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I can graph a rational function.
8-3 Rational Functions Unit Objectives: Graph a rational function Simplify rational expressions. Solve a rational functions Apply rational functions to real-world problems Todayβs Objective: I can graph a rational function.
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π π₯ = π(π₯) π(π₯) Rational Function: π π₯ and π π₯ are polynomials π¦= π₯ 2 π₯ 2 +1 π¦= π₯+2 π₯ 2 β4 Hole Asymptote Continuous Graph: No breaks in graph Discontinuous Graph: Breaks in graph
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Where the Denominator = zero Domain:
Discontinuities: Where the Denominator = zero Domain: All real numbers (β) except discontinuities Holes: Removable Same factor in numerator and denominator Vertical Asymptotes: Non-removable π¦= π₯+2 (π₯+2)(π₯β3) π¦= π₯+2 π₯ 2 βπ₯β6 = 1 (π₯β3) π¦= π₯+1 (π₯+1)(π₯+3) = 1 (π₯+3) Discontinuity: π₯=β1 or π₯=β3 π₯=β2 or π₯=3 Domain: All reals but π₯β β2, 3 All reals but π₯β β1,β3 1 2 β 1 5 Holes: π¦= π₯=β2 π¦= π₯=β1 V. Asymp: π₯=β3 π₯=3
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Horizontal Asymptotes: ππ₯ π ππ₯ π
ππ₯ π ππ₯ π Leading term of numerator and denominator (standard form) π<π π>π π=π No horizontal asymptote π¦=0 π¦= π π π¦= π₯+1 π₯ 2 β4 π¦= π₯ 3 +6 π₯+3 π¦= 3 π₯ π₯ 2 +5 1 1 ? 2 < 2 ? 2 = 3 ? 1 > π¦= 3 1 π¦=0 No horizontal asymptote =3 Range: All real numbers (β) except horizontal asymptote & holes
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Find and graph asymptotes & holes
Find and graph additional points β each side of v. asymptote Sketch graph Discontinuities: π₯=3 Hole: None V. Asymp.: π₯=3 Additional Points x y H. Asymp.: 2(0) 0β3 =0 β except π₯β 3 2 1 Domain: Range: π¦= =2 β except π¦β 2 2(4) 4β3 4 =8
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1 Graph: π¦= π₯β1 π₯ 2 β1 = π₯β1 (π₯β1)(π₯+1) Discontinuities: π₯=1 π₯=β1
= π₯β (π₯β1)(π₯+1) Discontinuities: Additional Points x y π₯=1 π₯=β1 Hole: π₯=1 1 β2 +1 β2 1 2 =β1 π¦= V. Asymp.: 1 0+1 =1 π₯=β1 H. Asymp.: Domain: Range: β except π₯β Β±1 β except π¦β 0, 0.5 π¦=
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p.521: 13-31 odd = π₯(π₯+3) (π₯+2)(π₯+3) Graph: π¦= π₯ 2 +3π₯ π₯ 2 +5π₯+6
= π₯(π₯+3) (π₯+2)(π₯+3) Discontinuities: Additional Points x y π₯=β2 π₯=β3 Hole: π₯=β3 β4 β4+2 β4 =2 π¦= 3 V. Asymp.: =0 π₯=β2 H. Asymp.: p.521: odd Domain: Range: β except π₯β β2, β3 1 1 π¦= =1 β except π¦β 0, 3
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