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Curve Sketching

Curve Sketching. TJ Krumins Rebecca Stoddard. Curve Sketching Breakdown. Find intercepts and asymptotes Take derivative Set up sign line Find critical points Take 2 nd derivative Set up sign line Find critical points Graph. The problem (dun…nun…nuh!). Y=x^3-3x^2+4 1) Find y’

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Curve Sketching

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  1. Curve Sketching TJ Krumins Rebecca Stoddard

  2. Curve Sketching Breakdown • Find intercepts and asymptotes • Take derivative • Set up sign line • Find critical points • Take 2nd derivative • Set up sign line • Find critical points • Graph

  3. The problem (dun…nun…nuh!) Y=x^3-3x^2+4 1) Find y’ 2) Find y’’ 3) Graph

  4. Look! No asymptotes! • Step 1: First Derivative • Y=x^3-3x^2+4 • Y’=3x^2-6x • Step 2:Factor • Y’=3x(x-2) • Step 3: Sign Line + - + x-2------------------______ 3x---_____________ ___________________F’(x) 0 2 Therefore, it is increasing when x<0 and increasing when x>2, but decreasing from 0<x<2

  5. Don’t forget MAX and MIN x=0 and x=2 • Step 4:Plug into the function • F(0)=4 • F(2)=0 • Therefore! (4,0) and (2,0) are either a max or a min -(4,0) is a max because it is increasing and then decreasing -(2,0) is a min because it is decreasing then increasing

  6. Now repeat those steps for the second derivative • Step 5: Second derivative • Y’= 3x^2-6x • Y”= 6x-6 • Step 6: Factor • Y”=6(x-1) • Step 7: Sign Line for the second derivative - + (x-1)------------__________ _________________f”(x) 1 Therefore, it is concave down when x<1 and concave up when x>1

  7. Step 8: Plug in x=1 for an inflection point F(1)=2 • Step 9: What do we now know? • Max at (0,4) • Min at (2,0) • Inflection point at (2,1) • it is increasing when x<0 and increasing when x>2, but decreasing from 0<x<2 • it is concave down when x<1 and concave up when x>1

  8. The problem (dun…nun…nuh!) Y=(x+3)/(x-2) 1) Find y’ 3) Graph

  9. Look! Vertical asymptotes at x=2 Horizontal at y=1 X intercept at x=-3 • Step 1: First Derivative • Y=(x+3)/(x-2) • Y’=(x-2)(1)-(x-3)(1)/((x-2)^2) • ((x-2)-(x-3))/((x-2)(x-2)) • Step 2: Simplify • -5/(x-2)(x-2) • Step 3: Sign Line - - -5------------------------ x-2------__________ x-2------__________ ____________________F’(x) 2 Therefore, it is decreasing for all real numbers

  10. Step 4: What do we now know? • Look! Vertical asymptotes at x=2 • Horizontal at y=1 • X intercept at x=-3 • it is decreasing for all real numbers

  11. The problem (dun…nun…nuh!) Y=x/(x-1) 1) Find y’ 3) Graph

  12. Look! Vertical asymptotes at x=1 Horizontal at y=1 X intercept at x=0 • Step 1: First Derivative • Y=(x)/(x-1) • Y’=(x-1)(1)-(x)(1)/(x-1)^2 • -1/((x-1)(x-1)) • Step 2: Sign Line - - -1------------------------ x-1------__________ x-1------__________ ____________________F’(x) 1

  13. Sample Problem: 1st Derivative: 2nd Derivative:

  14. X-Intercepts and Asymptotes • X-Intercept(s) no x-intercepts • Asymptotes • Vertical: • Horizontal:

  15. Sign Lines Sign Line of 1st Derivative (Increasing/Decreasing) Sign Line of 2nd Derivative (Concavity) - + - + 66------------------- x2+3 x+3--------- x+3--------- x+3--------- x-3--------------------------- x-3--------------------------- x-3--------------------------- - - + + -22x--------------- x+3--------- x+3--------- x-3------------------------- x-3------------------------- Y’ Y’’ -3 0 3 -3 0 3

