Mod. 3 Day 4 Graphs of Functions

Relative Extremes on a Graph Relative Maximum: A “peak” point on a graph. The highest point in a particular section of a graph. Relative Minimum: A “valley” point on a graph. The lowest point in a particular section of a graph.

Examples of graphs Rel. Max Rel. Max Rel. Min Rel. Min Rel. Min

Multiplicity When a factor is repeated, it touches the graph multiple times at the same root. Example: x² is the same as (x-0)², so the function would have two roots where x = 0

Simplifying Radicals Examples: 72 80 45 200 75 216 6 2 4 5 3 5 10 2 6 2 80 4 5 45 3 5 200 10 2 75 5 3 216 6 6

Quadratic Formula x= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Identify 𝑎,𝑏,𝑐 and plug them in the formula Simplify and solve You will end up with 2 solutions. If you get the same solution twice, that is multiplicity. If you have a negative value under the radical, that is an imaginary number Examples: 2𝑥 2 +5𝑥−12 𝑥=−4, 3 2 𝑥 2 +4 𝑥=±2𝑖

Imaginary Numbers 𝑖= −1 𝑖 2 =−1 Example: 5𝑖∙2𝑖 Solution: −10 𝑖= −1 𝑖 2 =−1 Example: 5𝑖∙2𝑖 Solution: −10 Example: −3𝑖(4+5𝑖) Solution: −12𝑖+15

(𝑎 + 𝑏𝑖) and (𝑎 – 𝑏𝑖) are conjugate pairs. Conjugates (𝑎 + 𝑏𝑖) and (𝑎 – 𝑏𝑖) are conjugate pairs. Example: (4+2𝑖) and (4−2𝑖) are conjugates Multiply/FOIL: 16 −8𝑖 +8𝑖 −4 𝑖 2 16+4 = 20

Multiply Complex Numbers Given: 2+𝑖 , create the missing conjugate and multiply (2+𝑖)(2−𝑖) = 4 − 𝑖² 4 + 1 5

Where do conjugates come from? Find the solutions to the equation: x² - 4x + 15=0 Can’t factor to solve, therefore use quadratic formula Use quadratic equation: 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Solution: 2 ± 𝑖 11 So the solutions are conjugates! 2 + 𝑖 11 and 2 − 𝑖 11

Real vs. Imaginary Solutions:

Real vs. Imaginary Solutions: How many turns? 4 Degree? Solutions 2 real solutions 2 imaginary solutions

Creating Factors and Foiling Given the following zeros, create the polynomial: 1, -5, 6 Step 1: Create the factors (x-1)(x+5)(x-6) Step 2: Foil the first two ( 𝑥 2 +4𝑥−5) Step 3: Multiply by the last factor 𝑥 2 +4𝑥−5 x−6 𝑥 3 −2 𝑥 2 −29𝑥+30

Creating Factors and Foiling Given the following zeros, create the polynomial: 2, -3i Step 1: Create the missing conjugate 3i Step 2: Create the factors (x-2)(x+3i)(x-3i) Step 2: Foil the conjugates 𝑥+3𝑖 𝑥−3𝑖 𝑥 2 +9 Step 3: Multiply by the last factor 𝑥 2 +9 x−2 𝑥 3 −2 𝑥 2 +9𝑥−18

Creating Factors and Foiling Given the following zeros, create the polynomial: 7, 5i, -5i Step 1: Create the factors (x-7)(x-5i)(x+5i) Step 2: Foil the conjugates 𝑥−5𝑖 𝑥+5𝑖 𝑥 2 +25 Step 3: Multiply by the last factor 𝑥 2 +25 x−7 𝑥 3 −7 𝑥 2 +25𝑥−175

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Multiplying Conjugates To Create A Polynomial Equation: Example: (x + 3i)(x-3i) Foil the conjugates 𝑥 2 −3𝑥𝑖+3𝑥𝑖−9 𝑖 2 𝑥 2 −9 𝑖 2 Polynomial Equation: 𝑥 2 +9 Example: (3x+4i)(3x-4i) 9𝑥 2 −12𝑥𝑖+12𝑥𝑖−16 𝑖 2 9𝑥 2 −16 𝑖 2 Polynomial Equation : 9𝑥2+16

Even Degree End Behavior of a Graph Odd Degree Positive Coefficient: as x → -∞ f(x) → -∞ as x → ∞ f(x) → ∞ Negative Coefficient: as x → ∞ f(x) → -∞ Positive Coefficient: as x → -∞ f(x) → -∞ as x → ∞ f(x) → ∞ Negative Coefficient: as x → -∞ f(x) → ∞ as x → ∞ f(x) → -∞

Roots, Solutions, or Zeros of Functions A function has as many complex solutions equal to the degree of the function. A real root, solution or zero of a graph can be found where the function crosses the x-axis. An imaginary function can not be seen on a graph.

Multiply Complex Numbers (5+ 2𝑖 )(5− 2𝑖 )= Solution: 25 − 4 𝑖 2 = 25 – 2𝑖 (3+7 2𝑖 )( 3−7 2𝑖 )= Solution: 9 − 49 4 𝑖 2 = 9 – 98𝑖