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Unformatted text preview: 134 CHAPTER 2 Linear, Quadratic, Polynomial, and Rational Functions x
1 he graph of f (x) = x2
In the interval [—1, 1]. For example,
the vertex of the
parabola shown above is the origin. Quadratic Functions and Parabolas SECTION OBJECTIVES
By the end of this section you should I be able to use the completingthesquare technique with quadratic
expressions; I be able to ﬁnd the vertex of a parabola;
I understand how the quadratic formula was diSCovered; I be able to solve quadratic equations. The Vertex of a Parabola The last section dealt with linear functions. Now we move up one level of
complexity to deal with quadratic functions. We begin with the deﬁnition: Quadratic functions
A quadratic function is a function f of the form f(x) = axz +bx +c, where a, b, and c are constants, with a, 7E O. The simplest quadratic function is the function f deﬁned by f (x) = 962;
this function arises by taking a = 1, b = 0, and c = O in the definition above. Parabolas can be deﬁned geometrically, but for our purposes it is simpler
to deﬁne a parabola algebraically: Paraboias
A parabola is the graph of a quadratic function. For example, the graph of the quadratic function f deﬁned by f (x) = x2 is
the familiar parabola shown here. For this function f we have f ( —x) = f (x).
In other words, f is an even function, which implies that the graph of f is
symmetric about the vertical axis. Note that the vertical axis intersects this parabola at the origin. As we will soon see, eve arabola is s etric about some line. The point where this line of symmetry intersects the parabola is sufﬁciently im
portant to deserve its own name: Vertex The vertex of a parabola is the point where the line of symmetry of the
parabola intersects the parabola. ) SECTION 2.2 Quadratic Functions and Parabolas 135 Suppose f, g, and h are deﬁned by for) = —x2, g(x) = —2x2, h(x) = —2x2 + 1. (a) Sketch the graphs of f, g, and h on the interval {—1, 1].‘
(b) What is the line of symmetry of the graphs of f, g, and h?
(C) Find the vertex of the graph of f , the graph of g, and the graph of h. (d) What is the maximum value of f? The maximum value of g? The maximum
value of h? SOLUTION (a) The graph of f is the reﬂection of the graph of the function at2 through the hor
izontal axis. The graph of g is then obtained from the graph of f by stretching
vertically by a factor of 2. The graph of h is then obtained from the graph of g
by shifting up by 1 unit. The graph of —2x2 + 1
on the interval [—1,1]. The graph of —2x2
on the interval [—1,1]. The graph of —x2
on the interval [—1, 1]. (b) As can be seen from the ﬁgures above, the line of symmetry for each of the
graphs of f, g, and h is the vertical axis. (c) The ﬁgures above show that the vertex of the graph of —x2 is the origin. the
vertex of the graph of —2x2 is also the origin, and the vertex of the graph of
—2x2 + 1 is the point (0, 1). In each case, the vertex is the intersection of the
line of symmetry (which for all three graphs is the vertical axis) with the graph. (d) As can be seen from the graphs above, the maximum value of f is 0, the max
imum value of g is 0, and the maximum value of h is 1. Note that all three
functions take on their maximum value when x : 0. Suppose a quadratic function f is deﬁned by f (x) = axz + c, where a and
c are constants with a 179 0. If a > 0, then f takes on its minimum value when
x = 0 (because ax2 is positive for all values of x except x = 0). Similarly,
if a < 0, then f takes on its maximum value when x = 0 (because ax‘a is
negative for all values of x except x = 0). In either case, the point (0, c) is
the vertex of the graph of f. EXAMPLE 1 The ancient Greeks
discovered that the
intersection of a cone and
an appropriately
positioned plane is a
parabola. For example, the func
tion f deﬁned by
f(x) =ﬂ8x2 + 5 takes
on its minimum value
When x = 0; the vertex
of the graph of f is
the point (0, 5). 136 CHAPTER 2 Linear, Quadratic, Polynomial, and Rational Functions Be sure that you
are thoroughly fa
miliar with this
