quadratic functions and modelsbanach.millersville.edu/~bob/math160/quadfunc/main.pdf · standard...

Quadratic Functions and ModelsMATH 160, Precalculus

J. Robert Buchanan

Department of Mathematics

Fall 2011

J. Robert Buchanan Quadratic Functions and Models

Objectives

In this lesson we will learn to:analyze the graphs of quadratic functions,write quadratic functions in standard form and use theresults to sketch the graphs of functions,find the minimum and maximum values of quadraticfunctions in real-world applications.


Polynomial Functions

The linear function (f (x) = ax + b), the constant function(f (x) = c), and the squaring function (f (x) = x2) are allexamples of polynomial functions.

DefinitionLet n be a nonnegative integer and let an, an−1, . . . , a2, a1, anda0 be real numbers with an 6= 0. The function given by

f (x) = anxn + an−1xn−1 + · · ·+ a2x2 + a1x + a0

is called a polynomial function of x with degree n.


Quadratic Functions

A quadratic function is a polynomial function in x of degree 2.

DefinitionLet a, b, and c be real numbers with a 6= 0. The function givenby

f (x) = ax2 + bx + c

is called a quadratic function.

The graph of a quadratic function is a roughly “U”-shaped curvecalled a parabola.


Upward Opening Parabola

f (x) = ax2 + bx + c, a > 0

x

y

vertex

axis


Downward Opening Parabola

f (x) = ax2 + bx + c, a < 0

x

y

vertex

axis


Anatomy of a Parabola

All parabolas are symmetric with respect to a line calledthe axis of symmetry or simply the axis.The point where the axis intersects the parabola is calledthe vertex.If f (x) = ax2 + bx + c then

the parabola opens upward when a > 0 andthe parabola opens downward when a < 0.


Standard Form

If a quadratic function is written in standard form we can easilydetermine the axis of symmetry, the vertex, and whether theparabola opens upward or downward.

Standard FormThe quadratic function

f (x) = a(x − h)2 + k , a 6= 0

is in standard form. The graph of f is a parabola whose axis isthe vertical line x = h and whose vertex is the point (h, k). Ifa > 0 the parabola opens upward, while if a < 0 the parabolaopens downward.


Example (1 of 4)

Write the following quadratic function in standard form, graph it,and determine the axis and vertex.

f (x) = 2x2 + 6x − 5

f (x) = 2(x2 + 3x)− 5= 2(x2 + 3x + 9/4)− 9/2− 5

= 2(

x +32

)2

− 192

axis: x = −3/2vertex: (−3/2,−19/2)

-5 -4 -3 -2 -1 1 2x

-10

-5

5

10

15

y


Example (2 of 4)

Write the following quadratic function in standard form, graph it,and determine the axis and vertex.

f (x) = −x2 − 4x + 1

f (x) = −(x2 + 4x) + 1= −(x2 + 4x + 4) + 4 + 1= −(x + 2)2 + 5

axis: x = −2vertex: (−2, 5)

-5 -4 -3 -2 -1 1x

-4

-2

2

4

y


Example (3 of 4)

For the following parabola find the vertex, the y -intercept, andthe x-intercepts.

f (x) = x2 + 5x +14

f (x) = x2 + 5x +254− 25

4+

14

=

(x +

52

)2

− 6

Vertex:(−5

2,−6

)

If x = 0 then y = f (0) =14

, the y -intercept. If y = 0 then(x +

52

)2

− 6 = 0

x +52

= ±√

6

x = −52±√

6

the x-intercepts.


Example (3 of 4)


f (x) = x2 + 5x +14

f (x) = x2 + 5x +254− 25

4+

14

=

(x +

52

)2

− 6

Vertex:(−5

2,−6

)If x = 0 then y = f (0) =

14

, the y -intercept.

If y = 0 then(x +

52

)2

− 6 = 0

x +52

= ±√

6

x = −52±√

6

the x-intercepts.


Example (3 of 4)


f (x) = x2 + 5x +14

f (x) = x2 + 5x +254− 25

4+

14

=

(x +

52

)2

− 6

Vertex:(−5

2,−6

)If x = 0 then y = f (0) =

14

, the y -intercept. If y = 0 then(x +

52

)2

− 6 = 0

x +52

= ±√

6

x = −52±√

6

the x-intercepts. J. Robert Buchanan Quadratic Functions and Models

Example (4 of 4)

Write the standard form of the equation of a parabola that hasvertex at (3,−1) and passes through the point (0, 2).

f (x) = a(x − h)2 + k= a(x − 3)2 − 1

2 = a(0− 3)2 − 12 = 9a− 1

a =13

Thus f (x) =13(x − 3)2 − 1.


Minimum and Maximum Values

If f (x) = ax2 + bx + c we may write the quadratic function instandard form by completing the square.

f (x) = a(

x +b2a

)2

+

(c − b2

4a

)Thus the coordinates of the vertex are

(− b

2a, f

(− b

2a

)).

Consider the function f (x) = ax2 + bx + c.

1 If a > 0, f has a minimum at x = − b2a

. The minimum

value is f(− b

2a

).

2 If a < 0, f has a maximum at x = − b2a

. The maximum

value is f(− b

2a

).


Example

The profit P (in hundreds of dollars) that a company makesdepends on the amount x (in hundreds of dollars) that thecompany spends on advertising according to the modelP = 230 + 20x − 0.5x2. What expenditure for advertising willyield maximum profit?

In this formula a = 0.5 and b = 20, thus profit is maximizedwhen

x = − b2a

= − 202(−0.5)

= 20.

Thus profit is maximized when $2,000 is spent on advertising.

The maximum profit will be

P = 230 + 20(20)− 0.5(20)2 = 430

or $43,000.


Example

The profit P (in hundreds of dollars) that a company makesdepends on the amount x (in hundreds of dollars) that thecompany spends on advertising according to the modelP = 230 + 20x − 0.5x2. What expenditure for advertising willyield maximum profit?In this formula a = 0.5 and b = 20, thus profit is maximizedwhen

x = − b2a

= − 202(−0.5)

= 20.

Thus profit is maximized when $2,000 is spent on advertising.

The maximum profit will be

P = 230 + 20(20)− 0.5(20)2 = 430

or $43,000.


Example

The profit P (in hundreds of dollars) that a company makesdepends on the amount x (in hundreds of dollars) that thecompany spends on advertising according to the modelP = 230 + 20x − 0.5x2. What expenditure for advertising willyield maximum profit?In this formula a = 0.5 and b = 20, thus profit is maximizedwhen

x = − b2a

= − 202(−0.5)

= 20.

Thus profit is maximized when $2,000 is spent on advertising.The maximum profit will be

P = 230 + 20(20)− 0.5(20)2 = 430

or $43,000.


Homework

Read Section 2.1.Exercises: 1, 5, 9, 13, . . . , 81, 85


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