Ch. 4 Pre-test1. Graph the function: y = – 4x2

Then label the vertex and axis of symmetry.

2. Write the quadratic function in standard form:

y = (x – 5)(x + 3)

3. Factor the expression: x2 – 7x – 18

4. Solve the equation: 11s2 – 44 = 0

5. Simplify the expression: 103245

4.1 Graph Quadratic Functions in Standard

FormKey Vocabulary:

- standard form of a quadratic function: y = ax2 + bx + c where a ≠ 0.

- The graph is a parabola.

- The lowest or highest point is the vertex: (x, y)

- The axis of symmetry divides the parabola in symmetric halves.

Graph a quadratic function:

2. Make a Table:

x y

-2 4

-1 1

0 0

1 1

2 4

3. Plot the points on the graph.

4. Draw a smooth curve through the points. Forms a parabola.

5. Identify the axis of symmetry.

6. Identify the vertex.

1. Identify a, b, c constants.

y = ax2 + bx + c

Key Concept

Sdf

asdf

Sadf

Asdf

asdf

The axis of symmetry on the parent graph is

x = 0.

The

Vertex

sdf

The vertex is (0 , 0).

Sa

asd The parent function is y = x2.

Graph y = ax2 + bx + c.

2. Make a Table:

x y

-2 4

-1 1

0 0

1 1

2 4

3. Plot the points on the graph.

4. Draw a smooth curve through the points. Forms a parabola.

5. Identify the axis of symmetry.

6. Identify the vertex.

1. Identify a, b, c constants.

EXAMPLE 1Graph y = 2x2. Compare the graph with y = x2.

Notice that both graphs have the same axis of symmetry ( x = 0) and the same vertex at (0,0).

Compare the two graphs:

Graph y = – . Compare with thex2 + 3 12

graph of y = x2.

Notice that both graphs have the same axis of symmetry x = 0. But different vertex points. One at (0, 0) and the other at (0, 3).

Graph y = ax2 + bx + c.

2. Make a Table:

x y

-2 4

-1 1

0 0

1 1

2 4

3. Plot the points on the graph.

4. Draw a smooth curve through the points. Forms a parabola.

5. Identify the axis of symmetry.

6. Identify the vertex.

1. Identify a, b, c constants.

GUIDED PRACTICEGraph the function. Compare the graph with the graph of y = x2.

1. y = – 4x2

Compared with y = x2 Same axis of symmetry and vertex, opens down, and is narrower.

2. y = – x2 – 5

Same axis of symmetry, vertex is shifted down 5 units, and opens down.

So far, no b term (b = 0) . . . now we add the b

term:Properties of the graph of y = ax2 + bx + c

a

b

2

Characteristics of the graph of y = ax2 + bx + c:

•The graph opens up if a > 0 (a is a positive #) and opens down if a < 0 (a is a negative #).

•The vertex has x-coordinate x = . Substitute to find y-coordinate.

•The axis of symmetry is x = .

•The y-intercept is c, so the point ( 0 , c ) is on the parabola.

a

b

2

a

b

2Asdf’

a

b

2

Graph y = ax2 + bx + c.

1. Identify a, b, c constants.

2. Find the vertex: x = a

b

2

Substitute x into function for y-coordinate.

Form for vertex point: (x, y)3. Identify the axis of symmetry: x-coordinate of vertex.

4. Graph the vertex point and axis of symmetry.

5. Make a table. (Use points on both sides of vertex.)

6. Graph the points from the table and connect the points with a smooth curve forming a parabola.

EXAMPLE 3Graph y = 2x2 – 8x + 6. Label the vertex and axis of symmetry.

1. Find the vertex:

x = b 2a =

(– 8) 2(2)

– – = 2

Now find the y - coordinate of the vertex:

y = 2(2)2 – 8(2) + 6 = – 2

So, the vertex is (2, – 2).

2. The axis of symmetry is: x = 2.

a = 2 b = -8 c = 6 2 > 0 so parabola opens up

3. Make a table:x y

-1 16

0 6

1

2

3

4

0

-2

0

6

4. Graph:

Graph y = ax2 + bx + c.

1. Identify a, b, c constants.

2. Find the vertex: x = a

b

2

Substitute x into function for y-coordinate.

Form for vertex point: (x, y)3. Identify the axis of symmetry: x-coordinate of vertex.

4. Graph the vertex point and axis of symmetry.

5. Make a table. (Use points on both sides of vertex.)

6. Graph the points from the table and connect the points with a smooth curve forming a parabola.

GUIDED PRACTICEGraph the function. Label the vertex and axis of symmetry.

4. y = x2 – 2x – 1

The vertex is (1, – 2).

The axis of symmetry x = 1.

Key Concept

Minimum and Maximum Values

Lindsey

Anderson

To find the minimum or maximum value of a quadratic function:

1. Is a > 0 or is a < 0. (i.e. is a negative or positive?)

a is negative – maximum

a is positive – minimum

2. Find the vertex: (substitute for y)

3. The max. or min. value is the y-coordinate of the vertex.

x = – b 2a

EXAMPLE 4Tell whether the function y = 3x2 – 18x + 20 has a minimum value or a maximum value. Then find the minimum or maximum value.

a > 0 (a is positive), so the function has a minimum value.

x = – b 2a = –

(– 18) 2(3)

= 3

y = 3(3)2 – 18(3) + 20 = –7

The minimum value is y = –7 at the point (3 , -7).

Homework 4.1: p. 240: 3-52 (EOP)

a = 3 b = -18 c = 20

Find the vertex: