The quadratic function y=ax2+bx+c whose graph passes through the points (1, 4), (2, 1) and (3, 4).

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Calculation: First, obtain the three system of three linear equation by substituting the coordinate points in y=ax2+bx+c y=ax2+bx+c(1) Consider the first coordinate point (1, 4). Here, x=1andy=4 Substitute the values of x and y in the equation (1) 4=a(1)2+b(1)+c 4=a+b+c Consider the second coordinate point (2, 1) Here,x=2andy=1 Substitute the values of x and y in the equation (1) 1=a(2)2+b(2)+c 1=4a+2b+c Consider the third coordinate point (3, 4). Here, x=3andy=4. Substitute the values of x an y in the equation (1) 4=a(3)2+b(3)+c 4=9a+3b+c The obtained equations are, a+b+c=4(2) 4a+2b+c=1(3) 9a+3b+c=4(4) Step 1: Reduce the system to two equation in two variables. Eliminate the variable c from (2) and (3) by multiplying (2) by -1 and adding with (3). (2)×−1:−a−b−c=−4 (3):4a+2b+c=13a+b=−3 That is, 3a+b=−3. Now, eliminate c from (3) and (4) by multiplying (3) by -1 and adding with (4). (3)×−1:−4a−2b−c=−1 (4):9a+3b+c=45a+b=3 That is, 5a+b=3 Hence, the system of two equation in two variable is obtained as, 3a+b=−3(5) 5a+b=3.(6) Step 2: Solve the resulting system of two equation in two variables. Multiply equation (5) by -1 and then add with (6) (5)×−1:−3a−b=3 (6):5a+b=32a=6 Apply the multiplication property of equality and simplify it. 2a2=62 a=3 Step 3: Use back-substribution in one of the equations in two variables to find the value of the second vatiable. Substitute a=3 in equation (5). 3(3)+b=−3 9+b=−3 [Multiply] −9+9+b=−3−9 [Subtract 9 from both the sides of the equation] b=−12 Step 4: Back-substitute the values found for two variables into one of the original equations to find the value of the third variable. Substitute a=3andb=−12 is (2), 3−12+c=4 −9+c=4 [Combine the like terms]

The quadratic function \displaystyle{y}={a}{x}^{2}+{b}{x}+{c} whose graph passes through the points (1, 4), (2, 1) and (3, 4).

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