crucial identity. EXAMPLE 2 EXAMPLE 3 Completing the Square We now know how to ﬁnd the vertex of the graph of quadratic functions of
the form axz + c. A technique called completing the square can be used to
put an arbitrary quadratic expression in a form that makes finding the vertex
easy. The key to this technique is the simple identity (x+t)2=x2+2tx+t2. y The next example illustrates the technique of completing the square. Write the expression x2 + ﬁx in the form
(x + t)2 + r
for some constants t and “r. SOLUTION When (x + t)2 is expanded to x2 + mm + t2, the th term must match
the 6x term in x’2 + 6x. Thus we must have 2t = 6, and hence we choose t = 3. When (x + 3)2 is expanded, we get x2 + 6x + 9. The x2 and 6x terms match the
corresponding terms in the expression x2 + ﬁx, but the expansion of (x + 3)2 has
an extra constant term of 9. Thus to get an equality we subtract 9, obtaining x2+6x=(x+3)2—9. The general formula for completing the square is shown below. You
should not need to memorize this formula. The key point is that the co
efﬁcient of the x term will need to be divided by 2, and then the appropriate
constant will need to be subtracted to get a correct identity. CO'mp‘len'ng the square x2+bx 2, (x+ 2f — (2)2 For example, if b 2 —10, then the identity above becomes x2e10x= (x75)2 #25. Note that the term that is subtracted is always positive because ($2 is pos itive regardless of whether b is positive or negative, but the (x + %)2 term
has a sign that matches the Sign of b.
The next example shows the usefulness of completing the square. Suppose f is the function defined by f(x) =x2 +6x+ 11. (a)
(b) (C) SECTION 2.2 Quadratic Functions and Parabolas 137 For what value of x does f (x) attain its minimum value? Sketch the graph of f on an interval of length 4 centered at the number where
f attains its minimum value. Find the vertex of the graph of f. SOLUTION (a) (b) (C) Use the result from the previous example to rewrite f (x) as follows:
f(x) : (x2 + 6x) + 11
=(x+3)2 —9+11
: (x + 3)2 + 2 The last expression shows that f takes on its minimum value when x = —3,
because (x + 3)2 is positive for all values of x except x = —3. The expression above implies that the graph of f is obtained from the graph of
x2 by shifting left 3 units and then shifting up 2 units. This produces the fol—
lowing graph on the interval [e 5. *1], which is the interval of length 4 centered
at —3:
3/
s The graph of x2 + 6x + 11
on the interval [e5, —l]. X
—5 —3 —1 The ﬁgure above shows that the vertex of the graph of f is the point (—3, 2).
We could have computed this even without the ﬁgure by noting that f takes on
its minimum value at —3 and that f (—3) = 2. Or we could have noted that the
graph of f is obtained from the graph of x2 by shifting left 3 units and then
shifting up 2 units, which moves the origin (which is the vertex of the graph of
x2) to the point (*3, 2). When the coefﬁcient of x2 is something other than 1, factor out that co efﬁcient from the x2 and x terms and then use the completing the square
identity. The following example illustrates this procedure. Suppose f is the function deﬁned by (a)
(b) (C) f(x) = —3x2 + 5x + 1. For what value of x does f (x) attain its maximum value? Sketch the graph of f on an interval of length 4 centered at the number where
f attains its maximum value. Find the vertex of the graph of f. Note that the complet
ing the square identity
is applied only to the
x2 + 6x part of this eX
pression, and that the
constant —9 is then
combined with the
constant 1 1. EXAMPLE 4 138 CHAPTER 2 Linear, Quadratic. Polynomial, and Rational Functions lhe new expres
sion for f has frac
tions making it look
more cumbersome
than the original ex
pression. However,
this new expression
for f allows us to
answer questions
about f more easily. / A quadratic func tion f deﬁned by f(x) = ax2 + bx +c
has a minimum value
(but no maximum
value) ifa > 0 and has a maximum Value (but no mini
mum valuen'fa < 0 \J SOLUTION (a) (b) (C) First we factor out the coefﬁcient —3 from the x2 and x terms and then apply the completing the square identity, rewriting x as ollows: f(x) = —s[x2 — ix] + 1 0;qu
f(x)=—3[(x—%) )——]+1 8’
