5.1 – Introduction to Quadratic Functions

Objectives: Define, identify, and graph quadratic functions. Multiply linear binomials to produce a quadratic

expression.Standard: 2.8.11.E. Use equations to represent curves.

cbxaxxf 2)(

The Standard Form of a Quadratic Function is:

(A quadratic function is any function that can be written in the form f(x)= ax2 + bx + c, where a ≠ 0.)

A Quadratic function is any function that can be written in the form f(x)= ax2 + bx + c, where a ≠ 0.

Ex 1a. Let f(x) = (2x – 5)(x - 2). Show that f represents a quadratic function. Identify a, b, and c when the function is written in the form f(x) = ax2 + bx + c.

FOIL First – Outer – Inner – Last OR

Distribute each term in the first set of parentheses to each term in the second set of parentheses!

(2x – 5)(x – 2) = 2x2 – 4x – 5x + 10 2x2 – 9x + 10

a = 2, b = -9, c = 10

p.278: #13-19 ODD

The graph of a quadratic function is called a parabola.

• Each parabola has an axis of symmetry, a line that divides the parabola into two parts that are mirror images of each other.

• The vertex of a parabola is either the lowest point on the graph or the highest point on the graph.

Axis of Symmetry

Vertex

Ex 2a. Identify whether f(x) = -2x2 - 4x + 1 has a maximum value or a minimum value at the vertex. Then give the approximate coordinates of the vertex.

• First, graph the function:

• Next, find the maximum value of the parabola (2nd, Trace):

• Finally, max(-1, 3).

III. Minimum and Maximum Values

• Let f(x) = ax2 + bx + c, where a ≠ 0. The graph of f is a parabola.– If a > 0, the parabola opens up and the vertex is

the lowest point. The y-coordinate of the vertex is the minimum value of f.

– If a < 0, the parabola opens down and the vertex is the highest point. The y-coordinate of the vertex is the maximum value of f.

a. f(x) = x2 + x – 6 b. g(x) = 5 + 4x – x2 c. f(x) = 2x2 - 5x + 2 d. g(x) = 7 - 6x - 2x2

Opens up, has minimum value

Opens down, has maximum value

Opens up, has minimum value

Opens down, has maximum value

5.2 Solving Quadratic Equations(an introduction)

Solve 5x2 – 19 = 231. Give exact solutions. Then give approximate solutions to the nearest hundredth.

2

2

2

5 19 231

5 250

50

50

x

x

x

x

07.7x 07.7x

Ex 2a. Solve 4(x+2)2 = 49

2

2

49( 2)

4

49( 2)

47

22

7 72 or 2

2 23 11

or 2 2

x

x

x

x x

x x

A rescue helicopter hovering 68 feet above a boat in distress drops a life raft. The height in feet of the raft above the water can be modeled by h(t) = -16t2 + 68, where t is the time in seconds after it is dropped. After how many seconds will the raft dropped from the helicopter hit the water?

68 ft2

2

2

2

2

0 16 68

16 68

68

1617

4

17

4

17

22.1 seconds

t

t

t

t

t

t

t

When the raft hits the water, the height will = 0, so h(t) = 0:

Since only positive values of time make sense, the answer is 2.1 seconds.

If ∆ABC is a right triangle with the right angle at C, then a2+ b2 = c2.

When you apply the Pythagorean Theorem, use the principal square root because distance and length cannot be negative.

a

b c

Homework:p 278 #20-26even, 27-32, 38-44even, 49p 287 #24-30even,36-42even, 50