Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: Example 8.2.3. Find the quadratic function which passes through the points (−1, 3), (2, 4), (5, −2). Solution. According to Definition 2.5, a quadratic function has the form f (x) = ax2 + bx + c where a = 0. Our goal is to find a, b and c so that the three given points are on the graph of f . If (−1, 3) is on the graph of f , then f (−1) = 3, or a(−1)2 + b(−1) + c = 3 which reduces to a − b + c = 3, an honest-to-goodness linear equation with the variables a, b and c. Since the point (2, 4) is also on the graph of f , then f (2) = 4 which gives us the equation 4a + 2b + c = 4. Lastly, the point (5, −2) is on the graph of f gives us 25a + 5b + c = −2. Putting these together, we obtain a system of three linear equations. Encoding this into an augmented matrix produces a−b+c = 3 1 −1 1 3 Encode into the matrix 4a + 2 b + c = 4 −− − − − − −→ 4 2 1 4 −−−−−−− 25a + 5b + c = −2 25 5 1 −2 7 Using a calculator,4 we find a = − 18 , b = 13 and c = 37 . Hence, the one and only quadratic which 18 9 7 fits the bill is f (x) = − 18 x2 + 13 x + 37 . To verify this analytically, we see that f (−1) = 3, f (2) = 4, 18 9 and f (5) = −2. We can...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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