Let f (x) = $A x^{2} + B x + C$ where A,B,C, are real numbers Find A,B,C if $f (1) = 6, f (0) = 3, f (- 1) = 2$

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Solution

$f (x) = A x^{2} + B x + C$

$f (0) = A (0) + B (0) + C = 3$

$C = 3$

$f (1) = A (1) + B (1) + C = 6$

$A + B + C = 6$

$A + B = 3$ $\dots (1)$

$f (- 1) = A (1) + B (- 1) + C = 2$

$A - B = - 1$ $\dots (2)$

add $(1)$ and $(2)$

$2 A = 2$

$A = 1$

$1 + B = 3$

$B = 2$

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Let f (x) = Ax2+Bx+C where A,B,C, are real numbers Find A,B,C if f(1)=6,f(0)=3,f(−1)=2

f(x)=Ax2+Bx+Cf(0)=A(0)+B(0)+C=3C=3f(1)=A(1)+B(1)+C=6A+B+C=6A+B=3 ⋯(1)f(−1)=A(1)+B(−1)+C=2A−B=−1 ⋯(2)add (1) and (2)2A=2A=11+B=3B=2

Let f (x) = $A x^{2} + B x + C$ where A,B,C, are real numbers Find A,B,C if $f (1) = 6, f (0) = 3, f (- 1) = 2$

$f (x) = A x^{2} + B x + C$

$f (0) = A (0) + B (0) + C = 3$

$C = 3$

$f (1) = A (1) + B (1) + C = 6$

$A + B + C = 6$

$A + B = 3$ $\dots (1)$

$f (- 1) = A (1) + B (- 1) + C = 2$

$A - B = - 1$ $\dots (2)$

add $(1)$ and $(2)$

$2 A = 2$

$A = 1$

$1 + B = 3$

$B = 2$