Find the quadratic function y=ax2+bx+c whose graph passes through the given points: (−1, −3), (3, 25), (−2, 5) [Hint: Substitute each point into y=ax2+bx+c to get a system of linear equations, then solve.]

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Find the quadratic function y=ax2+bx+c whose graph passes through the given points: (−1, −3), (3, 25), (−2, 5)   [Hint: Substitute each point into y=ax2+bx+c to get a system of linear equations, then solve.]

Ourst1977 · Accepted Answer

Step 1  Given:  (−1, −3), (3, 25), (−2, 5)  y=ax2+bx+c  Step 2  Substitute y=−3 and x=−1 as,  −3=a(−1)2+b(−1)+c  1) −3=a−b+c  Substitute y=25 and dx=3 as,  25=a(3)2+b(3)+c  2) 25=9a+3b+c  Substitute y=5 and x=−2 as,  5=a(−2)2+b(−2)+c  3) 5=4a−2b+c  Step 3  Solve the system of equation,  1) −3=a−b+c  2) 25=9a+3b+c  3) 5=4a−2b+c  Solve (1) for c as,  −3=a−b+c  c=−a+b−3  Substitute (−a+b−3) for c in (2) as,  25=9a+3b+(−a+b−3)  25=8a+4b±3  4) 8a+4b=28  Substitute (−a+b−3) for c in (3) as,  5=4a−2b+(−a+b−3)  5=3a−b−3  8=3a−b  5) 3a−b=8  Solve the equation (5) for b as,  3a−b=8  b=3a−8  Substitute 3a−8 for b in (4) as,  8a4(3a−8)=28  8a+12a−32=28  20a=60  a=3  Substitute 3 for a in (5) as,  3(3)−b=8  9−b=8  b=9−8  b=1  Substitute 3 for a and 1 for b in (1) as,  −3=a−b+c  −3=3−1+c  c=−3−2  c=−5  Hence, the value of

Find the quadratic function y = ax^{2} + bx +

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