1, 2) and line 2 2x y− = . 18. Prove that the line px qy C+ = , where 2 1( )p y y= − , 2 1( )q x x= − − , and

1 2 2 1C x y x y= − contains the points 1 1( , )x y and 2 2( , )x y . 19. Prove that only one straight line passes through the points 1 1( , )x y and

2 2( , )x y . Hint: First prove for the case where one of the points is the origin; then, reduce the general case to this special case.

20. The lines 4 3 25x y+ = and 4 3 75x y+ = are parallel. What is the distance

between them? Hint: “Draw” a perpendicular line through the origin and find the distance between its intersections with the given lines.

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21. Consider the surface of a sphere. Let us define a “straight line” on it as a “great circle” (that is, a circle on the surface of the sphere centered at the center of the sphere). Is Euclid’s First Postulate (that there is a unique straight line through any two points) true in this geometry? What about the Fifth Postulate (equivalent to the statement that for any line l and a point P outside of it there exists precisely one line through P parallel to l)?

6.3 Linear Functions A linear relation px qy C+ = in which 0q ≠ can be rewritten as y mx b= + , where

pmq

= − and Cbq

= .

The formula ( )f x mx b= + defines a linear function.

The natural domain of this function is all real numbers. m is called the slope of the line and b is its y-intercept (Figure 6-2).

b

O x

y y=mx+b

θ

Δx(“delta-x”) ( )tanym m

xθΔ

= =Δ

Δy(“delta-y”)

Figure 6-2. The slope m of a line y = mx + b is equal to xΔ

(“delta x”) over yΔ (“delta y”) (equal to the tangent of the angle the line makes with the x-axis)

The y = mx + b form of an equation for a straight line is called the slope-intercept form, where m is the slope and b is the y-intercept.

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Some teachers use the phrase “rise over run” to help their students understand how the slope is measured. But the “rise” yΔ (“delta y”) can actually be a “fall” when the slope is negative ( 0yΔ < ):

b

O x

y y=mx+b

θ

Δx(“delta-x”)

Δy(“delta-y”)

In any case, the slope is always equal to tanθ ; when 90 180 ( )2πθ θ π° < < ° < < ,

tanθ is negative.

A constant function ( )f x b= is a special case of a linear function; its graph is a horizontal line with the slope m = 0.

The slope of a vertical line is undefined (infinity). The line y = mx passes through the origin. If the slope m is positive, the line is located in the first and third quadrants:

x

y y = mx m > 0

If the slope is negative, the line is located in the second and fourth quadrants:

x

y y = mxm < 0

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Example 1 Rewrite the equation 2 3 7x y− = in slope-intercept form.

Solution

2 7 2 72 3 7 3 2 73 3 3

xx y y x y y x− +− = ⇒ − = − + ⇒ = ⇒ = −

−.

Example 2 What is the slope of the line 1x y+ = ?

Solution

1 1x y y x+ = ⇒ = − + , so the slope is 1− .

Example 3 Sketch the graph of 2 1y x= − .

Solution The y-intercept is 1− and the slope is 2:

y

x 0 -1

-2

3

1

2

-3

1 2 3 -1 -2 -3

2

1

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If a line passes through the points 1 1( , )x y and 2 2( , )x y and it is not vertical (that is,

1 2x x≠ ), then its slope 2 1

2 1

y ymx x−

=−

(Figure 6-3).

O x

y

1 1( , )x yθ

2 1

2 1

y ymx x−

=−

2 2( , )x y

2 1y y−

2 1x x−

Figure 6-3. The slope m of a line through the points

1 1( , )x y and 2 2( , )x y is 2 1

2 1

y ymx x−

=−

So, if 1 1( , )x y is a point on a non-vertical line, then for any other point ( , )x y on that

line 1

1

y y mx x−

= ⇒−

1 1( )y y m x x− = − or 1 1( )y m x x y= − + .

This form of an equation for a line is called the point-slope form.

Example 4

y

x0 -1

3

1

2

1 2 3 -1 -2 -3

Write an equation of a line shown on the graph above.

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Solution

The line passes through the point (1, 2) and has a slope of 12

. Its equation can be

written as 1 1 12 ( 1) 2 0.5 1.52 2 2

y x y x y x− = − ⇔ = − + ⇔ = + .

