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Analyzing Graph Behavior: Asymptotes, Intercepts, and Critical Points of a Function

This guide explores the critical features of the function f(x) = x^4 - 4x^3 + 2x^2 + 4x - 3, detailing its y-intercepts, x-intercepts, vertical and horizontal asymptotes, and regions of increase and decrease. It identifies critical points and examines local maxima and minima, along with inflection points using derivatives. The analysis concludes with sketching the graph of the function based on this information, providing a comprehensive understanding of its behavior across its domain.

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Analyzing Graph Behavior: Asymptotes, Intercepts, and Critical Points of a Function

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  1. A Model Solution and More

  2. Sketch the graph of y = Y- intercepts: For y-intercepts, set x = 0 The y-intercept is (0, -1) X- intercepts: For X-intercepts, set y = 0 The x-intercept is (-1,0) Asymptotes x – 1 = 0 gives a restriction of x = 1 is a vertical asymptote. and

  3. Asymptotes y = 1 is a horizontal asymptote. For critical points: For max/min points set y’ = 0 There are no critical points. But -2 ≠ 0

  4. Increasing/Decreasing Regions For increasing regions, y’>0 For decreasing regions, y’<0 < 0, for all x, x ≠1, the curve is always decreasing For Inflection Points: Check y” = 0 y” ≠ 0 for all x, x ≠ 1 there are no inflection points

  5. Given: f’(x) = x4-4x3+2x2+4x-3 f”(x) = 4x3 – 12x2 + 4x + 4 For Critical Points: Set f’(x) = 0, using the factor theorem f’(x) = (x+1)(x3-5x2+7x-3) = (x+1)(x-1)(x2 - 4x+3) = (x+1)(x-1)2(x-3) there are critical points at x = 1, -1, 3 For Max/Min: examine sign of f’(x) near the critical points -1 1 3 _ _ Sign of f’(x) + + There is a local max. at (-1,10) since y’ > 0 for all x in (-∞,-1) and y’ < 0 for all x in (-1,1). There is an Inflection pt. at (1,6) since y’ < 0 for all x in (-1,1) and y’ < 0 for all x in (1,3). There is a local min. at (3,1.5) since y’ < 0 for all x in (1,3) and y’ > 0 for all x in (3,∞) .

  6. Concavity using f ’’(x) f ”(-1) = -16, since f ”(x) < 0, therefore a local max f ”(1) = 0, since f ”(x) = 0, therefore not concave, suspect an inflection point –> check signs: since f ” > 0 for all x in (-1,1) and f ” < 0 for all x in (1,3) f ”(3) = 16, since f ”(x) > 0, therefore a local min There are no vertical asymptotes For Horizontal asymptotes – since is the dominant term in f(x), the function will tend towards y = as the end behaviour.

  7. Sketch the Graph of y = f(x) given the following information:

  8. Sketch the Graph of y = f(x) given the following information:

  9. Sketch the Graph of y = f(x) given the following information:

  10. Sketch the Graph of y = f(x) given the following information:

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