Chapter 6 - Polynomial Functions Algebra 2

Table of Contents Fundamental Theorem of Algebra Investigating Graphs of Polynomial Functions Transforming Polynomial Functions 6.8

6.6 - Fundamental Theorem of Algebra Algebra 2

You can use this information to write polynomial function when given in zeros. 6.6 Algebra 2 (bell work) Copy Notes Below

Write the simplest polynomial with roots –1,, and P(x) = (x + 1) ( x – ) (x – 4) 2 3 P(x) = (x 2 + x – )(x – 4) P(x) = x 3 – x 2 – 2x Example 1Writing Polynomial Functions Given Zeros

Write the simplest polynomial function with the given zeros. P(x) = (x – 0) ( x – ) (x – 3) 2 3 P(x) = x 3 – x 2 + 2x ,, P(x) = (x 2 – x)(x – 3)

Write the simplest polynomial function with the given zeros. P(x) = (x + 2)(x – 2)(x – 4) P(x) = (x 2 – 4)(x – 4) P(x) = x 3 – 4x 2 – 4x + 16 –2, 2, 4 6.6Optional

So a polynomial to the 3rd power can have 3 real roots, and has at least one complex Due to multiplicity (x-2) (x-2) (x-2) 6.6

Solve x 4 – 3x 3 + 5x 2 – 27x – 36 = 0 by finding all roots. The polynomial is of degree 4, so there are exactly four roots for the equation. Step 1 Use the rational Root Theorem to identify rational roots. ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36 Step 2 Graph y = x 4 – 3x 3 + 5x 2 – 27x – 36 to find the real roots. 6.6 Find the real roots at or near –1 and 4. Example 2Finding All Roots of a Polynomial Equation

1 –3 5 –27 –36–1–1 –1–1 1 –49–360 4–936 1 –4 9 – Step 4 Solve x = 0 to find the remaining roots. The polynomial factors into (x + 1)(x – 4)(x 2 + 9) = 0. x = 0 x 2 = –9 x = ±3i The fully factored form of the equation is (x + 1)(x – 4)(x + 3i)(x – 3i) = 0. The solutions are 4, –1, 3i, –3i. 6.6

Solve Each Equation by finding all roots. f(x) = y 4 – 16

Write the simplest function with zeros 2 + i,, and 1. Step 1 Identify all roots. By the Rational Root Theorem and the Complex Conjugate Root Theorem, the irrational roots and complex come in conjugate pairs. There are five roots: 2 + i, 2 – i,,, and 1. The polynomial must have degree 5. Step 2 Write the equation in factored form. P(x) = [x – (2 + i)][x – (2 – i)](x – )[(x – ( )](x – 1) 6.6Day 2Writing a Polynomial Function with Complex ZerosExample 3

P(x) = [x – (2 + i)][x – (2 – i)](x – )[(x – ( )](x – 1) 6.6 P(x) = x 5 – 5x 4 + 6x x 2 – 27x – 15

Math Joke Q: Why did the polynomial plant wilt? A: Its roots were imaginary. 6-6

The roots are 2i, -2i, 1 + 2, 1 – 2, and 3 P(x) = [ x - (2i)][x + (2i)][x - ( )][x - (1 - 2 )](x - 3) 6.6 P(x) = x 5 – 5x 4 + 9x 3 – 17x x + 12

HW pg – Day 1: 1-6, 18, 24, (Even) – Day 2: 7-9, 32, 33, 56 – Ch: 10, 23, 36, 37, 52-55

6.7 - Investigating Graphs of Polynomial Functions Algebra 2

6.7 Algebra 2 (bell work) Just Read

End behavior is a description of the values of the function as x approaches infinity (x +∞) or negative infinity (x –∞). 6.7

Identify the leading coefficient, degree, and end behavior. A. Q(x) = –x 4 + 6x 3 – x + 9 The leading coefficient is –1, which is negative. The degree is 4, which is even. As x –∞, P(x) –∞, and as x +∞, P(x) –∞. B. P(x) = 2x 5 + 6x 4 – x + 4 The leading coefficient is 2, which is positive. The degree is 5, which is odd. As x –∞, P(x) –∞, and as x +∞, P(x) +∞. 6.7Example 1Determining End Behavior of a Polynomial Function

Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient. As x –∞, P(x) +∞, and as x +∞, P(x) –∞. P(x) is of odd degree with a negative leading coefficient. 6.7

Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient. As x –∞, P(x) +∞, and as x +∞, P(x) +∞. P(x) is of even degree with a positive leading coefficient. 6.7

Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient. Odd Negative Even Positive 6.7

A turning point is where a graph changes from increasing to decreasing or from decreasing to increasing. A turning point corresponds to a local maximum or minimum. 6.7

Graph the function. f(x) = x 3 + 4x 2 + x – 6. Step 1 Identify the possible rational roots by using the Rational Root Theorem. ±1, ±2, ±3, ± – Step 2 Test all possible rational zeros until a zero is identified. Test x = 1. x = 1 is a zero, and f(x) = (x – 1)(x 2 + 5x + 6). 6.7Example 1Using Graphs to Analyze Polynomial Functions

Step 3 Write the equation in factored form. Factor: f(x) = (x – 1)(x + 2)(x + 3) Step 4 Plot other points as guidelines. The zeros are 1, –2, and –3. f(0) = –6, so the y-intercept is –6. Plot points between the zeros. Choose x = –, and x = –1 for simple calculations. 5 2 f( ) = 0.875, and f(–1) = –

x –∞, P(x) –∞, and as x +∞, P(x) +∞. Step 5 Identify end behavior. The degree is odd and the leading coefficient is positive so as Step 6 Sketch the graph of f(x) = x 3 + 4x 2 + x – 6 by using all of the information about f(x). 6.7

Math Joke Q: Why did the function f(x) = -x 2 finally get kicked out of class? A: In the end, its behavior was just too negative 6.7

Graph the function. f(x) = x 3 – 2x 2 – 5x + 6. Step 1 Identify the possible rational roots by using the Rational Root Theorem. ±1, ±2, ±3, ±6 1 –2 –5 6 – –3 8–2 1 –2 – –6–1 0–6 Step 2 Test all possible rational zeros until a zero is identified. Test x = –1.Test x = 1. x = 1 is a zero, and f(x) = (x – 1)(x 2 – x – 6). 6.7

Step 3 Write the equation in factored form. Factor: f(x) = (x – 1)(x + 2)(x – 3) Step 4 Plot other points as guidelines. The zeros are 1, –2, and 3. f(0) = 6, so the y-intercept is 6. Plot points between the zeros. Choose x = –1, and x = 2 for simple calculations. f(–1) = 8, and f(2) = –4. 6.7

x –∞, P(x) –∞, and as x +∞, P(x) +∞. Step 5 Identify end behavior. The degree is odd and the leading coefficient is positive so as Step 6 Sketch the graph of f(x) = x 3 – 2x 2 – 5x + 6 by using all of the information about f(x). 6.7

Graph the function. f(x) = –2x 2 – x + 6. Step 1 Identify the possible rational roots by using the Rational Root Theorem. ±1, ±2, ±3, ±6 –2 –1 6 –2 4 –6 3 0 Step 2 Test all possible rational zeros until a zero is identified. x = –2 is a zero, and f(x) = (x + 2)(–2x + 3). Test x = –2. Optional6.7

Step 3 The equation is in factored form. Factor: f(x) = (x + 2)(–2x + 3). Step 4 Plot other points as guidelines. f(0) = 6, so the y-intercept is 6. Plot points between the zeros. Choose x = –1, and x = 1 for simple calculations. The zeros are –2, and. 3 2 f(–1) = 5, and f(1) =

x –∞, P(x) –∞, and as x +∞, P(x) –∞. Step 5 Identify end behavior. The degree is even and the leading coefficient is negative so as Step 6 Sketch the graph of f(x) = –2x 2 – x + 6 by using all of the information about f(x). 6.7

