Curve

Sketching TJ Krumins

Rebecca Stoddard

Curve Sketching BreakdownCurve Sketching Breakdown

1)1) Find Find interceptsintercepts and and asymptotesasymptotes2)2) Take derivativeTake derivative3)3) Set up Set up sign linesign line4)4) Find Find critical pointscritical points5)5) Take 2Take 2ndnd derivative derivative6)6) Set up sign lineSet up sign line7)7) Find critical pointsFind critical points8)8) Graph Graph

The problem (dun…nun…nuh!)

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

1) Find y’

2) Find y’’

3) Graph

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

Look! No asymptotes!

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 derivativeNow repeat those steps for the second derivative

Step 5: Second derivativeStep 5: Second derivative Y’= 3x^2-6x Y’= 3x^2-6x Y”= 6x-6Y”= 6x-6

Step 6: FactorStep 6: Factor Y”=6(x-1)Y”=6(x-1)

Step 7: Sign Line for the second derivativeStep 7: Sign Line for the second derivative - +- +(x-1)------------__________(x-1)------------__________ _________________f”(x)_________________f”(x) 11

Therefore, it is concave down when x<1 and concave up when x>1

• Step 8: Plug in x=1 for an inflection pointF(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

• 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

Look! Vertical asymptotes at x=2

Horizontal at y=1

X intercept at x=-3

• 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

• 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

Look! Vertical asymptotes at x=1

Horizontal at y=1

X intercept at x=0

Sample Problem:

1st Derivative: 2nd Derivative:

X-Intercepts and Asymptotes

• X-Intercept(s)

no x-intercepts

• Asymptotes– Vertical:

– Horizontal:

Sign Lines

Y’ Y’’

-22x---------------x+3---------x+3---------x-3-------------------------x-3-------------------------

-3 0 3

- + +-

66-------------------x2+3x+3---------x+3---------x+3---------x-3---------------------------x-3---------------------------x-3---------------------------

-3 0 3

- - ++

Sign Line of 1st Derivative (Increasing/Decreasing)

Sign Line of 2nd Derivative (Concavity)

Reading Sign Lines

Y’ Y’’

-22x---------------x+3---------x+3---------x-3-------------------------x-3-------------------------

-3 0 3

- + +-

66-------------------x2+3x+3---------x+3---------x+3---------x-3---------------------------x-3---------------------------x-3---------------------------

-3 0 3

- - ++Plus sign “+” means that the function is increasing

Minus sign “-” means that the function is decreasing

Closed circle because the critical point is not a hole or asymptote

Critical Points: x=-3, x=0, x=3

Open circle because the critical point is either a hole or asymptote

Graph of

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-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

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.

Part D

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

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

- -

-5------------------------

x-2------__________

x-2------______________________________F’(x)

2

Show if the function is negative or positive at this point

Factors Critical Points

The circles are because the factors are in the denominator