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| Polinomial |
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| Find the third degree polinomial passes through point (2,3), (3,5), (4,-7), and (0,1) Solution : suppose the polinomial equation : y=a*x^3+b*x^2+c*x+d passes through point(2,3) means : 3=a*2^3+b*2^2+c*2+d 3=8a+4b+2c+d [1] passes through point(3,5) means : 5=a*3^3+b*3^2+c*3+d 5=27a+9b+3c+d [2] passes through point(4,-7) means : -7=a*4^3+b*4^2+c*4+d -7=64a+16b+4c+d [3] pasess through point (0,1)means : 1=a*0+b*0+c*0+d 1=d [4] from 3*[1] and 2*[2] and [4] : 6=24a+12b+6c+3 10=54a+18b+6c+2 ------------- - 4=30a+6b+1 3=30a+6b [5] from 4*[2] and 3*[3] and 4 : 20=108a+36b+12c+4 -21=192a+48b+12c+3 --------------- - -41=84a+12b+1 -42=84a+12b -21=42a+6b [6] from [5] and [6] : 3=30a+6b -21=42a+6b ------------ - -24=12a a=-2 [7] substitute to [5] 3=30*(-2)+6b 3=-60+6b 6b=63 b=10.5 [8] substitute [4],[7] and [8] to [1] 3=8*(-2)+4*10.5+2c+1 3=-16+42+2c+1 2c=3-1+16-42 2c=-24 c=-12 [9] from [4],[7],[8], and [9] : y=-2x^3+10.5x^2-12x+1 is the polinomial to be solved |
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