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Warm-up

• Factor the following

• x3 – 3x2 – 28x

• 3x2 – x – 4

• 16x4 – 9y2

• x3 + x2 – 9x - 9

Graphing Quadratic Functions

Section 2-1

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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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Two Real Solutions

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One Real Solution

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No Real Solutions

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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

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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)

Vertex (3, 1) and Point (6, 3)

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2( )y a x h k 23 (6 3) 1a

3 9 1a

2

9a

22( 3) 1

9y x

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Homework

• WS 3-2