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The parabola y=ax^2+bx-10 passes through the points (4.5,8) and (-2.5, 15). Determine the values of a and b

Sagot :

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[tex]y=ax^2+bx-10 \\ \\ (4.5,8) \\ 8=a \times 4.5^2 + b \times 4.5-10 \\ 8=20.25a+4.5b-10 \ \ \ |+10 \\ 18=20.25a+4.5b \\ \\ (-2.5,15) \\ 15=a \times (-2.5)^2+b \times (-2.5)-10 \\ 15=6.25a-2.5b-10 \ \ \ |+10 \\ 25=6.25a-2.5b[/tex]

The system of equations:
[tex]18=20.25a+4.5b \ \ \ |\times 5 \\ 25=6.25a-2.5b \ \ \ |\times 9 \\ \\ 90=101.25a+22.5b \\ \underline{225=56.25a-22.5b} \\ 90+225=101.25a+56.25a \\ 315=157.5a \ \ \ |\div 157.5 \\ a=2 \\ \\ 25=6.25a-2.5b \\ 25=6.25 \times 2-2.5b \\ 25=12.5 -2.5b \ \ \ |-12.5 \\ 12.5=-2.5b \ \ \ |\div (-2.5) \\ b=-5 \\ \\ \boxed{a=2} \\ \boxed{b=-5} \\ \boxed{y=2x^2-5x-10}[/tex]
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