# Lesson 2.2, page 273 Quadratic Functions - PowerPoint PPT Presentation

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Lesson 2.2, page 273 Quadratic Functions. Objectives Recognize characteristics of parabolas. Graph parabolas. Determine a quadratic function’s minimum or maximum value. Solve problems involving a quadratic function’s minimum or maximum value. Vocabulary.

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Lesson 2.2, page 273 Quadratic Functions

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## Lesson 2.2, page 273Quadratic Functions

Objectives

Recognize characteristics of parabolas.

Graph parabolas.

Determine a quadratic function’s minimum or maximum value.

Solve problems involving a quadratic function’s minimum or maximum value.

### Vocabulary

• Quadratic Function: a function that can be written in the form

f(x) = ax2 + bx + c, a≠ 0.

• Standard form: f(x) = a(x - h)2 + k

• Parabola: graph of quadratic function; U-shaped curve; symmetric

• Symmetry: when folded, two sides match exactly

### f(x) = a(x - h)2 + k

• Vertex: the point (h,k); highest or lowest point of parabola; on axis of symmetry

• a: describes steepness and direction of parabola

• Axis of symmetry: x = h; fold line that divides parabola into two matching halves

• Minimum: if a > 0, parabola opens up; vertex is minimum point or lowest point of parabola; k is the minimum value

• Maximum: If a < 0, parabola opens down; vertex the maximum point or lowest point of parabola; k is the maximum value

y

y

10

10

x

x

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### Minimum (or maximum) function value for a quadratic occurs at the vertex.

• Identify the vertex of each graph . Find the minimum or maximum value.

• A.B.

### ReviewSolving Quadratic Equations

Example: 3x2 – 2x = 21

• Get in standard form. 3x2 – 2x – 21 = 0

• Factor. (3x + 7)(x – 3) = 0

• Set each factor equal to zero.3x+7=0 or x–3 = 0

• Solve.x = -7/3 or x = 3

• Check.

### If no “B” term, can use square root property to solve for zeros.

For example, if x2 = 16, then x = ±4.

• Solve f(x) = 1 – (x – 3)2.

### When factoring or square root property won’t work, you can always use the QUADRATIC FORMULA

• Get in standard form: ax2 + bx + c = 0

• Substitute for a, b, and c in formula.

• Solve.

• Check.

This works every time!!!

### Try these. Solve using the Quadratic formula.

• 3x2 - 2x - 1 = 0x2 + 2x = -3

### Graphing Quadratic Functionsf(x) = a(x - h)2 + k

• Determine whether the parabola opens up (a > 0) or down (a < 0).

• Find the vertex, (h,k).

• Find x-intercepts by solving f(x) = 0.

• Find the y-intercept by computing f(0).

• Plot the intercepts, vertex, and additional points. Connect with a smooth curve.

Note: Use the axis of symmetry (x = h) to plot additional points.

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### Check Point 1, page 276

• Graph f(x) = -(x – 1)2 + 4.

y

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x

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### Check Point 2, page 276

• Graph f(x) = (x – 2)2 + 1.

### f(x) = ax2 + bx + c

• If equation is not in standard form, you may have to complete the square to determine the point (h,k).

Vertex =

y

10

x

-10

10

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### Check Point 3, page 279

• Graph f(x) = -x2 + 4x +1. Identify domain & range.

ax2 + bx + c = 0 or f(x) = ax2 + bx + c,

“x” is squared.

Graph would be a curve.

Solutions are at the x-intercepts. The “zeros” are the solutions or roots of the equation.

To solve: factor & use zero product property, take square root of both sides (if no B term), or use other methods.

### See Example 4, page 279

Check Point 4, page 280: f(x) = 4x2 - 16x + 1000

• Determine, without graphing, whether the function has a minimum value or a maximum value.

• Find the minimum or maximum value and determine where it occurs.

• Identify the function’s domain and range.

### Example 1- Application Problem

A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball, in meters, t seconds after the hit is given by the quadratic function

How long does it take for the baseball to reach its maximum height?

What is the maximum height obtained by the baseball?

Solution

Vertex:

Example 2 - Application

A farmer has 600 feet of fencing to enclose a rectangular field. If the field is next to a river, so no fencing is needed on one side, find the dimensions of the field that will maximize the area. Find the maximum area.

Area = (600  2x)(x)

A(x) = 2x2 + 600x

Find Vertex – to find max.

= 150

y = 2(150)2 + 600(150)

= 45000

Solution

x x

600 – 2x

Width (x) =

Length(600-2x) =

Max area =

150

300

45000