Given f(x) = $a x^{2} + b x + c$

If f(m) = a - b + c

f(n) = 4a + 2b + c

f(p) = a + b + c

Then, m + n + p = __

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Solution

$f (x) = a x^{2} + b x + c$

$f (- 1)$ = $a (- 1)^{2} + b (- 1) + c$ = a - b + c

$f (2)$ = $a (2)^{2} + b (2) + c$ = 4a + 2b + c

$f (1)$ = a + b + c

hence, m = -1, n = 2, p = 1

So, m+n+p = 2

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Given f(x) = ax2+bx+c If f(m) = a - b + c f(n) = 4a + 2b + c f(p) = a + b + c Then, m + n + p = __

f(x)=ax2+bx+c f(−1) = a(−1)2+b(−1)+c = a - b + c f(2) = a(2)2+b(2)+c = 4a + 2b + c f(1) = a + b + c hence, m = -1, n = 2, p = 1 So, m+n+p = 2

Given f(x) = $a x^{2} + b x + c$

If f(m) = a - b + c

f(n) = 4a + 2b + c

f(p) = a + b + c

Then, m + n + p = __

$f (x) = a x^{2} + b x + c$

$f (- 1)$ = $a (- 1)^{2} + b (- 1) + c$ = a - b + c

$f (2)$ = $a (2)^{2} + b (2) + c$ = 4a + 2b + c

$f (1)$ = a + b + c

hence, m = -1, n = 2, p = 1

So, m+n+p = 2