Section 2.2 Quadratic Functions. Graphs of Quadratic Functions.
5.1 – Introduction to Quadratic Functions
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5.1 – Introduction to Quadratic Functions
Objectives: Define, identify, and graph quadratic functions. Multiply linear binomials to produce a quadratic
expression.Standard: 2.8.11.E. Use equations to represent curves.
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The Standard Form of a Quadratic Function is:
(A quadratic function is any function that can be written in the form f(x)= ax2 + bx + c, where a ≠ 0.)
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A Quadratic function is any function that can be written in the form f(x)= ax2 + bx + c, where a ≠ 0.
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Ex 1a. Let f(x) = (2x – 5)(x - 2). Show that f represents a quadratic function. Identify a, b, and c when the function is written in the form f(x) = ax2 + bx + c.
FOIL First – Outer – Inner – Last OR
Distribute each term in the first set of parentheses to each term in the second set of parentheses!
(2x – 5)(x – 2) = 2x2 – 4x – 5x + 10 2x2 – 9x + 10
a = 2, b = -9, c = 10
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p.278: #13-19 ODD
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The graph of a quadratic function is called a parabola.
• Each parabola has 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.
Axis of Symmetry
Vertex
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Ex 2a. Identify whether f(x) = -2x2 - 4x + 1 has a maximum value or a minimum value at the vertex. Then give the approximate coordinates of the vertex.
• First, graph the function:
• Next, find the maximum value of the parabola (2nd, Trace):
• Finally, max(-1, 3).
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III. 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.
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a. f(x) = x2 + x – 6 b. g(x) = 5 + 4x – x2 c. f(x) = 2x2 - 5x + 2 d. g(x) = 7 - 6x - 2x2
Opens up, has minimum value
Opens down, has maximum value
Opens up, has minimum value
Opens down, has maximum value
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5.2 Solving Quadratic Equations(an introduction)
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Solve 5x2 – 19 = 231. Give exact solutions. Then give approximate solutions to the nearest hundredth.
2
2
2
5 19 231
5 250
50
50
x
x
x
x
07.7x 07.7x
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Ex 2a. Solve 4(x+2)2 = 49
2
2
49( 2)
4
49( 2)
47
22
7 72 or 2
2 23 11
or 2 2
x
x
x
x x
x x
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A rescue helicopter hovering 68 feet above a boat in distress drops a life raft. The height in feet of the raft above the water can be modeled by h(t) = -16t2 + 68, where t is the time in seconds after it is dropped. After how many seconds will the raft dropped from the helicopter hit the water?
68 ft2
2
2
2
2
0 16 68
16 68
68
1617
4
17
4
17
22.1 seconds
t
t
t
t
t
t
t
When the raft hits the water, the height will = 0, so h(t) = 0:
Since only positive values of time make sense, the answer is 2.1 seconds.
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If ∆ABC is a right triangle with the right angle at C, then a2+ b2 = c2.
When you apply the Pythagorean Theorem, use the principal square root because distance and length cannot be negative.
a
b c
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Homework:p 278 #20-26even, 27-32, 38-44even, 49p 287 #24-30even,36-42even, 50