Lesson 4 5 Summary of Curve Sketching Objectives

Lesson 4 -5 Summary of Curve Sketching

Objectives • Sketch or graph a given function

Vocabulary • Oblique – neither horizontal nor vertical • Slant asymptote – a line (y = mx + b) that the curve approaches as x gets very large or very small; found if the limit of [f(x) – (mx+b)] = 0 as x approaches ±∞

Graphing Checklist Domain – for which values is f(x) defined? Division by 0 or negatives under even roots x -intercepts – where is f(x) = 0? y -intercepts – what is f(0)? Symmetry Even functions y-axis – is f(-x) = f(x)? Origin – is f(-x) = -f(x)? Odd functions Period – is there a number p such that f(x + p) = f(x)? Trig functions Asymptotes Horizontal – does or exist? Limit as x→±∞ Vertical – for what is ? Division by 0 (and not removed by canceling)

Graphing Checklist (cont) Derivative Information: Critical numbers – where does f’(x) = 0 or DNE? Increasing – on what intervals is f’(x) ≥ 0? Decreasing – on what intervals is f’(x) ≤ 0? Local extrema – what are the local max/min? Use f’ or f’’ test. Concavity Up – where is f’’(x) > 0? Down – where is f’’(x) < 0? Inflection points – where does f change concavity?

Example 1 1 Graph ------- x² – 4 Domain: f’(x) = -2 x/(x² - 4)² f’’(x) = 2(3 x² + 4)/(x² - 4)³ x ≠ ± 2 x –intercepts: None, y ≠ 0 y –intercepts: y = -1/4 Yes No No Symmetry: Y-axis: Origin: Periodic: x = -2, 2 y = 0 Asymptotes H: V: Critical numbers: x = 0 (- , -2) -2 (-2, 0) 0 (0, 2) 2 (2, ) f(x) + va - -¼ - va + f’(x) + + 0 - - f’’(x) + - - - + Increasing: x < 0 Decreasing: x > 0 Max/Min: At x = 0, y = -1/4 is a relative max Concavity Up: |x|>2 Down: |x| < 2

Example 1 Graph y x

Mistake in notes • • Example 2 Domain: (-∞, -1) (1, ∞) |x| > 1 no y-intercept possible f(2) = 0 At least one x-intercept at x = 2 Symmetric about y-axis f(-2) = 0 as well f(x) = f(-x) Lim f(x) = -∞ as x→ 1+ lim f(x) = 4 as x→ ∞ Vertical asymptotes at x = ± 1 Horizontal asymptote at y = 4 • • • f’(x) ≥ 0 on (1, 3] [5, ∞) f(x) increasing on these intervals f’(x) ≤ 0 on [3, 5] f(x) decreasing on this interval f’(3) = 0, local max at x = 3 Critical value and extrema location f’(5) = 0, local min at x = 5 Critical value and extrema location Y-values unknown f’’(x) > 0 on (4, 6) f’’(4) = 0 f’’(6) = 0 f’’(x) < 0 on (1, 4) (6, ∞) Concave up on this interval IP points here because concavity changes Concave down on these intervals

Example 2 Graph f(x) = ? ? ? Domain: |x| > 1 x –intercepts: x-intercepts at x = +/- 2 y –intercepts: none Yes No No Symmetry: Y-axis: Origin: Periodic: x = -1, 1 y = 4 Asymptotes H: V: Critical numbers: x = 3, 5 x = -3, -5 Increasing: (1, 3] [5, ∞) [-5, -3] Decreasing: [3, 5] [-3, -1) (-∞, -5] Max/Min: At x = +/- 3, relative max and at x = +/- 5, relative min Concavity Up: (4, 6) & (-6, -4) Down: (1, 4) (6, ∞) & (-∞, -6) (-4, -1)

Example 2 Graph y x Just one example of many possibilities that fit the facts

Summary & Homework • Summary: – Lots of information goes into a good graph – Function helps with extrema values, intercepts and its Limits find asymptotes – Derivatives find locations of possible extrema (at critical values) and inflection points (f’’(x)=0) – Derivatives help give shape to the curve • Concavity • Increasing / Decreasing intervals • Homework: – pg 323 -324: 3, 18, 37, 58