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Objectives
• I can graph quadratic functions in vertex form.
• I can convert to vertex format by completing the square
• I can find the equation of a quadratic given the vertex point and another point.
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Applications
• Many applications that are used to model consumer behavior.
• Example: Revenue generated from manufacturing handheld video games
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Let a, b, and c be real numbers a 0. The function f (x) = ax2 + bx + cis called a quadratic function. The graph of a quadratic function is a parabola.Every parabola is symmetrical about a line called the axis (of symmetry).
The intersection point of the parabola and the axis is called the vertex of the parabola.
x
y
axis
f (x) = ax2 + bx + cvertex
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Quadratic Key Terms
• Vertex: The vertex is the point of intersection between the Axis of Symmetry and the parabola.
• Axis of Symmetry: This is the straight line that cuts the parabola into mirror images.
• Solutions (Also called Roots): These are the x-coordinates of the points where the parabola intersects the x-axis.
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y=ax2+bx+c
• One key point to graph is always the y-intercept. This is a point that has the x value equal to zero (0, b)
• If we let “x” be zero if the equation, then we see that the y-intercept is just the value of “c” in the equation.
• Example: y = 2x2 – 3x + 4• The y-intercept is (0, 4)
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Solutions versus Intercepts• Lets look at the differences in these vocabulary
terms:
• Solutions, roots, zeros, x-intercepts
• If the problem asks you to find solutions or roots your answer should be in the format x = ?? Or {2, 3, 5}. We just list the x-value
• If the problem asks you to find zeros or x-intercepts, your answer should be in ordered pair format: (2, 0) (-3, 0)
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The leading coefficient of ax2 + bx + c is a.
When the leading coefficient is positive, the parabola opens upward and the vertex is a minimum.
When the leading coefficient is negative, the parabola opens downward
and the vertex is a maximum.
x
y
f(x) = ax2 + bx + ca > 0 opens upward
vertex minimum
xy
f(x) = ax2 + bx + c
a < 0 opens
downward
vertex maximum
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5
y
x5-5
The simplest quadratic functions are of the form f (x) = ax2 (a 0) These are most easily graphed by comparing them with the graph of y = x2.
Example: Compare the graphs of
, and2xy 2
2
1)( xxf 22)( xxg
2
2
1)( xxf
22)( xxg
2xy
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Standard Vertex Format• The format for any parabola with an equation
ax2 + bx + c = 0 can be written in the following standard vertex format:
•y = a(x – h)2 + k• Where the Vertex is (h,k) and the Axis of
Symmetry is x = h
• If a > o, then the parabola opens upward
• If a < o, then the parabola opens downward
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Example 2: y = x2 – 4x + 5
• y = x2 – 4x + 5 (Now complete the square)• y = (x2 – 4x + ___) + 5 - _____• -4/2 = -2• (-2)2 = 4• y = (x2 – 4x + 4) + 5 – 4• y = (x – 2)2 + 1• Vertex is (2, 1), Axis of Sym: x = 2, a > 0, so
parabola turns upward
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Example 3: y = -5x2 + 80x - 319
• y = -5x2 + 80x - 319 (1st factor –5 from 1st two terms)
• y = -5(x2 – 16x) - 319 (Now complete square)• -16/2 = -8• (-8)2 = 64, • y = -5(x2 – 16x + 64) - 319 + 320 (WHY +320??)• y = -5(x – 8)2 + 1• Vertex is (8, 1), Axis of Sym: x = 8, a < 0, so
parabola turns downward
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x
y
4
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Example: Graph and find the vertex and x-intercepts of f (x) = –x2 + 6x + 7.
f (x) = – x2 + 6x + 7 original equation
f (x) = – ( x2 – 6x) + 7 factor out –1
f (x) = – ( x2 – 6x + 9) + 7 + 9 complete the square
f (x) = – ( x – 3)2 + 16 standard form
a < 0 parabola opens downward.
h = 3, k = 16 axis x = 3, vertex (3, 16).
Find the x-intercepts by solving–x2 + 6x + 7 = 0. (–x + 7 )( x + 1) = 0 factor
x = 7, x = –1 x-intercepts (7, 0), (–1, 0)
x = 3f(x) = –x2 + 6x + 7
(7, 0)(–1, 0)
(3, 16)