1 / 19

# Points & Polynomials - PowerPoint PPT Presentation

Points & Polynomials. Lecture 3D Pre AP and GT Precalculus. Agenda: Hodgepodge Day. Homework Questions? Difference Quotient Continuity Zero Product Property How many points? Viete Relations Intermediate Value Theorem Challenge Problem. Difference Quotient.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Points & Polynomials' - kelly-hill

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Points & Polynomials

Lecture 3D

Pre AP and GT Precalculus

Agenda: Hodgepodge Day

• Homework Questions?

• Difference Quotient

• Continuity

• Zero Product Property

• How many points?

• Viete Relations

• Intermediate Value Theorem

• Challenge Problem

• The slope of a graph’s secant line

• Difference Quotient of function y is symbolized with a prime after the function name:

• Difference Quotient can be used to find parabola vertex

• Find the difference quotient of h(t)=800t − 16t2

(this is the equation for the height of an object with an initial velocity of 800 mps as it returns to earth)

• Recall h(t)=800t − 16t2 and DQ = 800 −32t − 16Δt

• What is the highest elevation this projectile reached?

• When Δt=0 and h’(t)=0, a parabola its at its vertex

• So… 800 −32t − 16(0)=0 implies 800 = 32t so max height reached when t=25

• Max Height is 800(25)-16(252)=10,000

• Find the difference quotient of y= x2+5x−2

• y’ = 2x+5 +h

• What is the vertex of y= x2+5x−2?

• Let h=0

y’ = 2x + 5

• Set DQ to 0 and solve

2x + 5 = 0 → x = −2.5

• Use x to find y from original

x = −2.5 → y = (−2.5)2 + 5(−2.5) − 2 = − 8.25

Vertex is (−2.5, −8.25)

• Theorem: All polynomials are continuous

• This is not a polynomial

• Zero Product Property:

If a*b*c=0 then a=0, b=0, or c=0

• What does this mean for polynomials….

• If p(x)=x(x+2)(x-5)=0 then x=0, x+2=0, or x−5=0

• So 0, −2, and 5 are zeroes of the polynomial.

• Find a cubic polynomial which has zeroes 2, 3, -1

• Reflection: Is this the ONLY cubic with those zeroes?

• No there are many cubics with these zeroes

How many points does it take…To find the equation of an nth degree polynomial?

• How many points to find a line?

• 2 points – Point Slope Equation

• How many points to find a quadratic?

• 3 points – Simultaneous Equations

• Can any 3 points be used to find a quadratic?

• No, you can find a quadratic with any 3 non-collinear points

• How many points in general to find an nth degree polynomial?

• n+1 points

• Find Quadratic through (-1,19) (0,12) (3,3)

• General Form:

Example: Find Quadratic through (-1,25) (0,17) (2,7)

• Plug in Points:

• Solve System

Viete’s Formulae

• Polynomial Patterns

• Given

• Find Equation of Cubic with zeroes of 4, 2, -3

• General Form:

• If p(x) is continuous and if p(a) is positive and p(b) is negative then p(x) has a zero on the interval (a,b)

• Establish function is continuous

• Show that for point a that p(a) is positive

• Show that for point b that p(b) is negative

• Say by Intermediate Value Theorem, p(x) must have a zero on the interval (a,b)

• Note: IMVT only establishes existence, not value

• Given: Show that the function p(x)=x2+5x−2 has a zero between 0 and 1.

You Write:

• p(x) is a polynomial and must be continuous

• p(0)= −2 and p(1) = 5

• By the IMVT, p(x) must have a zero on the interval (0,1)

• Extension: use Quadratic Equation to find that exact zero!

• The quartic function

has four total roots (2 double roots).

What is p+q?

• Pg 150 #86,88

• Pg 152 #102

• Supplement on Web