HW pg – 2-13, 24, 25, 32-35, 43, 46, 47, – Ch: 14, 31, 42, 45, – Mark a point to the left/right and between all intercepts

6.8 - Transforming Polynomial Functions Algebra 2

Pg Algebra 2 (bell work)

For f(x) = x 3 – 6 Write the rule for each function and sketch its graph. g(x) = f(x) – 2 g(x) = (x 3 – 6) – 2 g(x) = x 3 – 8 To graph g(x) = f(x) – 2, translate the graph of f(x) 2 units down. This is a vertical translation. 6.8 Means find the polynomial Example 1Translating a Polynomial Function

For f(x) = x 3 – 6 Write the rule for each function and sketch its graph. h(x) = f(x + 3) h(x) = (x + 3) 3 – 6 To graph h(x) = f(x + 3), translate the graph 3 units to the left. This is a horizontal translation. 6.8 h(x) =x 3 +9x 2 +27x +12

Let f(x) = x 3 + 5x 2 – 8x + 1. Write a function g that performs each transformation. g(x) = –f(x) Reflect f(x) across the x-axis. g(x) = –(x 3 + 5x 2 – 8x + 1) g(x) = –x 3 – 5x 2 + 8x – 1 Check Graph both functions. The graph appears to be a reflection. 6.8Example 2Reflecting Polynomial Functions

Let f(x) = x 3 + 5x 2 – 8x + 1. Write a function g that performs each transformation. g(x) = f(–x) Reflect f(x) across the y-axis. g(x) = (–x) 3 + 5(–x) 2 – 8(–x) + 1 g(x) = –x 3 + 5x 2 + 8x + 1 Check Graph both functions. The graph appears to be a reflection. 6-8

Math Joke Q: Why didn’t the function recognize itself in the mirror? A: It had been completely transformed 6-8

Let f(x) = 2x 4 – 6x Graph f and g on the same coordinate plane. Describe g as a transformation of f. g(x) is a vertical compression of f(x). g(x) = f(x) 1 2 g(x) = (2x 4 – 6x 2 + 1) 1 2 g(x) = x 4 – 3x Example 3Compressing and Stretching Polynomial Functions

Let f(x) = 2x 4 – 6x Graph f and g on the same coordinate plane. Describe g as a transformation of f. g(x) is a horizontal stretch of f(x). g(x) = f( x) 1 3 g(x) = 2( x) 4 – 6( x) g(x) = x 4 – x

Write a function that transforms f(x) = 6x 3 – 3 in each of the following ways. Support your solution by using a graphing calculator. Compress vertically by a factor of, and shift 2 units right. 1 3 g(x) = f(x – 2) 1 3 g(x) = (6(x – 2) 3 – 3) 1 3 g(x) = 2(x – 2) 3 – 1 6-8Example 4Combining TransformationsDay 2

Write a function that transforms f(x) = 6x 3 – 3 in each of the following ways. Support your solution by using a graphing calculator. Reflect across the y-axis and shift 2 units down. g(x) = f(–x) – 2 g(x) = (6(–x) 3 – 3) – 2 g(x) = –6x 3 – 5 6-8

Write a function that transforms f(x) = 8x 3 – 2 in each of the following ways. Support your solution by using a graphing calculator. Compress vertically by a factor of, and move the x-intercept 3 units right. 1 2 g(x) = f(x – 3) 1 2 g(x) = (8(x – 3) 3 – g(x) = 4(x – 3) 3 – 1 g(x) = 4x 3 – 36x x – 1 6-8

Write a function that transforms f(x) = 6x 3 – 3 in each of the following ways. Support your solution by using a graphing calculator. Reflect across the x-axis and move the x-intercept 4 units left. g(x) = –f(x + 4) g(x) = –6(x + 4) 3 – 3 g(x) = –6x 3 – 72x 2 – 288x –

HW pg (Due Monday, ) – Day 1: 1-9, 14-18, 26, (Odd) – Day 2: 10-12, 22, 28, 29 – Use your calculator to confirm graphs – Use Pascal's for any binomial expansion If needed – Graph Original/New Graphs, or f(x)/g(x) on same graph