Find the equation of the quadratic function whose graph is a parabola containing the points (−2,−4), (3,1) and (2,4).

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Sadie Eaton · Accepted Answer

The quadratic function passes through the points (−2,−4), (3,1) and (2,4). Let the quadratic function be f(x)=ax2+bx+c Put (−2,−4) in f(x)=ax2+bx+c , we get −4=a(−2)2+b(−2)+c ⇒4a−2b+c=−4 →(1) Put (3,1) in f(x)=ax2+bx+c, we get 1=a(3)2+b(3)+c ⇒9a+3b+c=1 → (2) Put (2,4) in f(x)=ax2+bx+c, we get 4=a(2)2+b(2)+c ⇒4a+2b+c=4 → (3) Subtracting equation (1) from equation (3), we get ⇒(4a+2b+c)−(4a–2b+c)=4−(−4) ⇒4b=8 ⇒b=2 Now, [9 - Equation (1)] - [4 - Equation (2)] ⇒9(4a–2b+c)–4(9a+3b+c)=9(−4)–4(1) ⇒36a–18b+9c–36a–12b–4c=−36–4 ⇒−30b+5c=−40 ⇒−30(2)+5c=−40 ⇒−60+5c=−40 ⇒5c=20 ⇒c=4 Now put b=2 &amp;amp; c=4 in equation (1), we get Now put b=2 &amp;amp; c=4in equation (1)&amp;lt; we get 4a–2(2)+4=−4 ⇒4a=−4 ⇒a=−1 Now put a=−1, b=2 &amp;amp; c=4 in

Find the equation of the quadratic function whose graph is a parabola containing

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