Find the quadratic function f left parenthesis x right parenthesis equals ax squared plus bx plus cf(x)=ax2+bx+c for which f left parenthesis 1 right parenthesis equals negative 3f(1)=−3, f left parenthesis negative 2 right parenthesis equals 39f(−2)=39, and f left parenthesis 2 right parenthesis equals 11f(2)=11.

f(x) = 7x^2 -7x -3 We have 3 known points and 3 unknowns. So let's set up a series of equations to solve. f(1) = -3; a(1)^2 + b(1) + c = -3 a + b + c = -3 f(-2) = 39; a(-2)^2 + b(-2) + c = 39 4a - 2b + c = 39 f(2) = 11; a(2)^2 + b(2) + c = 11 4a + 2b + c = 11 Giving us the 3 equations to solve: (1) a + b + c = -3 (2) 4a - 2b + c = 39 (3) 4a + 2b + c = 11 Let's subtract equation (2) from equation (3) (4a + 2b + c = 11) - (4a - 2b + c = 39) = (4b = -28) 4b = -28 b = -7 Now let's add equations (2) and (3) together, then solve for c (4a - 2b + c = 39) - (4a + 2b + c = 11) = (8a + 2c = 50) 8a + 2c = 50 4a + c = 25 c = 25 - 4a Let's use the calculated value for b, and the equation for c and substitute into equation (1) above to calculate c. a + b + c = -3 a + (-7) + (25 - 4a) = -3 a + (25 - 4a) = 4 -3a + 25 = 4 -3a = -21 a = 7 Finally, let's calculate c c = 25 - 4a c = 25 - 4*7 c = 25 - 28 c = -3 So the final desired equation is f(x) = 7x^2 -7x -3 Let's verify with the known points. x = 1 f(1) = 7(1)^2 - 7(1) - 3 f(1) = 7 - 7 - 3 f(1) = -3 x = -2 f(-2) = 7(-2)^2 - 7(-2) - 3 f(-2) = 7(4) + 14 - 3 f(-2) = 28 + 14 - 3 f(-2) = 39 x = 2 f(2) = 7(2^2) - 7(2) - 3 f(2) = 7(4) - 14 - 3 f(2) = 28 - 14 - 3 f(2) = 11 All three points are verified, so the function is f(x) = 7x^2 -7x -3

Ithink its 126 hope its