Lecture 8: Graphing Quadratic Functions

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Assignments

These are the problems you should work before today, February 5.
Review problems, page 75, 4n+1, for n=0..22;
Section 1.1, page 92, 4n+1, for n=0...24;
Section 1.2, page 102, 1-6, and 4n, for n=3...23;
Section 1.3, page 113, 4n+1, for n=0...19.

These are the problems you should work before February 12.
Section 1.4, page 120, problems 6n+1, for n=0...15;
(skip 1.5) 1.6, page 149, problems 6n+1, for n=0...16.

These are the problems you should work before February 19.
Section 1.7, page 160, 4n+1, for n=0,...,24.
Section 1.8, page 171, problems 6n+1, for n=0...8;

These are the problems you should work before February 26.
The first test, covering Chapters P and 1 and sections 2.1 and 2.2 is on February 26.
Section 2.1, page 191, problems 6n+1, for n=0...12.
Section 2.2, page 025, problems 8n+1, for n=0...11.

Brief review of the lecture

Outline for Monday, February 1

 

Analyzing Quadratic Equations

A. What does it mean to analyze an equation?

  1. What is its shape? Which way does it open?
  2. Does it have x-intercepts and if so how many?

B. Consider the equation y=x2-4x+7

C. Next consider y=2x2-4x+6

D. Finally consider the most general quadratic function y=ax2+bx+c, where a, b, and c are constants.

E. Determine which of the six types a quadratic function is based on a,b, and c.

 
Below this, as you know by now, is material left from last time. We'll eventually cover compount interest. Without using the compound interest formula how would you answer the following question?
Q : Suppose you have invested $1000 for 8 years at 6% annual interest, compounded yearly. How much would you have after 8 years?

Start with $1000.

After 1 year you would have 1000 +1000(0.006)=1000(1+0.06)=1000(1.06).

After 2 years you would have 1000(1.06)+1000(1.06)(0.06)=1000(1.06)(1+0.06) =1000(1.06)^2

After 8 years you would have $1000(1.06)^8

In this situation the formula is B=P(1+r)^t

Q : Suppose we change the problem so that interest is now compounded twice a year, everything else is the same.

Start with $1000.

After 6 months you would have 1000+1000(0.03)=1000(1.03)

After 12 months you would have 1000(1.03)^2

After 8 years you would have 1000(1.03)^(16)

In the situation the formula is B=(1+r/2)^(2t)

We leave it to you to find a formula for any time period and any number of compoundings.

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