Curve Sketching
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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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Curve sketching

Curve Sketching

TJ Krumins

Rebecca Stoddard


Curve sketching breakdown

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


The problem dun nun nuh

The problem (dun…nun…nuh!)

Y=x^3-3x^2+4

1) Find y’

2) Find y’’

3) Graph


Curve sketching

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


Curve sketching

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


Now repeat those steps for the second derivative

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


Curve sketching

  • 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


The problem dun nun nuh1

The problem (dun…nun…nuh!)

Y=(x+3)/(x-2)

1) Find y’

3) Graph


Curve sketching

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


Curve sketching

  • 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


The problem dun nun nuh2

The problem (dun…nun…nuh!)

Y=x/(x-1)

1) Find y’

3) Graph


Curve sketching

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


Sample problem

Sample Problem:

1st Derivative:

2nd Derivative:


X intercepts and asymptotes

X-Intercepts and Asymptotes

  • X-Intercept(s)

    no x-intercepts

  • Asymptotes

    • Vertical:

    • Horizontal:


Sign lines

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


Reading sign lines

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


Graph of

Graph of


1996 ab 1 frq

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.


Basic info from graph

Basic Info From Graph

Found in intervals above x-axisInflection Points of derivative

[-3,-2) increasingx=-2…maximum

(-2,4) decreasingx=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


Part a

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)


Part b

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


Part c

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.


Part d

Part D


The end

 THE END 


Intercepts and asymptotes

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


Oblique asymptotes

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


Setting up a sign line

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


Critical points

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.


Parts of a sign line

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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