Example 5 Write an equation of the line that passes through the points ( 1, 3)− and (2, 5)− .

Solution

2 1

2 1

( 5) 3 8 82 ( 1) 3 3

y ymx x− − − −

= = = = −− − −

. Using the point-slope form,

( )8 8 8 8( 1) 3 ( 1) 3 33 3 3 3

y x x x= − − − + = − + + = − − + ⇔ 8 13 3

y x= − + .

Different forms of an equation for a straight line are convenient in different situations. To summarize:

Standard form: px qy C+ =

Slope-intercept form: y mx b= +

Point-slope form: 1 1( )y y m x x− = − or 1 1( )y m x x y= − +

Two-point form, 2 1x x≠ : 1 1( )y m x x y= − + , where 2 1

2 1

y ymx x−

=−

Horizontal line: y = b

Vertical line: x = a

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Recall from Section <...> that the lines 1px qy C+ = and 2qx py C− + = are

perpendicular to each other. The slope of the first line is p mq

− = ; the slope of the

second line is ( ) 1q qp p m−

− = = − . Therefore,

any two lines 1y mx b= + and 21y x bm

= − + are perpendicular to each

other. In other words, if two lines with slopes 1m and 2m are perpendicular to each other,

then 21

1mm

= − .

Example 6 Write an equation of the line that passes through the origin and is perpendicular to the line from Example 5.

Solution In Example 5 we found that the slope of the line that passes through the points

( 1, 3)− and (2, 5)− is 83

m = − . The slope of the perpendicular line is

1 3 38 8m

⎛ ⎞− = − − =⎜ ⎟⎝ ⎠

. The answer is 38

y x= .

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Exercises In Questions 1-4 write the slope-intercept equation of the line shown in the graph. 1. 2.

y

x0 -1

-2

3

1

2

-3

1 2 3 -1 -2 -3

y

x0 -1

-2

3

1

2

-3

1 2 3 -1 -2 -3

3. 4.

y

x0 -1

-2

3

1

2

-3

1 2 3 -1 -2 -3

y

x0 -1

-2

3

1

2

-3

1 2 3 -1 -2 -3

In Questions 5-8 sketch by hand the graph of the given linear function and label the y-intercept. 5. 2y x= − 6. 2 3y x= − 7. 6 2y x= − +

8. 1 52 2

y x= +

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In Questions 9-12 write an equation of the line that passes through the given point and is parallel to the given line. 9. Point (0, 0) , line 5 1y x= + . 10. Point (1, 2) , line 3 1y x= − . 11. Point ( 2, 1)− − , line 4 3y x= − . 12. Point (0, 3)− , line y x= − . In Questions 13-16 find the slope of the line that passes through the two given points, write its equation, sketch its graph, and label its y-intercept. 13. ( 3, 1)− − and (3,1) 14. (1, 0) and (5, 3) 15. (2, 2)− and ( 4, 2)− 16. ( 3, 3)− and (1, 3) In Questions 17-20 write an equation of the line that passes through the given point and is perpendicular to the given line. 17. Point (0, 0) , line 1y x= − + . 18. Point (2,1) , line 2 1y x= + . 19. Point ( 4, 1)− − , line 3 2y x= − .

20. Point (2, 2) , line 1 32

y x= − + .

21. What is the distance from the origin to the line 3 52 2

y x= − + ?

22. Given a triangle ABC with the vertices (2, 3)A , (1, 3)B − , and ( 4, 1)C − − ,

write an equation of the line that contains the altitude from A to BC.

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23. The point (3, 4)P lies on a circle centered at the origin. What is the y-intercept of the line tangent to the circle at P? Hint: A tangent line to a circle is perpendicular to the radius at the point of tangency.

