Problem 39 - Algebra and Trigonometry

Question

Find the quadratic function \(f(x)=a x^{2}+b x+c\) for which \(f(-2)=-4, f(1)=2,\) and \(f(2)=0\)

Step-by-Step Solution

Answer

The quadratic function is \(f(x)= -x^{2} + x + 2\).

Step 1: Formulate the System of Equations

Using the given points, three equations can be formed by replacing x and f(x) in the function \(f(x)=a x^{2}+b x+c\):1) For the first point \(-2,-4\), we get \(a(-2)^2 + b(-2) + c = -4 \)2) For the second point \(1,2\), we get \(a(1)^2 + b(1) + c = 2\)3) For the third point \(2,0\), we get \(a(2)^2 + b(2) + c = 0\)

Step 2: Simplify the Equations

We simplify above equations to:1) \(4a - 2b + c = -4\)2) \(a + b + c = 2\)3) \(4a + 2b + c = 0\)

Step 3: Solve the System of Equations

Solve this system of equations using a methodology such as elimination or substitution. For simplicity, we will add first and third equation and substract the second equation from the result. This gives us a new equation that is: \(a= -1\). Now, we will substitute \(a= -1\) into the second and third equations to find b and c.

Step 4: Find a, b, and c

By substituting \(-1\) for \(a\) into the second equation, we find \(b=1\) and by substituting both \(-1\) for \(a\) and \(1\) for \(b\) into the third equation, we find \(c= 2\).

Step 5: Substitute a, b, and c into the Quadratic Function

Substitute \(a= -1\), \(b= 1\), and \(c= 2\) into the quadratic function \(f(x)= a x^{2}+b x+c\). This yields the function \(f(x)= -x^{2} + x + 2\) as the final output.