2.5 – Rational Functions. Ex. 1 Graph 5 x – 2 Ex. 1 Graph 5 x – 2.

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Presentation transcript:

2.5 – Rational Functions

Ex. 1 Graph 5 x – 2

Ex. 1 Graph 5 x – 2

Ex. 1 Graph 5 x – 2 x – 2 = 0

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2 (vertical asymptote)

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2 (asymptote)

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2 (asymptote)

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2 (asymptote) *Graph on Calc.

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2 (asymptote) *Graph on Calc. Type: y = 5/(x – 2)

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2 (asymptote) *Graph on Calc. Type: y = 5/(x – 2) 2 nd Table, 3pts on each curve

Ex. 1 Graph 5 x – 2 x – 2 = 0 x = 2 (asymptote) *Graph on Calc. Type: y = 5/(x – 2) 2 nd Table, 3pts on each curve

Ex. 2 Graph x + 1 x 2 + 3x + 2

Ex. 2 Graph x + 1 x 2 + 3x + 2 x + 1 (x + 2)(x + 1)

Ex. 2 Graph x + 1 x 2 + 3x + 2 x + 1 (x + 2)(x + 1) 1 x + 2

Ex. 2 Graph x + 1 x 2 + 3x + 2 x + 1 (x + 2)(x + 1) 1 x + 2 x = -2

Ex. 2 Graph x + 1 x 2 + 3x + 2 x + 1 (x + 2)(x + 1) 1 x + 2 x = -2 x = -1

Ex. 2 Graph x + 1 x 2 + 3x + 2 x + 1 (x + 2)(x + 1) 1 x + 2 x = -2 x = -1 Graph y = (x+1)/(x 2 +3x+2)

Ex. 2 Graph x + 1 x 2 + 3x + 2 x + 1 (x + 2)(x + 1) 1 x + 2 x = -2 x = -1 *Graph y = (x+1)/(x 2 +3x+2) *Then 2 nd Table for 3 ordered pairs per curve.

Ex. 2 Graph x + 1 x 2 + 3x + 2 x + 1 (x + 2)(x + 1) 1 x + 2 x = -2 x = -1 *Graph y = (x+1)/(x 2 +3x+2) *Then 2 nd Table for 3 ordered pairs per curve.

Ex. 3 Determine any asymptotes and intercepts. Then graph and find the domain. f(x) = 3x 2 – 3 x 2 – 9

Ex. 3 Determine any asymptotes and intercepts. Then graph and find the domain. f(x) = 3x 2 – 3 = 3(x + 1)(x – 1) x 2 – 9 (x + 3)(x – 3)

Ex. 3 Determine any asymptotes and intercepts. Then graph and find the domain. f(x) = 3x 2 – 3 = 3(x + 1)(x – 1) x 2 – 9 (x + 3)(x – 3) Vertical Asymptotes at x = -3 & x = 3

Ex. 3 Determine any asymptotes and intercepts. Then graph and find the domain. f(x) = 3x 2 – 3 = 3(x + 1)(x – 1) x 2 – 9 (x + 3)(x – 3) Vertical Asymptotes at x = -3 & x = 3 Graph, use Table to find limits for Horizontal.

Ex. 3 Determine any asymptotes and intercepts. Then graph and find the domain. f(x) = 3x 2 – 3 = 3(x + 1)(x – 1) x 2 – 9 (x + 3)(x – 3) Vertical Asymptotes at x = -3 & x = 3 Graph, use Table to find limits for Horizontal. Limits show Horizontal Asymptote at y = 3.

Domain: {x | x ≠ -3, x ≠ 3}

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2)

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2) =

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2) = 3p

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2) = 3p 2p – 4 = 3p

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2) = 3p 2p – 4 = 3p -2p -2p

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2) = 3p 2p – 4 = 3p -2p -2p -4

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2) = 3p 2p – 4 = 3p -2p -2p -4 =

Proportions: 1 fraction = 1 fraction Ex. 1 Solve each equation. a. p _ = 2 p – 2 3 2(p – 2) = 3p 2p – 4 = 3p -2p -2p -4 = p

b. w + w = 4w – 3 w – 1 w – 1

b. (w-1)w + w (w-1) = 4w – 3 (w-1) w – 1 w – 1

b. (w-1)w + w (w-1) = 4w – 3 (w-1) w – 1 w – 1 w + w(w – 1) = 4w – 3 w + w 2 – w = 4w - 3

b. (w-1)w + w (w-1) = 4w – 3 (w-1) w – 1 w – 1 w + w(w – 1) = 4w – 3 w + w 2 – w = 4w – 3 w 2 = 4w – 3

b. (w-1)w + w (w-1) = 4w – 3 (w-1) w – 1 w – 1 w + w(w – 1) = 4w – 3 w + w 2 – w = 4w – 3 w 2 = 4w – 3 w 2 – 4w + 3 = 0

b. (w-1)w + w (w-1) = 4w – 3 (w-1) w – 1 w – 1 w + w(w – 1) = 4w – 3 w + w 2 – w = 4w – 3 w 2 = 4w – 3 w 2 – 4w + 3 = 0 (w – 3)(w – 1) = 0

b. (w-1)w + w (w-1) = 4w – 3 (w-1) w – 1 w – 1 w + w(w – 1) = 4w – 3 w + w 2 – w = 4w – 3 w 2 = 4w – 3 w 2 – 4w + 3 = 0 (w – 3)(w – 1) = 0 w = 3, w = 1

b. (w-1)w + w (w-1) = 4w – 3 (w-1) w – 1 w – 1 w + w(w – 1) = 4w – 3 w + w 2 – w = 4w – 3 w 2 = 4w – 3 w 2 – 4w + 3 = 0 (w – 3)(w – 1) = 0 w = 3, w = 1