24. In Question <...> on page <...> we introduced the golden rectangle: it is a

rectangle of such a shape that if you cut off a square from it, the remaining rectangle has the same shape (aspect ratio) as the original one:

1

1

φ

φ-1

1

A

M O C

B

1 11

ϕϕ

−=

The Greek letter ϕ (phi, pronounced as “fie”) — is commonly used to denote the golden ratio. Suppose a golden rectangle is placed in the Cartesian plane with O at the origin, A at (0, 1), and C at (φ, 0). Write the equations of the lines AC and BM and prove that they are perpendicular to each other:

A

M O C

B

25. In linear algebra, the branch of mathematics that studies vector spaces and

linear transformations on them, a function is called linear if, for any 1x and

2x , 1 2 1 2( ) ( ) ( )f x x f x f x+ = + and for any real number k and for any x, ( ) ( )f kx kf x= . Are any of the linear functions ( )f x ax b= + on (which

is a one-dimensional vector space) also “linear” in the linear algebra sense? Which ones?

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6.4 Parabolas

A function from to given by a formula 2( )f x ax bx c= + + , where 0a ≠ , is called a quadratic function.

The simplest quadratic function is 2( )f x x= . The graph of this function is a curve called a parabola (Figure 6-4). As we will see in the next section, the graph of any quadratic function is a parabola, perhaps stretched or shrunk, shifted and/or reflected over the x-axis.

y

x0

1

1

2y x=

Figure 6-4. A parabola

The parabola in Figure 6-4 is symmetric with respect to the y-axis.

Any parabola is symmetric with respect to one line, called the axis of the parabola. The point where its axis intersects the parabola is called the vertex of the parabola.

The vertex of the parabola in Figure 6-4 is the origin. The parabola in Figure 6-4 appears to have a horizontal tangent line at the vertex. How do we know that the parabola does not have a “corner” or even a “cusp” at the vertex? Something like this:

? ?

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To show that the graph of 2y x= indeed has a tangent line at the vertex, let’s take a point (x, y) on the parabola and draw a line through that point and the origin:

(x, y)

A line passing through two points of a curve is called a secant line. The slope of this

line is 2y x x

x x= = . If we move the point (x, y) along the parabola toward the origin,

x becomes smaller, and the slope of the secant line gets closer and closer to 0. We can say that the “limit” of the secant line, as x approaches 0, is the horizontal tangent line through the origin.

A parabola is a beautiful shape. First, if we take a cone and cut it with a plane parallel to one side of the cone, the shape of the cross-section is a parabola (Figure 6-5). In fact, the name parabola means “a conic section” in ancient Greek; the name is due to the Greek geometer Apollonius (about 262 BC - about 190 BC) who studied conic sections.

Figure 6-5. A cross-section of a cone with a plane

parallel to its side is parabola

Second, any parabola is the locus of points (the set of all points) equidistant from a certain straight line, called the directrix, and a certain point, called the focus of the

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parabola (Figure 6-6). The focus lies on the axis of the parabola, and the directrix is perpendicular to the axis.

F l

Figure 6-6. The locus of points equidistant from a given point F and a line l is a parabola. l is called the directrix of the

parabola and F is called its focus.

Third, if you make a mirror in the shape of a parabola, any ray of light emitted from the focus of the parabola will be reflected in the direction parallel to the axis (Figure 6-7). No wonder parabolically shaped reflectors are used in the headlights of cars and in spotlights.

F

Figure 6-7. A parabola reflects all the rays from its focus F in the same direction, parallel to the axis of the parabola.

Example 1 Find the focus and the directrix for the graph of 2y x= .

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Solution The focus of this parabola lies on the y-axis, and the directrix is a horizontal line. Suppose the focus is at (0, d). Then the directrix must be the line y d= − , because the vertex (0, 0) must be equidistant from the focus and the directrix:

y

x0

1

2y x=

d

y =-d

(x, y)

y+d

d

We need to find d. Let us take any point (x, y) on the parabola. It must be equidistant from the focus and the directrix. The distance from the point (x, y) to the directrix is y d+ . The distance from the point (x, y) to the focus (by the distance formula) is

2 2( 0) ( )x y d− + − = 2 2( )x y d+ − . We must have:

2 2( )y d x y d+ = + − ⇒ 2 2 2( ) ( )y d x y d+ = + − ⇒ (recall that 2 2 2( ) 2a b a ab b± = ± + ) 2 2 2 2 22 2y yd d x y yd d+ + = + − + ⇒ ( 2y and 2d terms cancel out) 22 2yd x yd= − ⇒ 24yd x=

Since (x, y) is on the parabola, 2 14 14

y x d d= ⇒ = ⇒ = . The focus is 10,4

⎛ ⎞⎜ ⎟⎝ ⎠

and the directrix is 14

y = − .

