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Using Derivatives for Curve Sketching
And
Finding Asymptotes
Thanks to Greg Kelly, Hanford High School, Richland, Washington
Photo by Vickie Kelly, 1995
Old Faithful Geyser, Yellowstone National Park
Using Derivatives for Curve Sketching
Curves have local Maximums and Minimums that are like Hills and Valleys
They may be like the wild terrain in his Picture !!
Photo by Vickie Kelly, 2007
Yellowstone Falls, Yellowstone National Park
Using Derivatives for Curve Sketching
Mammoth Hot Springs, Yellowstone National Park
Our Job is to use our mathematical tools to sketch curves
We have a formidable Arsenal of maths tools!!
In the past, one of the important uses of derivatives was as an aid in curve sketching. Even though we usually use a calculator or computer to draw complicated graphs, it is still important to understand the relationships between derivatives and graphs.
Curve Sketching Tools1. Inspection2. Intercepts and roots3. Gradients4. 2nd derivative5. Asymptote analysis6. Plus Some
First derivative:
y is positive Curve is rising.
y is negative Curve is falling.
y is zero Possible local maximum or minimum.
Second derivative:
y is positive Curve is concave up.
y is negative Curve is concave down.
y is zero Possible inflection point(where concavity changes).
f’(x)f’’(x)
f(x)=2x3-4x-4
f’(x)=0
f(x) decreasingas f’(x)<0
f(x) increasingas f’(x)>0
f ’’(x)<0Concave up
f ’’(x)>0Concave down
f ‘’(x)=0Pt Inflection
Example:Graph 23 23 4 1 2y x x x x
There are roots at and .1x 2x
23 6y x x
0ySet
20 3 6x x
20 2x x
0 2x x
0, 2x
First derivative test:
y0 2
0 0
21 3 1 6 1 3y negative
21 3 1 6 1 9y positive
23 3 3 6 3 9y positive
Possible extreme at .0, 2x
We can use a chart to organize our thoughts.
Example:Graph 23 23 4 1 2y x x x x
There are roots at and .1x 2x
23 6y x x
0ySet
20 3 6x x
20 2x x
0 2x x
0, 2x
First derivative test:
y0 2
0 0
maximum at 0x
minimum at 2x
Possible extreme at .0, 2x
Example:Graph 23 23 4 1 2y x x x x
23 6y x x First derivative test:
y0 2
0 0
NOTE: In any Exam, it is not sufficient to simply draw the chart and write the answer. You must give a written explanation!
There is a local maximum at (0,4) because for all x in and for all x in (0,2) .
0y( ,0) 0y
There is a local minimum at (2,0) because for all x in(0,2) and for all x in .
0y(2, )0y
Because the second derivative atx = 0 is negative, the graph is concave down and therefore (0,4) is a local maximum.
Example:Graph 23 23 4 1 2y x x x x
There are roots at and .1x 2x
23 6y x x Possible extreme at .0, 2x
Or you could use the second derivative test:
6 6y x
0 6 0 6 6y
2 6 2 6 6y Because the second derivative atx = 2 is positive, the graph is concave up and therefore (2,0) is a local minimum.
inflection point at 1x There is an inflection point at x = 1 because the second derivative changes from negative to positive.
Example:Graph 23 23 4 1 2y x x x x
6 6y x
We then look for inflection points by setting the second derivative equal to zero.
0 6 6x
6 6x
1 x
Possible inflection point at .1x
y1
0
0 6 0 6 6y negative
2 6 2 6 6y positive
Make a summary table:
x y y y
1 0 9 12 rising, concave down
0 4 0 6 local max
1 2 3 0 falling, inflection point
2 0 0 6 local min
3 4 9 12 rising, concave up
Finding Asymptotes
x f(x)
2 0.5
1 1
0.5 2
0.1 10
0.01 100
0.001 1000
x f(x)
-2 -0.5
-1 -1
-0.5 -2
-0.1 -10
-0.01 -100
-0.001 -1000
As x → 0–, f(x) → -∞.As x → 0+, f(x) → +∞.
A rational function is a function of the form f(x) = ,
where P(x) and Q(x) are polynomials and f(x) = 0.)(
)(
xQ
xP
f(x) =x
1
Example: f (x) = is defined for all real numbers except x = 0. x
1
-8 -6 -4 -2 2 4 6 8
-8
-6
-4
-2
2
4
6
8
x
y
As x → 0+, f(x)→ +∞.
As x → 0–, f(x)→ -∞.
x
x = a
as x → a –
f(x) → + ∞
x
x = a
as x → a –
f(x) → – ∞
x
x = a
as x → a +
f(x) → + ∞
x
x = a
as x → a +
f(x) → – ∞
The line x = a is a vertical asymptote of the graph of y = f(x), if and only if f(x) → + ∞ or f(x) → – ∞ as x → a + or as x → a –.
