Download Section 5.1 - Monroe County Schools

Section 5.1
Introduction to Quadratic Functions
Quadratic Function
• A quadratic function is any function that can
be written in the form f(x) = ax² + bx + c,
where a ≠ 0.
• It is defined by a quadratic expression, which
is an expression of the form as seen above.
• The stopping-distance function, given by:
d(x) = ⅟₁₉x² + ¹¹̸₁₀x, is an example of a quadratic
function.
Quadratic Functions
• Let f(x) = (2x – 1)(3x + 5). Show that f
represents a quadratic function. Identify a, b,
and c.
• f(x) = (2x – 1)(3x + 5)
• f(x) = (2x – 1)3x + (2x – 1)5
• f(x) = 6x² - 3x + 10x – 5
• f(x) = 6x² + 7x – 5
a = 6, b = 7, c = - 5
Parabola
• The graph of a quadratic function is called a
parabola. Parabolas have an axis of symmetry,
a line that divides the parabola into two parts
that are mirror images of each other.
• The vertex of a parabola is either the lowest
point on the graph or the highest point on the
graph.
Domain and Range of Quadratic
Functions
• The domain of any quadratic function is the
set of all real numbers.
• The range is either the set of all real numbers
greater than or equal to the minimum value of
the function (when the graph opens up).
• The range is either the set of all real numbers
less than or equal to the maximum value of
the function (when the graph opens down).
Minimum and Maximum Values
• Let f(x) = ax² + bx + c, where a ≠ 0. The graph
of f is a parabola.
• If a > 0, the parabola opens up and the vertex
is the lowest point. The y-coordinate of the
vertex is the minimum value of f.
• If a < 0, the parabola opens down and the
vertex is the highest point. The y-coordinate
of the vertex is the maximum value of f.
Minimum and Maximum Values
• f(x) = x² + x – 6
• Because a > 0, the
parabola opens up and
the function has a
minimum value at the
vertex.
• g(x) = 5 + 4x - x²
• Because a < 0, the
parabola opens down
and the function has a
maximum value at the
vertex.
Section 5.2
Introduction to Solving Quadratic
Equations
Solving Equations of the Form x² = a
• If x² = a and a ≥ 0, then x = √a or x = - √a, or
simply x = ± √a.
• The positive square root of a, √a is called the
principal square root of a.
• Simplify the radical for the exact answer.
Solving Equations of the Form x² = a
• Solve 4x² + 13 = 253
• 4x² + 13 = 253
- 13 - 13
4x²
= 240
Simply the Radical
√60 = √(2 ∙ 2 ∙ 3 ∙ 5)
√60 = 2√(3 ∙ 5)
√60 = ± 2√15 (exact answer)
4x²
= 240
4
4
x²
= 60
x = √60 or x = - √60 (exact answer)
x = 7.75 or x = - 7.75 (approximate answer)
Properties of Square Roots
• Product Property of Square Roots:
• If a ≥ 0 and b ≥ 0: √(ab) = √a ∙ √b
• Quotient Property of Square Roots:
• If a ≥ 0 and b > 0: √(a/b) = √(a) ÷ √(b)
Properties of Square Roots
• Solve 9(x – 2)² = 121
• 9(x – 2)² = 121
9
9
(x – 2)² = 121/9
√(x – 2)² = ±√(121/9)
x – 2 = ±√(121/9)
x–2
+2
= √(121/9)
+2
x = 2 + √(121/9) or 2 - √(121/9)
x = 2 + [√(121) / √ (9)] or 2 – [√(121) / √(9)]
x = 2 + (11/3) or 2 – (11/3)
x = 17/3 or x = - 5/3
Pythagorean Theorem
• If ∆ABC is a right triangle with the right angle
at C, then a² + b² = c²
A
c
a
C
B
b
Pythagorean Theorem
• If ∆ABC is a right triangle with the right angle
at C, then a² + b² = c²
A
c
2.5
C
B
5.1
2.5² + 5.1² = c²
6.25 + 26.01 = c²
32.26 = c²
√(32.26) = c
5.68 = c
Section 5.3
Factoring Quadratic Expressions
Factoring Quadratic Expressions
• When you learned to multiply two expressions
like 2x and x + 3, you learned how to write a
product as a sum.
