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Polynomial Function
A polynomial function of degree n, where n is a nonnegative integer, is a function defined by an expression of the form
where an, an-1, …, a1, and a0 are real numbers, with an ≠ 0.
11 1 0,
n nn nx a x a x a x a
f
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Quadratic Function
A function is a quadratic function if
2( ) ,x ax bx c f
where a, b, and c are real numbers, with a ≠ 0.
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Simplest Quadratic
x (x)
– 2 4
– 1 1
0 0
1 1
2 4
2 3
2
– 2
3
– 2
4
– 3
– 4
– 3– 4
4
range[0, )
2x xfdomain (−, )
x
y
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Simplest Quadratic
Parabolas are symmetric with respect to a line. The line of symmetry is called the axis of the parabola. The point where the axis intersects the parabola is the vertex of the parabola.
Vertex
Vertex
Axis
Axis
Opens up
Opens down
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Applying Graphing Techniques to a Quadratic Function
The graph of g(x) = ax2 is a parabola with vertex at the origin that opens up if a is positive and down if a is negative. The width of the graph of g(x) is determined by the magnitude of a. The graph of g(x) is narrower than that of (x) = x2 if a> 1 and is broader (wider) than that of (x) = x2 if a< 1. By completing the square, any quadratic function can be written in the form
the graph of F(x) is the same as the graph of g(x) = ax2 translated hunits horizontally (to the right if h is positive and to the left if h is negative) and translated k units vertically (up if k is positive and down if k is negative).
2( ) ( ) .F x a x h k
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Example 1 GRAPHING QUADRATIC FUNCTIONS
Solution
a.
Graph the function. Give the domain and range.
2 4 2 (by plotting points)x x x fx (x)
– 1 3
0 – 2
1 – 5
2 – 6
3 – 5
4 – 2
5 3
2
3
– 2
– 6
Domain (−, )
Range[– 6, )
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Example 1 GRAPHING QUADRATIC FUNCTIONS
Solution
b.
Graph the function. Give the domain and range.
212
x xg
Domain (−, )
Range(–, 0]
2
3
– 2
– 6
212
y x
2y x
212
x xg
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Example 1 GRAPHING QUADRATIC FUNCTIONS
Solution
c.
Graph the function. Give the domain and range.
214 3
2F x x
Domain (−, )
Range(–, 3]
212
x xg
214 3
2x x F
(4, 3)
3
– 2
– 6 x = 4
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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution Express x2– 6x + 7 in the form (x– h)2 + k by completing the square.
Graph by completing the square and locating the vertex.
2 6 7x x x f
2 6 7x x x f Complete the square.
2
212
96 3
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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution Express x2 – 6x + 7 in the form (x– h)2 + k by completing the square.
Graph by completing the square and locating the vertex.
2 6 7x x x f
2 9 96 7x x x f Add and subtract 9.
2 6 9 79x xx f Regroup terms.
23 2x x f Factor; simplify.
This form shows that the vertex is (3, – 2)
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Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution
Graph by completing the square and locating the vertex.
2 6 7x x x f
Find additional ordered pairs that satisfy the equation. Use symmetry about the axis of the parabola to find other ordered pairs. Connect to obtain the graph.Domain is (−, ) Range is [–2, )
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Graph of a Quadratic FunctionThe quadratic function defined by (x) = ax2 + bx + c can be written as
where
2, 0,y x a x h k a f
and .2b
h k ha
f
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Graph of a Quadratic FunctionThe graph of has the following characteristics.1.It is a parabola with vertex (h, k) and the vertical line x = h as axis.2.It opens up if a > 0 and down is a < 0.3.It is broader than the graph of y = x2 if a< 1 and narrower if a> 1.4.The y-intercept is (0) = c. 5.If b2 – 4ac > 0, the x-intercepts are If b2 – 4ac = 0, the x-intercepts is
If b2 – 4ac < 0, there are no x-intercepts.
2 42
b b aca
2ba
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Example 4 FINDING THE AXIS AND THE VERTEX OF A PARABOLA USING THE VERTEX FORMULA
Solution Here a = 2, b = 4, and c = 5. The axis of the parabola is the vertical line
Find the axis and vertex of the parabola having equation (x) = 2x2 +4x + 5 using the vertex formula.
4
2 2 21
bx h
a
The vertex is (– 1, (– 1)). Since (– 1) = 2(– 1)2 + 4 (– 1) +5 = 3, the vertex is (– 1, 3).
