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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.
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
Upward Opening Parabola
f (x) = ax2 + bx + c, a > 0
x
y
vertex
axis
J. Robert Buchanan Quadratic Functions and Models
Downward Opening Parabola
f (x) = ax2 + bx + c, a < 0
x
y
vertex
axis
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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
J. Robert Buchanan Quadratic Functions and Models
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
J. Robert Buchanan Quadratic Functions and Models
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
J. Robert Buchanan Quadratic Functions and Models
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
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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. 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.
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.
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.
J. Robert Buchanan Quadratic Functions and Models
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
).
J. Robert Buchanan Quadratic Functions and Models
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
).
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models
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.
J. Robert Buchanan Quadratic Functions and Models