Example 2 Find an equation for the parabola that is the locus of points equidistant from the x-axis and the point (0, 4).

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Solution Since the focus lies on the y-axis, the y-axis is the axis of the parabola. The vertex of the parabola lies halfway between the x-axis and the directrix, at (0, 2). This means our parabola is the “standard” parabola 2y x= , stretched or shrunk by some factor and shifted up by 2 units: 2 2y ax= + . When x = 1, y = a + 2, so the point (1, 2)a + lies on the parabola. The distance from that point to the directrix is | 2 |a + , and the distance from the focus is:

( ) ( )2 21 0 ( 2) 4a− + + − = 21 ( 2)a+ − = 21 4 4a a+ − + = 2 4 5a a− + We must have: 2 2( 2) 4 5a a a+ = − + ⇒ 2 24 4 4 5a a a a+ + = − + ⇒

18 18

a a= ⇒ = .

An equation for this parabola is 21 28

y x= + .

Exercises 1. Look up the Greek geometer Apollonius on the Internet. When and where

did he live? What is his major work? In Questions 2-5, sketch the graph of the given quadratic function and label three points on the graph with the exact coordinates. 2. 22y x= 3. 2y x= − 4. 2 3y x= + 5. 21 4y x= − 6. Show that the graph of 2y ax= is symmetrical with respect to the y-axis.

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7. The points 1 , 32

⎛ ⎞⎜ ⎟⎝ ⎠

and (2, 3) lie on the graph of 22 3 1y x x= − + — a

parabola. Find the axis of this parabola, then its vertex, then sketch the graph of 22 3 1y x x= − + . Hint: Since the given points are at the same horizontal level, the axis must be halfway between them.

8. Find the x-intercepts of the parabola 2 5 6y x x= − + , then find its axis and its

vertex, then sketch its graph. Hint: 2 5 6 ( 2)( 3)x x x x− + = − − . 9. Find the focus and the directrix of the parabola 23 1y x= + . 10. Find the focus and the directrix of the parabola 24 2y x= − . 11. The graph of the function 2 4 3y x x= − + is a parabola that has the same

shape as the graph of 2y x= , but its axis is the line 2x = . Find the coordinates of its vertex, then its focus and directrix.

12. Write an equation for the locus of points on the Cartesian plane that are

equidistant from the point (0, 3) and the line 1y = − . 13. Write an equation for the locus of points on the Cartesian plane that are

equidistant from the point (0, 2)− and the line 4y = .

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14.

F

A

M D l

In the picture above, F is the focus of the parabola, l is its directrix, the point A is on the parabola, and AD l⊥ . A is equidistant from the focus and the directrix, so AF = AD. We want to show that the ray FA is reflected by the parabola in the direction parallel to the axis (and perpendicular to the directrix) of the parabola. In other words, it must be reflected along DA . If AM is tangent to the parabola at A, then FAM must be equal to the angle of reflection, which is the same as DAM . Show that this is true. Hint: Show that the perpendicular bisector of FD shares only one point with the parabola (point A) so it is tangent to the parabola.

15. As we saw in Example 1, the focus of the parabola 2y x= is 10,4

⎛ ⎞⎜ ⎟⎝ ⎠

and the

directrix is 14

y = − . Find, in terms of x, the slope of the line tangent to the

parabola at the point A(x, y). Hint: see the diagram in Question 14: AM is the tangent line; AM FD⊥ ; find the slope of FD , then the slope of AM .

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16. Suppose that the points A, B, and C lie on the graph of 2y x= and have x-coordinates of x d− , x, and x d+ , respectively. Show that the area of

ABC does not depend on x, only on d (more precisely, the area of 3ABC d= ). Hint:

y

x

A

2y x=

B C

P R Q The area of ABC is equal to the area of the trapezoid PACR minus the areas of the trapezoids PABQ and QBCR.