Example: Show that the line x = 2 is a vertical asymptote
of the graph of f(x) = .2)2(
4
x
x f(x)
1.5 16
1.9 400
1.99 40000
2 -
2.01 40000
2.1 400
2.5 16
x→2–,
f (x) → + ∞
x→2+,
f (x) → + ∞
This shows that x = 2 is a vertical asymptote.
y
x100
0.5x = 2
x→2– { f (x) → + ∞
{{x→2+
{f (x) → + ∞
Set the denominator equal to zero and solve. Solve the quadratic equation x2 + 4x – (x – 1)(x + 5) = 0
Therefore, x = 1 and x = -5 are the values of x for which f may have a vertical asymptote.
A rational function may have a vertical asymptote at
x = a for any value of a such that Q(a) = 0.)(
)(
xQ
xP
Example:Find the vertical asymptotes of the graph of f(x) = .)54(
12 xx
-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-8
-6
-4
-2
2
4
6
8
x
y
As x →1– , f(x) → – ∞
As x →1+, f(x) → + ∞
As x → -5–, f(x) → + ∞
As x →-5+, f(x) → – ∞
x = 1 is a vertical asymptote.
x = -5 is a vertical asymptote.
Example 1: Vertical Asymptote
1. Find the roots of the denominator. 0 = x2 – 4 = (x + 2)(x –2)
Possible vertical asymptotes are x = -2 and x = +2.
2. Since the (x + 2) will cancel from the numerator and denominator, x = -2 is not a VA. The table on the graphing calculator will show that as x → -2–, f(x) → -0.25 and as x → -2+, f(x) → -0.25; another reason why x = -2 is not a vertical asymptote.
x → +2–, f(x) → – ∞ and x →+2+, f(x) → + ∞. So, x = 2 is a vertical asymptote.
f is undefined at x = -2
A hole in the graph of f at (-2, -0.25) shows a removable singularity.
x = 2
Example: Find the vertical asymptotes of the graph of f(x) = .
)4(
)2(2
x
x
x
y
(-2, -0.25)
y
y = b
as x → + ∞ f(x) → b –
y
y = b
as x → – ∞ f(x) → b –
y
y = b
as x → + ∞ f(x) → b +
y
y = b
as x → – ∞ f(x) → b +
The line y = b is a horizontal asymptote of the graph of y = f(x) if and only if
x f(x)
10 0.1
100 0.01
1000 0.001
0 –-10 -0.1
-100 -0.01
-1000 -0.001
As x becomes unbounded positively, f(x) approaches zero from above; therefore, the line y = 0 is a horizontal asymptote of the graph of f. As f(x) → – ∞, x → 0 –.
Example: Show that the line y = 0 is a horizontal asymptote of the graph of the function f(x) = .
x
1
x
y
f(x) =x
1
y = 0
Nobody wants to plot points to determine a horizontal
asymptote!Luckily you can determine the horizontal asymptotes by remembering just 3 little things:
1. If the degree of the num degree of the denom, then y = 0 is a HA. 2. If the degree of the num degree of the denom, there is NO HA. 3. If the degree of the num degree of the denom, then the liney = LC of num is the HA. LC of denom
2
2
2( )
3 1
xf x
x
Example: has a HA at .
2
3y
y
x
The degree of the num = the degree of the denom
so the graph has a HA at y = 1.
Therefore, f(x) has y = 1 as a horizontal asymptote.
Example: Determine the horizontal asymptotes of the
graph of
f(x) = .)1( 2
2
x
x
y = 1
Example 1: Horizontal Asymptote
Finding Asymptotes for Rational Functions
• Check to factor and cancel.
• Vertical asymptotes occur at x values that make the denominator equal zero.
• If n > d, then there are no horizontal asymptotes.
• If n < d, then y = 0 is a horizontal asymptote.
• If n = d, then y = LCm is a horizontal asymptote. LCn
Given a rational function: f (x) = N(x) D(x)
Factor the numerator and denominator.
Since (x + 1) will cancel, there is a hole in the graph at x = -1 .
x = 2 will be a vertical asymptote.
Since the polynomials have the
same degree, y = 3 will be a
horizontal asymptote.
Example: Find all horizontal and vertical asymptotes of f (x) = . 2
3632
2
xx
xx
y = 3
x = 2
x
y
3( 1)( 1)
( 2)( 1)
x x
x x
Slant Asymptote
A slant asymptote occurs when the
degree of the numerator is exactly 1 more than the degree of the denominator.
A slant asymptote is found by using long division.
The slant asymptote is y = 2x – 5.
As x → + ∞, →
0+. 3
14
x
Example: Find the slant asymptote for f(x) = .3
12 2
x
xx
x
yx = -3
y = 2x - 5
Divide: 3
12 2
x
xx3
1452
xx
Therefore as x → ∞, f(x) is more like the line y = 2x – 5.
3
14
xAs x → – ∞, →
0–.
