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Quadratic Functions and Parabolas
highpointoftheparabola.Thex-interceptsoftheparabolaaregiven by
lowpointoftheparabola;ifa 0,then theparabolaopensdownward,andthevertexis the
iscalledthevertex.Ifa 0,then theparabolaopensupward,andthevertexis the
wherex -
.Thepointofthe parabola
,andthevalueoff(x)decreasesasxincreasesfor x -
for x -
,thevalueoff(x)increasesasx increases
foreveryrealnumber xdifferentfrom -
f(x)
.Ifa 0,thenf -
,andthevalueoff(x)increasesasxincreasesfor x -
for x -
,thevalueoff(x)decreasesasx increases
f(x)foreveryrealnumber xdifferentfrom -
f -
.Ifa 0, then
It thenfollowsthat thegraphoffissymmetricabout theverticalstraightlinex -
uponcompletingthesquarein x. Thus
Firstnote that
discusssomebasicpropertiesoffandits graph.
quadraticfunction.Thegraphoffisanexampleofthetypeofcurvecalledaparabola.We now
allrealnumbersandwith theproperty that foreveryrealnumber x.Wecallf a
Hereleta,b,cbegivenrealnumberssuch thata 0.Letfbethefunction withdomain theset of
2
2
2
2
2
2
2 2
2
2 2
2
a
b ac
a
b
a x
a
ac b
a
b
a x
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
ac b
a
b
f(x) a x
a
b
c a
a
b
x
a
b
a x
x c
a
b
f(x) ax bx c a x
f(x) ax bx c
thelowpoint.Wenowdetermineafewmorepointson thisparabola.
They-interceptisgivenbyf(0) 16 .Sincea 0,theparabolaopensupward,andthevertex is
x 8,x 2.
(x-8)(x 2) 0
andthey-coordinateofthevertexisf(3) 3 - 6 (3) 16 25 .Thex-interceptsaregiven by
Herea 1,b -6,c -16.Thex-coordinateofthevertexisgivenby x -
(b) y f(x) x 6 16.
this parabola.
Sincea 0,theparabolaopensdownward,andthevertexisthehighpoint.Figure 1 isasketch of
f(x) - 7
in the sketching.
They-interceptisgivenbyf(0) 8.Wenowdetermineafewmorepointson theparabolato aid
x -4,x 2.
(-x-4)(x-2) 0
andthey-coordinateofthevertexisf(-1) -(-1) - 2 (-1) 8 9.Thex-interceptsaregiven by
2a
b
Herea -1,b -2,c 8.Thex-coordinateofthevertexisgivenby x -
(a)y f(x) -x -2x 8.
Example1.Sketcheachofthefollowingparabolas,showingthevertexoftheparabolaineach case.
applicationsofquadratic functions.
quadraticformula.Wenowpresentsomeexamplesofsketchinggraphsofquadraticfunctions and
ornox-intercepts.Oftenitiseasier tofindthex-interceptsbyfactoringrather thanusing the
Itshouldbepointedout that theparabolamayhave 2 x-intercepts,just 1 x- intercept,
ofthequadratic equation
Formula(1)for thex-interceptsof theparabolaisknownasthequadraticformulafor the solutions
(1) x
sothesex-interceptsaregivenby the formula
2
2
2
2
2
2
2
2
2
2
x
x
ax bx c
a
b b ac
a
b ac
a
b
x
a
b ac
a
b
x
f(x) 11
x 3
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ x
y
Vertex (-1,9)
y f(x) -x -2x 8.
2
Example Here let Г be the parabola which is the graph of the equation
2
y ax
where
a 0
. This is a special case of the quadratic function
y ax bx c
2
where b=0 and c=0. In the calculus textbooks this parabola has the equation
x
2
= 4py.
Thus
a
p
or
a
p
The vertex V of Г is the origin (0,0). The focus F of Г is the point
4 a
, or (0,p).
The directrix L of Г is the horizontal straight line defined by the equation
a
y
or
y p
Recall now that Г is the set of all points P= (x, y) such that the distance between P and
the focus F is equal to the distance between P and the directrix L. We now check this
using the equation
x 4 py
2
for Г, the focus F given by the point (0,p), and the directrix L the horizontal straight line
defined by the equation
y p
First let P= (x,y) be a point of Г. The distance between P and the focus F is given by
2 2
2 2
2 2
2
4 2
0
y py p
py y py p
x y p
2
y p
y p
y p
This is the distance between P and the horizontal straight line L defined by the equation
y p
Next let P= (x,y) be a point such that the distance between P and the point F=(0,p) is
equal to the distance between P and the horizontal straight line L defined by the equation
y p
. Then
2
2 2 2 2 2
2 2 2
2 2
x py
x y py p y py p
x y p y p
x y p y p
Hence P is on the parabola Г.
We now consider the optical property of a parabola. This property is used, for example,
in designing searchlights with light source at the focus. To illustrate this property, we use
the parabola Г defined by the equation
2
y px
with p>0. This is a horizontal parabola opening to the right with vertex at the origin and
with focus the point F= (p,0).
