1 / 32

# Curve Sketching - PowerPoint PPT Presentation

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’

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Curve Sketching

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Curve Sketching

TJ Krumins

Rebecca Stoddard

### 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!)

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

1) Find y’

2) Find y’’

3) Graph

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

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

• 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

• 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…nuh!)

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

1) Find y’

3) Graph

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

• 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…nuh!)

Y=x/(x-1)

1) Find y’

3) Graph

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

1st Derivative:

2nd Derivative:

### X-Intercepts and Asymptotes

• X-Intercept(s)

no x-intercepts

• Asymptotes

• Vertical:

• Horizontal:

### 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

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

### 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

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

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

• 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

(-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.

##  THE END 

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

### 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

• 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

• 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

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