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^33x^2+4 1) Find y’
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Curve Sketching
TJ Krumins
Rebecca Stoddard
Y=x^33x^2+4
1) Find y’
2) Find y’’
3) Graph
Look! No asymptotes!
+  +
x2______
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
(4,0) is a max because it is increasing and then decreasing
(2,0) is a min because it is decreasing then increasing
 +
(x1)__________
_________________f”(x)
1
Therefore, it is concave down when x<1 and concave up when x>1
F(1)=2
Y=(x+3)/(x2)
1) Find y’
3) Graph
Look! Vertical asymptotes at x=2
Horizontal at y=1
X intercept at x=3
 
5
x2__________
x2__________
____________________F’(x)
2
Therefore, it is decreasing for all real numbers
Y=x/(x1)
1) Find y’
3) Graph
Look! Vertical asymptotes at x=1
Horizontal at y=1
X intercept at x=0
 
1
x1__________
x1__________
____________________F’(x)
1
1st Derivative:
2nd Derivative:
no xintercepts
Sign Line of 1st Derivative (Increasing/Decreasing)
Sign Line of 2nd Derivative (Concavity)

+

+
66
x2+3
x+3
x+3
x+3
x3
x3
x3


+
+
22x
x+3
x+3
x3
x3
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
x3
x3
x3


+
+
22x
x+3
x+3
x3
x3
Y’
Y’’
3
0
3
3
0
3
Critical Points: x=3, x=0, x=3
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 xyplane provided, draw a sketch that shows the general shape of the graph of the function f on the open interval 0<x<2.
Found in intervals above xaxisInflection 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
X=2….Maximum
Xintercepts 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 xaxis)
(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+x6
(X+3)(x2)
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
Parts of a sign line
Show if the function is negative or positive at this point
 
5
x2__________
x2__________
____________________F’(x)
2
The circles are because the factors are in the denominator
Factors
Critical Points