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# 2.1 a and b - PowerPoint PPT Presentation

2.1 a and b. Finding the vertex and y-intercept from standard form Graphing in standard form. These properties can be generalized to help you graph quadratic functions. Helpful Hint. When a is positive, the parabola is happy (U). When the a negative, the parabola is sad ( ). U.

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### 2.1 a and b

Finding the vertex and y-intercept from standard form

Graphing in standard form

When a is positive, the parabola is happy (U). When the a negative, the parabola is sad ( ).

U

The Axis of Symmetry is the same as the x-coordinate of the vertex.

The axis of symmetry is given by . quadratic functions.

Example 2A: Graphing Quadratic Functions in Standard Form

Consider the function f(x) = 2x2 – 4x + 5.

a. Determine whether the graph opens upward or downward.

Because a is positive, the parabola opens upward.

b. Find the axis of symmetry.

Substitute –4 for b and 2 for a.

The axis of symmetry is the line x = 1.

Consider the function f(x) = 2x2 – 4x + 5.

c. Find the vertex.

The vertex lies on the axis of symmetry, so the x-coordinate is 1. The y-coordinate is the value of the function at this x-value, or f(1).

f(1) = 2(1)2 – 4(1) + 5 = 3

The vertex is (1, 3).

d. Find the y-intercept.

Because c = 5, the intercept is 5.

Consider the function f(x) = 2x2 – 4x + 5.

e. Graph the function.

Graph by making a table of values with the x-coordinate of the vertex in the center.

The axis of symmetry is given by . quadratic functions.

Example 2B: Graphing Quadratic Functions in Standard Form

Consider the function f(x) = –x2 – 2x + 3.

a. Determine whether the graph opens upward or downward.

Because a is negative, the parabola opens downward.

b. Find the axis of symmetry.

Substitute –2 for b and –1 for a.

The axis of symmetry is the line x = –1.

Consider the function f(x) = –x2 – 2x + 3.

c. Find the vertex.

The vertex lies on the axis of symmetry, so the x-coordinate is –1. The y-coordinate is the value of the function at this x-value, or f(–1).

f(–1) = –(–1)2 – 2(–1) + 3 = 4

The vertex is (–1, 4).

d. Find the y-intercept.

Because c = 3, the y-intercept is 3.

Consider the function f(x) = –x2 – 2x + 3.

e. Graph the function.

Graph by making a table of values with the x-coordinate of the vertex in the center.

b. quadratic functions. The axis of symmetry is given by .

Check It Out! Example 2a

For the function, (a) determine whether the graph opens upward or downward, (b) find the axis of symmetry, (c) find the vertex, (d) find the y-intercept, and (e) graph the function.

f(x)= –2x2 – 4x

a. Because a is negative, the parabola opens downward.

Substitute –4 for b and –2 for a.

The axis of symmetry is the line x = –1.

Check It Out! quadratic functions. Example 2a

f(x)= –2x2 – 4x

c. The vertex lies on the axis of symmetry, so the x-coordinate is –1. The y-coordinate is the value of the function at this x-value, or f(–1).

f(–1) = –2(–1)2 – 4(–1) = 2

The vertex is (–1, 2).

d. Because c is 0, the y-intercept is 0.

Check It Out! quadratic functions. Example 2a

f(x)= –2x2 – 4x

e. Graph the function.

Graph by making a table of values with the x-coordinate of the vertex in the center.

b. quadratic functions.The axis of symmetry is given by .

The axis of symmetry is the line .

Check It Out! Example 2b

For the function, (a) determine whether the graph opens upward or downward, (b) find the axis of symmetry, (c) find the vertex, (d) find the y-intercept, and (e) graph the function.

g(x)= x2 + 3x – 1.

a. Because a is positive, the parabola opens upward.

Substitute 3 for b and 1 for a.

c. quadratic functions. The vertex lies on the axis of symmetry, so the x-coordinate is . The y-coordinate is the value of the function at this x-value, or f().

f( ) = ( )2 + 3( ) – 1 =

The vertex is ( , ).

Check It Out! Example 2b

g(x)= x2 + 3x – 1

d. Because c = –1, the intercept is –1.

Check It Out! quadratic functions. Example2

g(x)= x2 + 3x – 1

e. Graph the function.

Graph by making a table

of values with the x-coordinate

of the vertex in the center.

Lesson Quiz: Part I quadratic functions.

Consider the function f(x)= 2x2 + 6x – 7.

1. Determine whether the graph opens upward or downward.

2. Find the axis of symmetry.

3. Find the vertex.

4. Identify the maximum or minimum value of the function.

5. Find the y-intercept.

upward

x = –1.5

(–1.5, –11.5)

min.: –11.5

–7

Lesson Quiz: Part II quadratic functions.

Consider the function f(x)= 2x2 + 6x – 7.

6. Graph the function.

By making a table