17. Archimedes (3rd century BC) was the first to calculate the area of a segment

of a parabola below a chord. He represented the area as an infinite sum of triangles inscribed in the parabola. For the first triangle, ABC , he chose the point B directly under the midpoint of AC :

A

C

A

C B He then similarly inscribed two triangles, one under AB , another under BC . Then four triangles under each of the four chords, and so on. Show that the

area of the segment of the parabola below AC is equal to 43

of the area of

ABC . Hint: See Question 16. What is the area of the parabola 2y x= under the horizontal line y = 1?

18. 1R is the region in the Cartesian plane where 2y x≥ ; 2R is the region where

2x y≥ . What is the area of 1 2R R∩ ? Hint: See Question 17.

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6.5 Quadratic Functions and the Quadratic Formula In the previous section we defined a quadratic function as a function of the form

2( )f x ax bx c= + + , where 0a ≠ . The graph of any quadratic function is a parabola. Quadratic functions come up frequently in science, applied mathematics, engineering, and even everyday life. For example, the trajectory of a fly ball is a parabola (if we don’t take into account the curvature of the earth’s surface and air resistance). In many applications we are interested in the zeros (roots) of a function (When will the ball hit the ground?) or in its minimum or maximum value (When or where will the ball reach its highest point above the ground and at what altitude?). In other words, we might need to solve a quadratic equation 2 0ax bx c+ + = , and/or to find the coordinates of the vertex of a parabola 2y ax bx c= + + . (We will sometimes say informally “the parabola 2y ax bx c= + + ,” meaning the graph of

2y ax bx c= + + .) In this section we will derive formulas for the xy-coordinates of the vertex of a parabola and the formula for solving a quadratic equation. Let us first find the x-coordinate of the axis of a parabola, which is the same as the x-coordinate of its vertex. Recall that a parabola is symmetric with respect to its axis. So if 2( )f x ax bx c= + + , 1( ) 0f x = and 2( ) 0f x = , the axis of the parabola will be halfway between 1x and 2x :

y

x x1 x2

(xv, yv)

1 2

2vx x

x+

=

2y ax bx c= + +

The graph of 2( )g x ax bx= + is the same parabola as the graph of

2( )f x ax bx c= + + , only shifted vertically by c units (up or down). Therefore, these two parabolas have the same axis, and the x-coordinates of their vertices are the

same. ( ) bg x ax xa

⎛ ⎞= +⎜ ⎟⎝ ⎠

, so (0) 0g = and 0bga

⎛ ⎞− =⎜ ⎟⎝ ⎠

. The axis of the graph of g is

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the line 0

2 2

bbaxa

⎛ ⎞+ −⎜ ⎟⎝ ⎠= = − . The x-coordinate of the vertex of this parabola, as

well as the parabola 2y ax bx c= + + , is 2vbxa

= − . To find the y-coordinate, we

simply plug vx into the function:

2v v vy ax bx c= + + =

2

2 2b ba b ca a

⎛ ⎞ ⎛ ⎞− + − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2

24 2b ba ca a

− + =

2 2

4 2b b ca a− + =

2

4b ca

− +

The vertex of the parabola 2y ax bx c= + + is the point 2

,2 4b b ca a

⎛ ⎞− − +⎜ ⎟⎝ ⎠

.

If you forget these formulas, you can always reconstruct them for the specific numeric values of a, b, and c.

Example 1 Find the coordinates of the vertex of the parabola 22 7 3y x x= − + .

Solution 1. Using the formulas:

7 72 2 2 4vbxa

−= − = − =

⋅;

2 49 2534 4 2 8vby ca

= − + = − + = −⋅

.

2. If you forgot the formulas:

2 72 7 3 2 32

y x x x x⎛ ⎞= − + = − +⎜ ⎟⎝ ⎠

. vx is halfway between 0 and 72

, so 74vx = .

27 7 49 49 49 252 7 3 2 3 34 4 16 4 8 8vy ⎛ ⎞= − + = − + = − + = −⎜ ⎟

⎝ ⎠.

