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SFM Productions Presents: Another semi-chilly non-snow day in your Pre-Calculus journey! 2.6 Rational Functions Homework for section 2.6/2.7 p190 p201 # 21-29, 41-45, 57, 61, 65 #49, 51 (this is something we did at the beginning of the year) Rational Functions: functions expressed as a ratio. N x f (x) D x N for numerator D for denominator What is the Domain of a Rational Function? All x-values except those that make the denominator zero. As 1 f (x) x x f (x) 0 5 Y As x 4 f (x) 0 3 2 1 X -5 -4 -3 -2 -1 0 1 2 3 4 5 As x 0 -1 f (x) -2 -3 -4 -5 As x 0 f ( x ) 10 9 8 7 6 5 4 3 2 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 -2 -3 -4 -5 -6 -7 -8 -9 -10 Y 25 Y 20 15 10 5 X 1 2 3 4 5 6 7 8 9 10 X -25 -20 -15 -10 -5 0 5 10 15 20 25 -5 -10 -15 -20 -25 All these graphs are the same…it just looks like the graph disappears…in actuality, the graph keeps getting closer and closer and closer and closer and closer and closer to both the x and y axes. As 1 f (x) 2 x f (x) 0 5 Y As 4 2 1 X -4 -3 -2 -1 0 -1 -2 -3 -4 -5 x 0 f (x) 3 -5 x 1 2 3 4 5 NOTE: It is only when very far away from the origin that the graph approaches an asymptote… 5 x f (x) 3x2 1 5 Y 4 3 2 1 X -10 -8 -6 -4 -2 0 -1 -2 -3 -4 -5 2 4 6 8 10 You CAN cross the asymptote at numbers that are not far from x=0. 2x 1 f (x) x 1 10 Asymptotes do not have to be just on either the x or y axis… Y 9 8 7 6 5 4 3 2 1 X -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 1 2 3 4 5 6 7 8 9 10 But how are the asymptotes determined?...... Vertical Asymptote m VA when D(x) = 0 Horizontal Asymptote 3 different possibilities for HA The degree of N(x) is less than that of D(x). y=0 The degree of N(x) is equal to that of D(x). y=a/b The degree of N(x) is more than that of D(x). no HA 2x f (x) 3x2 1 5 Y 4 3 2 1 X -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 -3 -4 -5 HA is y = 0 2 x2 f (x) 3x2 1 5 Y 4 3 HA is: 2 y = 2/3 1 X -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 1 2 3 4 5 2 x3 f (x) 3x2 1 5 HA is: Non existent Y 4 3 However, there is a slant asymptote: 2 1 X -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 1 2 3 4 5 The equation of the slant asymptote is equal to the quotient of the long division. 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. So, what is the point of all this?...... It’s so you can sketch the graph of a rational function without the use of a calculator……muah, ha ha ha Rules of for graphing Rational Functions 1. Set x = 0. 2. Find the zeros of function by setting N(x) = 0. Plot the x-intercepts. Find the zeros of the denominator by setting D(x) = 0. Sketch the VA. Find and sketch any HA by comparing the degree of N(x) and D(x). 3. 4. 5. 6. Plot the y-intercepts. Plot at least 1 point on either side of x-intercepts and VA. Draw nice smooth curves and then say ahhhhhh… f (x) x2 x 2 2x 8 10 Y 8 6 4 2 X -10 -8 -6 -4 -2 0 -2 -4 -6 -8 -10 2 4 6 8 10 f (x) x2 2 x 3 10 Y 8 6 4 2 X -10 -8 -6 -4 -2 0 -2 -4 -6 -8 -10 2 4 6 8 10 5 f (x) x 4 x 2 x 12 10 Y 8 6 4 2 X -10 -8 -6 -4 -2 0 -2 -4 -6 -8 -10 2 4 6 8 10 x 2 16 f (x) x 4 10 *: what is unique about this problem? Y 8 6 4 2 X -10 -8 -6 -4 -2 0 -2 -4 -6 -8 -10 2 4 6 8 10 Can you work backwards??? 10 Y 8 6 4 2 X -10 -8 -6 -4 -2 0 -2 -4 -6 -8 -10 2 4 6 8 10 Go! Do!