The General Quadratic Function

Students will be able to graph functions defined by the

general quadratic equation.

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The General Quadratic Function

• The general quadratic function may be written as: f(x) = ax2 + bx + c or y = ax2 + bx + c , where a, b, and c are real numbers and a ≠ 0.

• Why can’t a be equal to 0?

• Because if a were equal to 0, then there would be no x2 term. Then we would not have a quadratic equation.

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Vertex of the Parabola• When we have a quadratic equation

written as y = ax2 + bx + c, the vertex of the parabola is the lowest or highest point on the graph. The x-

coordinate of the vertex is .

• The y-coordinate of the vertex can be found by substituting the x-value for x in the original equation and finding y.

2

bx

a

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Axis of Symmetry

• The axis of symmetry for the parabola is the vertical line that passes through the vertex.

• The equation of that line is

2

bx

a

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The y-intercept• Another point that helps in graphing

the parabola is the y-intercept.• To find the y-intercept, substitute 0

into the equation in place of x.• The value that you get for y is the y-

intercept.• You can then use the graph and

reflect the y-intercept across the axis of symmetry to find another point on the graph.

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The Vertex Form

• To find the vertex form of the quadratic equation, find the vertex and substitute it into the following form:

where (h, k) is the vertex of the parabola.

2y a x h k

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Example

Given the equation find each of the following:

1. The coordinates of the vertex2. The axis of symmetry3. The y-intercept4. The reflection of the y-intercept5. The vertex form of the equation6. The graph of the equation

22 4 5 y x x

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Example: Vertex• Find the coordinates of the vertex

for the graph of .

• Find the x-coordinate for the vertex.

• Find the y-coordinate for the vertex.

• Thus, the vertex is (1, 3)

2

bx

a 4

2 2

22 4 5 y x x

1

22 4 5 y x x 3 22 51 14

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Example: Axis of Symmetry and y-intercept

• For this equation: The axis of symmetry is:

• The y-intercept is:

22 4 5 y x x

2

bx

a

22 0 4 0 5y

1x

5y

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Example: Vertex Form

• Since the vertex for this equation:

was (1, 3) and a = 2, we substitute those into the following:

with vertex (h, k)

to get the following vertex form:

22 4 5 y x x

2( )y a x h k

22 1 3y x

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12

10

8

6

4

2

-2

5

Example• When we graph this

quadratic equation, the parabola opens up.

• The vertex is (1, 3).• The axis of

symmetry is x = 1.• The y-intercept is 5.• The reflected point is

(2, 5).• So the graph is:

22 4 5 y x x

(1, 3)

axis of symmetr

y