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If a parabola opens upward and its vertex is above the y-axis, or the parabola opens downward and its vertex is below the x-axis, then it does not intersect the x-axis —

y

x

y

x

— so the equation 2 0ax bx c+ + = has no solutions in real numbers. This happens

when 0a > and 2 2 4 0

4 4b b acca a

− +− + = > or 0a < and

2 4 04

b aca

− +< . To have two

zeros, we must have 0a > and 2 4 0

4b ac

a−

> or 0a < and 2 4 0

4b ac

a−

< . Either way,

we must have 2 4 0d b ac= − > . If 0d = , the vertex of the parabola is on the x-axis, and the two zeros “merge into one.”

The expression 2 4d b ac= − is called the discriminant of the quadratic function 2ax bx c+ + . The quadratic equation 2 0ax bx c+ + = has two distinct solutions in real numbers if and only if 0d > .

If 0d < , there are no solutions in real numbers. If 0d = , the two solutions merge into one. One way to derive the formula for the solutions of a quadratic equation (called the quadratic formula) is the technique called completing the square (Figure 6-8).

If a quadratic equation 2 0ax bx c+ + = has solutions in real numbers,

these solutions are 2

14

2b b acx

a− + −

= and 2

24

2b b acx

a− − −

= .

Notice the discriminant under the square root in the quadratic formula; this confirms our earlier finding that two distinct solutions in real numbers exist if and only if the discriminant is positive, that they merge into one solution when the discriminant is zero, and that there are no real-number solutions when the discriminant is negative.

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Also notice that 1 2

2 2 vx x b x

a+

= − = , the x-coordinate of the vertex of the parabola, as

expected.

2ax bx c+ + = [factoring out a]

2 b ca x xa a

⎛ ⎞+ + =⎜ ⎟⎝ ⎠

[adding and subtracting 2

24ba

to complete the square]

2 22

2 24 4b b b ca x xa a a a

⎛ ⎞+ + − + =⎜ ⎟

⎝ ⎠ [collecting the first three and the last two terms]

2 2

2

4 02 4b b aca xa a

⎛ ⎞−⎛ ⎞+ − = ⇔⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2 2

2

42 4b b acxa a

−⎛ ⎞+ = ⇔⎜ ⎟⎝ ⎠

2 42 2b b acxa a

−+ = ± ⇔

2 42

b b acxa

− ± −=

Figure 6-8. Deriving the quadratic formula by “completing the square”

Example 2 Solve the equation 23 2 1 0x x− − = .

Solution

By the quadratic formula, 2 4 4 3 ( 1) 2 16

2 3 6x

± − ⋅ ⋅ − ±= =

⋅. The solutions are

12 4 1

6x += = and 2

2 4 16 3

x −= = − .

Example 3 Solve the equation 2 5 7 0x x+ + = .

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Solution The discriminant of this quadratic function is 25 4 7 25 28 3 0− ⋅ = − = − < , so there are no real solutions.

If 21 2( ) ( )( )f x ax bx c a x x x x= + + = − − , then, clearly, 1x and 2x are the zeros of f.

The converse is also true: if 1x and 2x are the zeros of f, then ( )f x can be factored as 1 2( ) ( )( )f x a x x x x= − − . (This is true even when 1x and 2x are the same number.) When we open the parentheses and collect the like terms, we get

2 21 2 1 2( )ax bx c ax a x x x ax x+ + = − + + . This must be true for all x, so we must have

1 2 1 2( ) ba x x b x xa

− + = ⇒ + = − and 1 2 1 2cax x c x xa

= ⇒ = . This fact is known

as Viète’s Theorem (a.k.a. Vieta’s Theorem).

1x and 2x are the solutions to the quadratic equation 2ax bx c+ + , if and

only if 1 2bx xa

+ = − and 1 2cx xa

= .

This is consistent, of course, with the quadratic formula: if 2

14

2b b acx

a− + −

= and

2

24

2b b acx

a− − −

= , then

1 2 22b bx xa a

⎛ ⎞+ = − = −⎜ ⎟⎝ ⎠

and

( )( )

( )

2 2

1 2 2

4 4

2

b b ac b b acx x

a

− + − − − −= =

( ) ( )

( )( )

22 2 2 2

2 2 2

4 4 44 42

b b ac b b ac ac ca a aa

− − − − −= = = .