  16. Reading Sign Lines Open circle because the critical point is either a hole or asymptote Closed circle because the critical point is not a hole or asymptote Minus sign “-” means that the function is decreasing - + - + Plus sign “+” means that the function is increasing 66------------------- x2+3 x+3--------- x+3--------- x+3--------- x-3--------------------------- x-3--------------------------- x-3--------------------------- - - + + -22x--------------- x+3--------- x+3--------- x-3------------------------- x-3------------------------- Y’ Y’’ -3 0 3 -3 0 3 Critical Points: x=-3, x=0, x=3

  17. Graph of

  18. 1996 AB 1 FRQ The figure above shows the graph of f’(x), the derivative of a function f. The domain of f is the set of all real numbers x such that -3<x<5. a]For what values of x does f have a relative maximum? Why? b]For what values of x does f have a relative minimum? Why? c]On what intervals is the graph of f concave upward? Use f’ to justify your answer. d]Suppose that f(1)=0. In the xy-plane provided, draw a sketch that shows the general shape of the graph of the function f on the open interval 0<x<2.

  19. Basic Info From Graph Found in intervals above x-axis Inflection Points of derivative [-3,-2) increasing x=-2…maximum (-2,4) decreasing x=4….minimum (4,5) increasing Found in areas between max’s/min’s of derivative graph [-3,-1) concave down (-1,1) concave up (1,3) concave down (3,5] concave up

  20. Part A X=-2….Maximum X-intercepts of the derivative are max’s and min’s of the function and x=-2 is the point where the function changes from increasing to decreasing (which is seen through the intervals where the derivative is above or below the x-axis)

  21. Part B • X=4…minimum • X-intercepts of the derivative are max’s and min’s of the function. • x=4 is the point where the function changes from decreasing to increasing • This change from decreasing to increasing is a minimum because the graph dips down before rising up, similar to a U-shape

  22. Part C (-1,1) & (3,5] Max’s and min’s of the derivative are points of inflection of the function. A minimum in the function occurs in the interval (3,5] & concavity changes at each inflection point. Therefore, these intervals are concave up.

  23. Part D

  24.  THE END 

  25. Find where x=0 in the original function Do this by factoring (unless already factored) Y=x^2+x-6 (X+3)(x-2) X intercepts at x=-3 and x=2 Vertical asymptotes: where the denominator equals zero or where there is a negative under a radical Horizontal Asymptotes: Power on bottom is bigger y=0 Power on top is oblique Powers are equal: Ratio of the coefficients Intercepts and Asymptotes

  26. Oblique Asymptotes • Oblique is where power on the top is greater than the power on the bottom • To solve these use long division (divide the numerator by the denominator). • The answer will be a line (if done correctly) and will be the oblique asymptote

  27. Setting up a sign line • Draw a line and label it accordingly • List all factors on the left most column • List all critical points underneath the line • Label accordingly • For each critical point draw a circle • Draw open circles for the factors that are in the denominator • Draw closed circles for the factors that are in the numerator • Where x is positive in the factor draw ---- leading up to the circle, then a solid line following it. (do the opposite if it is negative) • For each interval, if the number of --- lines is even draw a + sign over that interval. • If the interval has a odd amount of ---- lines than draw a negative sign over the interval • These will tell you either if the graph is increasing/decreasing (first derivative) or if it is concave up/concave down (second derivative) Parts of a sign line

  28. Critical Points • Max and min: Found on the first derivative sign line. Once the x is found plug back into the original function to find the y value • X-intercepts: See intercepts slide • Vertical asymptotes: See asymptotes slide • Inflection points: Found on the sign line of the second derivative. Once the x value is found plug back into the original function to find the y value.

  29. Parts of a Sign Line Show if the function is negative or positive at this point - - -5------------------------ x-2------__________ x-2------__________ ____________________F’(x) 2 The circles are because the factors are in the denominator Factors Critical Points

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