Given the Quadratic Function f(x)=ax^2+bx+c meets the following conditions f(0)=2; f(1)=-1; its graph intersects the x-axis and cut the segment with a length of 2 sqrt(2), please determine the expression of the Quadratic Function.

thank you so much in advance!!

yasbib555 Sep 18, 2018

#1**+1 **

\(f(0)=2 \Rightarrow c=2 \\ \\ \text{Then we have }a+b+2=-1 \Rightarrow b=-3-a\)

\(\text{ so we have }f(x) = a x^2 -(3+a)x + 2\)

\(\text{This has zeros at }\\ x = \dfrac{(3+a)\pm \sqrt{(3+a)^2-8a}}{2a}\)

\(\text{subtracting we have}\\ \dfrac{\sqrt{(3+a)^2-8a}}{a}=\dfrac{\sqrt{9-2a+a^2}}{a} = 2\sqrt{2}\)

\(\dfrac{9-2a+a^2}{a^2}= 8 \\ \\ 7a^2+2a-9=0 \\ \\ a=\dfrac{-2\pm \sqrt{4+252}}{14}=\dfrac{-2\pm 16}{14} = 1,~-\dfrac 9 7\)

\(f(x) = x^2 -4x+2 \text{ or }-\dfrac 9 7 x^2 - \dfrac{12}{7}+2\)

.Rom Sep 18, 2018