1. 2.4 Curve Sketching (Conclusion)

2. After identifying relative maximum and minimum points, concavity, and points of inflection, examine other characteristics of a graph. Graph intercepts Check for asymptotes

3. Page 170 of our textbook provides the following summary of curve sketching techniques. 1.Compute f’(x) and f’’(x). 2.Find all relative extreme points. a.Set f’(x) = 0 and solve for x. Suppose that x = a is a solution. Substitute x = a into f(x) to find f(a), plot the point (a, f(a)), and draw a small horizontal tangent line through the point. Compute f’’(a). i.If f’’(a)>0, draw a small concave up arc with (a, f(a)) as its lowest point. The curve has a relative minimum at x = a.

4. ii.If f’’(a)<0, draw a small concave down arc with (a, f(a)) as its peak. The curve has a relative maximum at x = a. iii.If f’’(a) = 0, examine f’(x) to the left and right of x = a in order to determine if the function changes from increasing to decreasing or vice versa. If a relative extreme point is indicated, draw the appropriate arc as in parts i and ii. b.Repeat the preceding steps for each of the solutions to f’(x) = 0.

5. 3.Find all the inflection points of f(x). a.Set f’’(x) = 0 and solve for x. Suppose that x = b is a solution. Compute f(b) and plot the point (b, f(b)). b.Test the concavity to the right and left of b. If the concavity changes at x = b, then (b, f(b)) is an inflection point. 4.Consider other properties of the function and complete the sketch. a.If f(x) is defined at x = 0, the y-intercept is (0, f(0)).

6. b.Does the partial sketch suggest that there are x-intercepts? If so, they are found by setting f(x) = 0 and solving for x. (Solve only in easy cases or when a problem essentially requires you to calculate the x-intercepts.) c.Observe where f(x) is defined. Sometimes the function is given only for restricted values of x. Sometimes the formula for f(x) is meaningless for certain values of x. d.Look for possible asymptotes.

7. i.Examine the formula for f(x). If some terms become insignificant as x gets large and if the rest of the formula gives the equation of a straight line, then that straight line is an asymptote. ii.Suppose that there is some point a such that f(x) is defined for x near a but not at a (e.g., 1/x at x = 0). If x gets arbitrarily large (in the positive or negative sense) as x approaches a, then the vertical line x = a is an asymptote for the graph. c.Complete the sketch.

8. Find the x-intercepts of y = 4 – 2x – x 2. Set y = 0, 4 – 2x – x 2 = 0 x 2 + 2x – 4 = 0 a = 1, b = 2, c = -4

9. y = 4 – 2x – x2

10. Graph f(x) = x 3 – 6x 2 + 12x - 5 1. Find the relative extreme points…f’(x) = 0

11. Potential relative extreme point at (2, f(2) = 3)

12. 2. Check concavity at x = 2

13. Second derivative does not tell us if we have a relative max (concave down) or min (concave up) at x = 2. Now what do we do? Use first derivative to look at slopes near x = 2 to the left and right. Pick x values 1 to the left and 3 to the right of x = 2. The slope is 3 (positive) on both sides of x = 2. So, x = 2 cannot be a relative extreme point.

14. 3. Find inflection points…f’’(x) = 0 There is a potential point of inflection at x = 2. How can we confirm it?

15. Check concavity at points near x = 2 to the left and the right. Pick x = 1 to the left and x = 3 to the right. Concavity differs on the left and right sides of x = 2, so (2, f(2) = 3) is a point of inflection.

16. 4. Find y-intercept and maybe x-intercepts Set x = 0 to find y-intercept. f(x) = x 3 – 6x 2 + 12x - 5 f(0) = 0 3 – 6(0) 2 + 12(0) – 5 = -5 There is a y-intercept at (0, -5). Do we find x-intercepts? -Are they required? -Are they easy to compute for the cubic function?

17. 5. What about asymptotes? Vertical asymptotes can appear where points are omitted from the function’s domain because they would cause division by zero. Function f(x) = x 3 – 6x 2 + 12x – 5 is a polynomial. Domain is all real numbers. No vertical asymptotes.

18. Horizontal asymptotes appear if either of the following limits exist. Neither limit exists…no horizontal asymptotes.

19. f(x) = x 3 – 6x 2 + 12x - 5 (2,3) Inflection point (0,-5) y-intercept