Rational Functions Part Two Mathematics 140 - Precalculus with trigonometry Fall 2016

Hints for successful graphing Rational Functions 0. Use your calculator to sketch the graph 1. Set x = 0. Plot the y-intercepts. 2. Find the zeros of function by setting N(x) = 0. Plot the x-intercepts. 3. Find the zeros of the denominator by setting D(x) = 0. Sketch the VA. 4. Find and sketch any HA by comparing the degree of N(x) and D(x). 5. Plot at least 1 point on either side of x-intercepts and VA. 6. Draw nice smooth curves and then say ahhhhhh…

Let’s examine the homework problems

For Review : The Analysis of f From the calculator we have the following graph The Analysis of f Domain : R − { 1 } Vertical Asymptote at x = 1 Let’s Find the Horizontal Asymptotes Using the TI-89 Calculator limit(y1(x), x, ∞) = 2 − limit(y1(x), x, −∞) = 2 + We have a HA at y = 22

For Review : (continued) From the calculator we have the following graph To find the Zeros set the Numerator function to 0. Solve 2x + 1 = 0. We have a zero at ( ½ , 0) To find the y-intercept find f(0). So the y-intercept is ( 1, 0) Range R − { 2 }

Example 1 : The Analysis of f From the calculator we have the following graph The Analysis of f Domain : R Vertical Asymptote None Let’s find the Horizontal Asymptotes Using the TI-89 Calculator limit(y1(x), x, ∞) = 0 + limit(y1(x), x, −∞) = 0 − You CAN cross the asymptote at numbers that are not far from x=0.

Example 1 : (continuing) From the calculator we have the following graph Zeros − Solve 5x = 0 gives That ( 0, 0) is the only zero To find the y-intercept find f(0). So the y-intercept is ( 0, 0) Using the calculator, we have a Min at (-.58, -1.44) and a Max At (.58, 1.44) Range (-1.44, 1.44)

Let’s talk about Asymptotes Vertical Asymptote → VA when D(x) = 0 Horizontal Asymptote →3 different possibilities for HA The degree of N(x) is less than that of D(x). The HA is at y=0 The degree of N(x) is equal to that of D(x). The HA is at y=a/b with a and b the lead coefficients of N(x) and D(x) respectively 3. The degree of N(x) is more than that of D(x). There is no HA The Calculator is a help with Horizontal Asymptotes

Example 2 : The Anaylsis of F From the calculator we have the following graph The Anaylsis of F Domain : R because 3x2 + 1 ≠ 0 Vertical Asymptote None Comparing the degree of numerator And denominator polynomial, we have that the Horizontal Asymptote is at y = ⅔ Using the TI-89 Calculator limit(y1(x), x, ∞) = ⅔ − limit(y1(x), x, −∞) = ⅔ − We have that the Horizontal Asymptote is at y = ⅔

Example 2 : (continuing) From the calculator we have the following graph Zeros − Solve 2x2 = 0 gives That ( 0, 0) is the only zero To find the y-intercept find f(0). So the y-intercept is ( 0, 0) Range [ 0, ⅔ )

Example 3 : The Analysis of F From the calculator we have the following graph Domain : R because 3x2 + 1 ≠ 0 (ever) Vertical Asymptote None Zeros − Solve 2x3 = 0 gives That ( 0, 0) is the only zero Range : R There are no HA! Use your calculator to verify However, there is a slant asymptote!!! The equation of the slant asymptote is equal to the quotient of the long division.

Facts about Slant Asymptotes The slant asymptote is present only if the degree of N is exactly 1 more than the degree of D. The equation of the slant asymptote is the quotient of the long division of D into N. Slant asymptote are often called oblique asymptotes.

Example 3 : Continued From the calculator we have the following graph The equation of the slant asymptote is equal to the quotient of the long division. So we have y = ⅔x for the slant asymptote.

Problem 1 :

Problem 2 :

Problem 3 :

Problem 4 : *: what is unique about this problem?

Problem 5 : Can you work backwards???