Question

5. The mapping f: xax² + bx + c defined on the set of real numbers is such that f(0) = -4,f(1)-1 and/(-1)= -5. Find a, b and c. 5 . The mapping f : xax² + bx + c defined on the set of real numbers is such that f ( 0 ) = -4 , f ( 1 ) -1 and / ( - 1 ) = -5 . Find a , b and c .​

Answer 1

It looks like you're saying

$f(x) = ax^2 + bx + c$

and you're asked to find $a,b,c$ given $f(0)=-4$, $f(1)=-1$, and $f(-1)=-5$.

Evaluate $f$ at the three given points:

$x=0 \implies f(0) = \boxed{c = -4}$

$x=1 \implies f(1) = -1 = a + b + c \implies a+b = 3$

$x=-1 \implies f(-1) = -5 = a - b + c \implies a - b = -1$

$(a+b) + (a-b) = 3 + (-1) \implies 2a = 2 \implies \boxed{a=2}$

$a-b = -1 \implies \boxed{b=3}$

and the mapping is $f(x) = 2x^2 + 3x - 4$.