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1 . 4 x 9 x g(x) 2 2 Example 2 Sketch the graph of the function Solution Observe that g is an even function, and hence its graph is symmetric with respect to the y- axis. I. Intercepts The x-intercepts occur when 0 = x 2 -9 = (x+3)(x-3), i.e. when x=-3 or x=3. The y-intercept occurs at II. Asymptotes Vertical asymptotes occur where the denominator of g(x) is zero: 0 = x 2 -4 = (x+2)(x-2), i.e. when x=-2 and x=2. The horizontal asymptote on the right is given by: The horizontal asymptote on the left is given by: Thus g has the horizontal asymptote y=1 on both the left and the right. . ) ( 4 9 4 9 0 g y . 1 4 9 2 2 lim x x x . 1 4 9 2 2 lim x x x

1 Example 2 Sketch the graph of the function Solution Observe that g is an even function, and hence its graph is symmetric with respect to the y-axis

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Page 1: 1 Example 2 Sketch the graph of the function Solution Observe that g is an even function, and hence its graph is symmetric with respect to the y-axis

1

.4x9x

g(x) 2

2

Example 2 Sketch the graph of the function

Solution Observe that g is an even function, and hence its graph is symmetric with respect to the y-axis.

I. Intercepts

The x-intercepts occur when 0 = x2-9 = (x+3)(x-3), i.e. when x=-3 or x=3.

The y-intercept occurs at

II. Asymptotes Vertical asymptotes occur where the denominator of g(x) is zero: 0 = x2-4 = (x+2)(x-2), i.e. when x=-2 and x=2.

The horizontal asymptote on the right is given by:

The horizontal asymptote on the left is given by:

Thus g has the horizontal asymptote y=1 on both the left and the right.

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49

0gy

.14

92

2

lim

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x

.14

92

2

lim

x

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x

Page 2: 1 Example 2 Sketch the graph of the function Solution Observe that g is an even function, and hence its graph is symmetric with respect to the y-axis

2

III. First Derivative

By the quotient rule, the derivative of is:

Since the denominator of g /(x) is always positive, g /(x) has the same sign as its numerator 10x. Hence g /(x) is positive for x>0 while g /(x) is negative for x<0. Thus g is increasing for x>0 while g is decreasing for x<0. We depict this information on a number line.

g only has one critical point: x=0 where g /(x) = 0 because the numerator of g /(x) vanishes. By the First Derivative Test, x=0 is a local minimum. Note that x=-2 and x=2 are not critical points of g because g has vertical asymptotes at these numbers and they are not in the domain of g.

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decr -2 decr 0 i n cr 2 i n crl oca lm i n

si gn g ’(x )x

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4x9x

g(x) 2

2

Page 3: 1 Example 2 Sketch the graph of the function Solution Observe that g is an even function, and hence its graph is symmetric with respect to the y-axis

3

IV. Vertical Tangents and Cusps

g has neither vertical tangents nor cusps.

V. Concavity and Inflection Points

By the quotient rule, the derivative of is:

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Observe that the numerator of g //(x) is always negative. Hence g //(x) is positive for –2<x<2 and negative for x<-2 or x>2. Therefore the graph of g is concave up for –2<x<2 and concave down for x<-2 or x>2. Note that g has vertical asymptotes x=-2 and x=2, and these numbers are not in the domain of g. Hence g has no inflection points.

con c down -2 con c u p 2 con c dow nx

si gn g“(x )- - - - - - + + + - - - - -

Page 4: 1 Example 2 Sketch the graph of the function Solution Observe that g is an even function, and hence its graph is symmetric with respect to the y-axis

4

VI. Sketch of the graph

We summarize our conclusions and sketch the graph of g.

x-intercepts: -3, +3 y-intercept: 9/4 vertical asymptotes: x=-2 and x=+2

horizontal asymptote: y=1 on the left and right g is an even function

increasing: 0<x decreasing: x<0 local min: x=0

concave up: -2<x<2 concave down: x<-2 and x>2 no inflection points

.4x9x

g(x) 2

2