We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
176
1
avatar+283 

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!!

 Sep 18, 2018
 #1
avatar+4789 
+1

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

 

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

 

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

 

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

 

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

 

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

.
 Sep 18, 2018

31 Online Users

avatar
avatar
avatar
avatar