=—3(x—%)2+%+1
has—ans The last expression shows that f takes on its maximum value when x = %, because —3 (x — %)2 is negative for all values of x except x= %. The expression above for f implies that the graph of f is obtained from the
graph of x2 by shifting right; — units, then stretching vertically by3 a factor of 3, then reﬂecting through the horizontal axis and then shifting up— Sunits. This
7 1? produces the following graph on the interval [—3, 7], which is the interval of
length 4 centered at5 — x The graph of —3x2 + 5x + 1 on the interval
7 167 The ﬁgure above shows that the vertex of the graph of f is the point (g, % ). We could have computed this even without the ﬁgure by noting that f takes on its
maximum value at 5 5and that ﬁg ) = —.Or we could have noted that the graph
of f is obtained from the graph of x2 by shifting right: A units, then stretching
vertically by3 a factor of 3, then reﬂecting through the horizontal axis and then
shifting up— gunits, which moves the origin (which 1s the vertex of the graph of x2) to the1 point (g, E). Notice the relationship in the examples above between the minimum or maximum value of a quadratic function and the vertex of its graph. Speciﬁ
cally, the ﬁrst coordinate of the vertex of the graph of f is the number where
this minimum or maximum value is attained, and the second coordinate of
the vertex of the graph of f is the minimum or maximum value of f. This
relationship holds for all quadratic functions. Vertex at minimum or maximum value Suppose f is a quadratic function that takes on its minimum ot’maximum '
value at t. Then the point (t, f(t)) is the vertex of the graph 'of f. SECTION 2.2 Quadratic Functions and Parabolas 139 The Quadratic Formula Having worked through some examples, we will now follow the same pattern and complete the square with an arbitrary quadratic function. This will allow us to derive the quadratic formula for solving the equation f (x)‘= 0.
Consider the quadratic function f(x) =ax2+bx+c, where a 179 0. Factor out the coefﬁcient a from the ﬁrst two terms and then
complete the square, rewriting f(x) as follows: f(x) =ax2+bx+c =a[x2+%x] +c
10 b2
=d[(x+E)Z—E]+C Suppose now that we want to ﬁnd the numbers x such that f (x) = 0.
Setting the last expression for f (x) equal to O, we have
I9 2 b2 — 46w
a( > — — = x+—
4a 2a 0' which is equivalent to the equation a 2_
(“$2 = _—b 4;“. Regardless of the value of x, the left side of the last equation is a positive
number or 0. Thus if the right side is negative, the equation does not hold
for any real number x. In other words, if b2 — 46LC < 0, then the equation
f (x) = 0 has no real solutions. If 192 — 46w 2 0, then we can take the square root of both sides of the last
equation, getting ' x + L = + be — 40'“: Here i is used to in
2191 _ 2a, ' dicate that we can
. . . choose either the lus
which is eqmvalent to 51' or minus 31' p
—b i b2 — 4:16 gn 811'
x =
2.5; By completing the square, we have derived the quadratic formula! 140 CHAPTER 2 linear, Quadratic, Polynomial, and Rational Functions The quadratic formula often is useful in problems that do not initially
seem to involve quadratic functions, as illustrated by the following example. E X ,\ M PL]: 3 Find two numbers whose sum equals 7 and whose product equals 8. SOLITI'ION Let's call the two numbers '3 and t. We want
s+t'=7 and 3t=8. Solving the ﬁrst equation for s, we have s = 7 — t. Substituting this expression for 3
into the second equation gives (7 — t)t = 8, which is equivalent to the equation t3—7t+3=0. Using the quadratic formula to solve this equation for it gives _7:\/72—48_7:\/ﬁ
~f—T You should verify Let's choose the solution t = 743/17. Plugging this value of 1: into the equation 3 = 7—t thatif we had cho then gives 5 = 7J1‘?_ _ pm 2
sen t ‘ 2 ’ than Thus two numbers whose sum equals .7 and whose product equals 8 are I"? we would have ended Hm
up with the same 2 ' pair 0f numbers. REMARK To check that this solution is correct, note that
7—i/17 7+«/'17 _ lg _ —+ 2 2 2 t and 7 and
7—m_7'+¢ﬁ:72—m2 49—17 32 8 2 .2 4 4 4 ...
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