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Functions and Their Graphs
Continuation
TYPES OF FUNCTIONS
5. Quadratic FunctionsA quadratic function is a function of the form f(x) = ax2 +bx +c where a, b and c are real numbers and a ≠ 0.Domain: the set of real numbersGraph: parabolaExamples: parabolas parabolas
opening upward opening downward
The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below.
If the coefficient of x2 is positive, the parabola opens upward; otherwise, the parabola opens downward.
The vertex (or turning point) is the minimum or maximum point.
Graphs of Quadratic Functions
Graphing Parabolas • Given f(x) = ax2 + bx +c
1. Determine whether the parabola opens upward or downward. If a > 0, it opens upward. If a < 0, it opens downward.
2. Determine the vertex of the parabola. The vertex is
3. The axis of symmetry is
The axis of symmetry divides the parabola into two equal parts such that one part is a mirror image of the other.
a
bac
a
b
4
4,
2
2
a
bx
2
Graphing Parabolas • Given f(x) = ax2 + bx +c4. Find any x-intercepts by replacing f (x) with 0. Solve the
resulting quadratic equation for x.
5. Find the y-intercept by replacing x with zero.
6. Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.
Example 9.• The function f(x) = 1 - 4x - x2 has its vertex at
_____.
• A. (2,11)• B. (2,-11)• C.( -2,-3)• D.(-2,1)
Example 10a.• Identify the graph of the given function: y = 3x2 - 3.
Example 10b.• Identify the graph of the given function: 4y = x2.
Example 10c.• Identify the graph of the given function: y = (x - 2)(x – 2).
Minimum and Maximum: Quadratic Functions
• Consider f(x) = ax2 + bx +c.
1. If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)).
2. If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).
Example 11a.• The maximum value of the function
f(x) = -3x2 – 2x + 4 is ____.
• A. 13/3• B. 3/13• C. 9• D. 13
Example 11b.
• The function f(x) = x2 – 8x + 16 has _____.
• A. a minimum value at x = -4• B. a maximum value at x = -4• C. a minimum value at x = 4• D. a maximum value at x = 4
Example 12.• The revenue of a charter bus company depends on the number of
unsold seats. If the revenue R(x) is given by R(x) = 5000 + 50x – x2, where x is the number of unsold seats, find the maximum revenue and the number of unsold seats that corresponds to the maximum revenue.
• Solution:
The revenue function, R(x) = 5000 + 50x – x2 is a quadratic function with a = -1, b = 50 and c = 5000. Since a = -1 < 0, the R(x) has a maximum that occurs at x = -b/(2a) and the maximum value is R(-b/(2a)).
25)1(2
)50(
2
a
bx
6255)25()25(500005)25( 2 R
6. Absolute Value FunctionsAn absolute value function f is defined by
Domain: the set of real numbersGraph: v-shapedExamples: y = -|x| y = |x| y = x - |x|
0,
0,)(
xifx
xifxxf
7. Rational FunctionsA Rational Function is a function in the form:
where p(x) and q(x) are polynomial functions and q(x) ≠ 0.
Examples:
)(
)()(
xq
xpxf
xy
2
4
x
xy
A polynomial function is a function of the form:
on
nn
n axaxaxaxf 1
11
All of these coefficients are real numbers
n must be a positive integer
The degree of the polynomial is the largest power on any x term in the polynomial.
an ≠ 0
8. Polynomial Functions
Examples:Graphs of Polynomial Functions
y x3 4xy x3 2x2 x 4
POLYNOMIAL FUNCTIONS AND EQUATIONS
The Remainder Theorem
• If P(x) is a polynomial and r is a real number, then if P(x) is divided by x – r, the remainder is P( r ).
Example 1.• Find the remainder when (2x3 – 3x2 - 4x - 17)
is divided by (x – 3). • Solution:• Applying the Remainder Theorem, we have
P(3) = 2(3)3 – 3(3)2 – 4(3) – 17
= 54 – 27 – 12 – 17 = -2 • The remainder is -2.
Example 2.• Use the Remainder Theorem to find the
remainder obtained by dividing the polynomial by the given binomial that follows it.
• a. ;
• b. ;
• c. ; • Answers: a. -1.5 b. 2 c. 2
12841632 2345 xxxxx2
1x
375 23 xxx 1x
26732 234 xxxx 1x
• A consequence of the remainder theorem is the Factor Theorem.
• It enables us to determine whether a specific expression of the form (x – r) is a factor of a given polynomial.
The Factor Theorem
• If P(x) is a polynomial and r is a real number, then P(x) has (x – r) as a factor if and only if P( r ) = Q.
Example 3.• Show that (x – 4) is a factor of
(2x3 – 6x2 – 5x – 12).
Solution: Applying the Factor Theorem, we have• P(x) = 2x3 – 6x2 – 5x – 12• P(4) = 2(4)3 – 6(4)2 – 5(4) – 12
= 2(64) – 6(16) – 20 – 12
= 128 – 96 – 20 – 12 = 0
Therefore, by the factor theorem, (x – 4) is a factor of (2x3 – 6x2 – 5x – 12).
Example 4.• Use the Factor Theorem to show that the
given binomial is a factor of the polynomial.
• a. ; • b. ;
4343 23 xxx4423 xxx
1x1x
Converse of the Factor Theorem
• If (x – r) is a factor of f(x) then
f(r ) = R = 0, then r is a zero of f(x).
Example 5.• Determine whether (x + 1) is a factor
of 5x4 + x3 – 4x2 – 6x – 10.
• Solution:• P(x) = 5x4 + x3 – 4x2 – 6x – 10• P(-1) = 5(-1)4 + (-1)3 – 4(-1)2 – 6(-1) – 10
= 5 – 1 – 4 + 6 – 10
= - 4
Since P( -1) 0, then (x + 1) is not a factor of P(x) = 5x4 + x3 – 4x2 – 6x – 10.
Exercises:
• Determine whether the linear expression is a factor of P(x):
• a. x – 2 ; P(x) = 4x3 – 7x2 + x – 2 • b. x + 3 ; P(x) = 2x4 + 5x3 + 11x + 6• c. x + 2 ; P(x) = x4 + 2x3 – 12x2 – 11x + 6
References:
• http://rechneronline.de/function-graphs/• http://www.coolmath.com/graphit/