Sometimes you can spot easy factorization for a quadratic function based on Viète’s Theorem, especially when the coefficients are integers. Then you know the zeros right away.

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Example 4 Solve the equation 2 3 2 0x x− + = .

Solution 1 2 3+ = and 1 2 2⋅ = , so 2 3 2 ( 2)( 1)x x x x− + = − − . The solutions are 2x = and

1x = .

Exercises 1. Look up the history of the quadratic formula on the Internet. Roughly when

was it discovered? 2. When and where did François Viète (of Viète’s Theorem fame) live? In Question 3-6, find the zeros of a quadratic function (if any) and the coordinates of the vertex of its graph. 3. 23( 4) xf xx = − − 4. 2( 6) 3xf x x− += + 5. 2( 1) xf xx = + + 6. 2( 3) 3 6xf xx −= + 7. 2( 3) 3 6xf xx −= + 8. Factor 2 5 14x x− − . Hint: First find the zeros using the quadratic

formula. 9. Can the graphs of two different quadratic functions intersect in more than

two points? If yes, give an example; if not, explain why not. 10. Write an equation 2x px q+ + such that 3x = − and 7x = are its two

solutions.

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11. Find the sides of a rectangle whose area is 40 and whose perimeter is 25. 12. Find the distance between the points A and B at which the line y x=

intersects the parabola 2y x= . Then find the area of the region bounded by the parabola and the chord AB . Hint: See Questions <...> and <...> in Section <...>.

13. A line tangent to the parabola 2y x= passes through the point (0, 2)− . Find

the slope of the line and the point of tangency. Hint: The slope must be such that the line touches the parabola in exactly one point.

14. Factor 4 2 1x x+ + into two quadratic terms. Hint: Set 2x u= and find the

zeros using the quadratic formula. 15. Using the quadratic formula, find the Golden Ratio ϕ as the positive solution

to the equation 11 1x

x=

−. Using Viète’s Theorem, show that the other

solution is 1ϕ

− .

16. A flare is shot up from a boat over the sea with an initial vertical velocity of

v = 400 ft/sec. Its vertical position over the water after t seconds is 2

2gtvt − ,

where 32g ≈ ft/sec2 is the acceleration due to gravity. For how long will the flare fly? What altitude will it reach?

17. Find on YouTube a trailer for the film October Sky. In the film, a group of

kids interested in rocketry are accused of starting a forest fire with their model rocket, but Homer Hickam, the main character, learns how to calculate the trajectories of rockets and shows that their rocket couldn’t possibly fly that far. (The film is based on a true story; Homer Hickam eventually became a NASA engineer.) Suppose a rocket, when launched straight up, reaches an altitude of approximately 100 meters. If the same rocket is launched at 45 degrees to the surface (the optimal angle for the greatest distance), approximately how far from the launching place will the rocket land? Hint: The trajectory of a rocket is approximately a parabola; see Question 16.

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18. Suppose 1x , 2x , and 3x are solutions to the cubic equation 3 2 0ax bx cx d+ + + = . Express 1 2 3x x x+ + and 1 2 3x x x in terms of a, b, c,

and d.

6.6 Review Concepts, terms, methods, and formulas introduced in this chapter:

Linear relation: px + qy = C

Intercept form of a linear relation: 1x ya b+ =

Parallel lines Perpendicular lines

1px qy C+ = is perpendicular to 2( )q x py C− + = Linear function Slope-intercept form of an equation of a line: y mx b= + Point-slope form of an equation of a line: 1 1( )y y m x x− = −

1y mx b= + and 21y x bm

= − + are perpendicular to each other

Quadratic function: 2( )f x ax bx c= + + Parabola:

Focus

Axis

Directrix Vertex

Coordinates of the vertex: 2vbxa

= − , 2

4vby ca

= − +

Quadratic formula for zeros of a quadratic function: 2

1,24

2b b acx

a− ± −

=

Discriminant: 2 4d b ac= − Viète’s Theorem for zeros 1x and 2x of a quadratic function 2ax bx c+ + :

1 2bx xa

+ = − ; 1 2cx xa

=

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Linear and Quadratic Functions - skylit.com · 1 6 Linear and Quadratic Functions 6.1 Prologue...

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