## Friday, December 22, 2006

### Precalc 2.7 Graphs of Rational Functions

2.7 Graphs of Rational Functions

A. Guidelines for Graphing Rational Functions

Let f(x) = N(x)/D(x), where N(x) and D(x) are polynomials with no common factors.

1. Find and plot the y-intercept (if any) by evaluating f(0).

2. Set the numerator equal to zero and solve the equation N(x) = 0. The real solutions represent the x-intercepts of the graph. Plot these intercepts.

3. Set the denominator equal to zero and solve the equation D(x)=0. The real solutions represent the vertical asymptotes. Sketch these asymptotes using dashed vertical lines.

4. Find and sketch the horizontal asymptotes of the graph using a dashed horizontal line.

5. Plot at least one point between and one point beyond each x-intercept and vertical asymptote.

6. Use smooth curves to complete the graph between and beyond the vertical asymptotes.

7. Test for Symmetry (origin, x-axis, y-axis)

Example:

g(x) = (x2 + 1)/x

1. g(0) = (0 + 1)/0 = undefined so no y-intercept

2. x2 + 1 = 0

x2 = -1

x = plus and minus the square root of negative one

Therefore no real x-intercepts

3. D(x) = 0 so x=0 vertical asymptote

4. n = 2 and m = 1 so n>m, the graph has no horizontal asymptote.

5. points to plot (-3, -3.3), (-2, -2.5), (-1, -2), (0, error), (1,2), (2,2.5), (3,3.3)

6. Sketch the graph using steps #1 - 5

7. Test for symmetry (origin, x-axis, y-axis) - none

B. Slant Asymptote

If the degree of the numerator of a rational function is EXACTLY ONE MORE than the degree of the denominator, the graph of the function has a slant (or oblique) asymptote. (n = m + 1)

From last example:

g(x) = (x2 +1)/x

n = 2 and m = 1 so slant asymptote

divide them out using long division and you get “x + 1/x”

Dropping the remainder, you get the slant asymptote y=x.

Add this asymptote line to the graph.

Example 2:

f(x) = (2x2 - 5x + 5)/ (x2 - 2)

1. f(0) = -2.5 so (0, -2.5) is a y-intercept

2. 2x2 - 5x + 5 = 0

using the quadratic equation you have 5/4 + i √15 and 5/4 - i √15

So the roots are imaginary so there are no x-intercepts

3. D(x) = 0

x2 - 2 = 0

x2 = 2

x = √2 and x = - √2

4. n = 2 and m = 2 so n = m so (2/1) = 2 = y

So the horizontal asymptote is y = 2.

5. Finding points (-2, 11.5), (-1, -12), (0, -2.5), (1, -2), (2, 1.5)

6. Using these points and asymptotes, graph the polynomial function.

Application:

Page Design: you have a rectangular page with a width of x units and a height of y units. It has a margin of 1″ on both sides of the x length and 2″ on the y length.

The inner page then has a width of x - 1 - 1 or x - 2

and a height of y - 2 - 2 or y - 4. The page contains 30 square inches of print.

Therefore it has an area of A = xy.

The smaller inner rectangle has an area of 30 square inches of print.

30 = (x - 2)(y - 4)

30/(x-2) = y - 4

30/(x - 2) + 4 = y

(30 + 4(x-2)) / (x-2) = y

(30 + 4x - 8)/ (x - 2) = y

(4x + 22)/ (x - 2) = y

Recall: A = xy so plugging y in

A = x ((4x + 22)/(x-2))

What is the domain? Since the margins on the left and right are each 1 inch so x>2.

Sketching this:

You see that the Area is minimum when x = 5.87

Homework #23; pg. 204; #1,2,31-39 odd, 47-53 odd, 70, 75