The function f is defined by f(x)=ax^(2)+bx+c, where a,b, and c are constants and 1

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The function f is defined by f(x)=ax^(2)+bx+c, where a,b, and c are constants and 1 < a < 4. The graph of y=f(x) in the xy-plane passes through points. (11,0) and (-2,0). If a is an integer, what could be the value of a+b ?

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Identify Roots of Quadratic: Since the graph of f(x) passes through the points (11,0) and (-2,0), we can say that x=11 and x=-2 are roots of the quadratic equation ax^2+bx+c=0. We can use these roots to create two equations based on the fact that f(11)=0 and f(-2)=0.Substitute Roots into Equation: Substituting x=11 into the equation gives us a(11)^2 + b(11) + c = 0. This simplifies to 121a + 11b + c = 0.Create System of Equations: Substituting x=-2 into the equation gives us a(-2)^2 + b(-2) + c = 0. This simplifies to 4a - 2b + c = 0.Eliminate Variable c: We now have a system of two equations with three unknowns: 1) 121a + 11b + c = 0 2) 4a - 2b + c = 0 We need one more equation to solve for a, b, and c. However, we are only asked to find the value of a+b, not the individual values of a, b, and c.Solve for b in terms of a: We can eliminate c by subtracting the second equation from the first equation: (121a + 11b + c) - (4a - 2b + c) = 0 - 0 This simplifies to 117a + 13b = 0.Find a+b: We can solve for b in terms of a by rearranging the equation: b = -117a/13 b = -9aDetermine Possible Values for a: Now we can find a+b by substituting b = -9a into the expression: a + b = a - 9a a + b = -8aCalculate Possible Values for a+b: Since 1 < a < 4 and a is an integer, the possible values for a are 2 and 3.Final Answer: If a = 2, then a + b = -8(2) = -16. If a = 3, then a + b = -8(3) = -24.Both -16 and -24 are possible values for a+b given the constraints on a. However, since the question asks for ＂what could be the value of a+b＂, we can provide any one of the possible values.

The function $f$ is defined by $f(x)=a x^{2}+b x+c$ , where $a, b$ , and $c$ are constants and \( 1

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