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Review Preview What is the degree of the following Polynomials?

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Presentation on theme: "Review Preview What is the degree of the following Polynomials?"β€” Presentation transcript:

1 Review Preview What is the degree of the following Polynomials?
𝑓 π‘₯ = 2π‘₯ 2 βˆ’4 π‘₯ 6 +8π‘₯ 𝑓 π‘₯ = βˆ’4π‘₯ 3 +2π‘₯βˆ’3 List each zero and its multiplicity. 3. y=(π‘₯+2) π‘₯βˆ’3 2 (π‘₯βˆ’4) 4. 𝑓 π‘₯ = π‘₯+7 3 (π‘₯βˆ’2)

2 U5D2 Polynomial functions
January 4th, 2016

3 Turning points Turning points are the points on the graph where the graph changes from increasing to decreasing, and vice versa. Example: Graph 𝒙 πŸ‘ βˆ’πŸ 𝒙 𝟐 to determine its turning points *** A polynomial of degree n has at most n-1 turning points. Similarly, if the graph has n-1 turning points, the polynomial is at least n. Example: how many turning points will a cubic function have? A quartic? If a polynomial has 5 turning points, it must be of what degree?

4 How many turning points will the graph 𝑓 π‘₯ =3(π‘₯βˆ’7) (π‘₯+3) 2 have?
You try! How many turning points will the graph 𝑓 π‘₯ =3(π‘₯βˆ’7) (π‘₯+3) 2 have?

5 Example: What is the end behavior of 𝒇 𝒙 = 𝒙 𝟐 π’™βˆ’πŸ ?
Where is the graph β€œtraveling” to? Definition: For large values of x, either positive or negative, the graph of the polynomial resembles the graph of the power function. Example: What is the end behavior of 𝒇 𝒙 = 𝒙 𝟐 π’™βˆ’πŸ ?

6 Graphing Polynomial Functions
Graph 𝑓 π‘₯ = 2π‘₯+1 π‘₯βˆ’3 2 . What are the zeros? What is the domain and Range? What are the Intervals of Increase And Decrease? What is the End Behavior?

7 Graphing Polynomial Functions
Graph 𝑓 π‘₯ = π‘₯ 2 (π‘₯βˆ’2)(π‘₯+2). What are the zeros? What is the domain and Range? What are the Intervals of Increase And Decrease? What is the end behavior?

8 Classwork: Pg. 189: #s Graph, List the zeros, domain and range, intervals of increase and decrease, and end Behavior.

9 homework Graph, find the zeros, state the domain and range and intervals of increase and decrease, and the end behavior 𝑓 π‘₯ = π‘₯+1 π‘₯βˆ’2 π‘₯+4 𝑓 π‘₯ = π‘₯ 2 π‘₯βˆ’3 (π‘₯+4)


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