• Factoring reverses the process, allowing you
to write a sum as a product.
• To factor an expression containing two or
more terms, factor out the greatest common
factor (GCF) of the two expressions.
Factoring Quadratic Expressions
•
•
•
•
•
•
3a² - 12a
3a² = 3a ∙ a
12a = 3a ∙ 4
The GCF = 3a
3a(a) – 3a ∙ 4
(3a)(a – 4)
3x(4x + 5) – 5(4x + 5)
The GCF = 4x + 5
(3x – 5)(4x +5)
Factoring x² + bx + c
• To factor an expression of the form:
• ax² + bx + c where a = 1, look for integers r and
s such that r ∙ s = c and r + s = b.
• Then factor the expression.
• x² + bx + c = (x + r)(x + s)
Factoring x² + bx + c
• x² + 7x + 10
x² - 7x + 10
(5+2) = 7 & (5∙2) = 10 (-5-2) = -7 & (-5∙(-2)) = 10
(x + 5)(x + 2)
(x – 5)(x – 2)
Factoring the Difference of Two
Squares
•
•
•
•
•
a² - b² = (a + b)(a – b)
(x + 3)(x – 3)
x² + 3x - 3x - 9
x² - 9
x² - 3²
Factoring Perfect-Square Trinomials
a² + 2ab + b² = (a + b)²
(x + 3)²
(x + 3)(x + 3)
x² + 3x + 3x + 9
X² +2(3x) + 9
a² - 2ab + b² = (a – b)²
(x – 3)²
(x – 3)(x – 3)
x² - 3x – 3x + 9
x² - 2(3x) + 9
Zero-Product Property
• If pq = 0, then p = 0 or q = 0.
•
•
•
•
2x² - 11x = 0
x(2x – 11) = 0 (Factor out an x)
x = 0 or 2x – 11 = 0
x = 11 or x = 11/2
Section 5.4
Completing the Square
Completing the Square
• When a quadratic equation does not contain a
perfect square, you an create a perfect square
in the equation by completing the square.
• Completing the square is a process by which
you can force a quadratic expression to factor.
Specific Case of a Perfect-Square
Trinomial
• x² + 8x + 16 = (x + 4)²
• Understand: (½)8 = 4 → 4² = 16
Examples of Completing the Square
•
•
•
•
•
•
x² - 6x
(½)(-6) = -3
(-3)² → = 9
The perfect-square
x² - 6x + 9 =
(x – 3)²
x² + 15x
(½)(15) = (15/2)
(15/2)² → = (15/2)²
The perfect square
x² + 15x + (15/2)² =
[x + (15/2)]²
Solving a Quadratic Equation by
Completing the Square
•
•
•
•
•
•
•
•
x² + 10x – 24 = 0
+ 24 +24
x² +10x = 24
x²+10x+(5)²=24+(5)²
x² + 10x + 25 = 49
(x + 5)² = 49
x+5=±7
x = - 12 or x = 2
2x² + 6x = 7
2(x² + 3x) = 7
x² + 3x = (7/2)
x²+3x+(3/2)²=(7/2)+(3/2)²
x²+3x+(3/2)² =(7/2)+(9/4)
[x + (3/2)]² = (23/4)
x + (3/2) = ±√(23/4)
x = - (3/2) + √(23/4) (0.90)
X = - (3/2) - √(23/4) (-3.90)
Vertex Form
• If the coordinates of the vertex of the graph of
y = ax² + bx + c, where a ≠ 0, are (h,k), then
you can represent the parabola as:
• y = a(x – h)² + k, which is the vertex form of a
quadratic function.
Vertex Form
•
•
•
•
•
•
Given g(x) = 2x² + 12x + 13
2(x² + 6x) + 13
2(x² + 6x + 9) + 13 – 2(9)
2(x + 3)² - 5
2[x – (-3)]² + (- 5) (Vertex Form)
The coordinates (h,k) of the vertex are (-3, -5)
and the equation for the axis of symmetry is
x = - 3.