Determine the second grade function if f(-1)=1, f(0)=1, f(1)=3.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The second grade function is described by the expression:

f(x)=ax^2 + bx + c

f(-1)=1

f(-1)=a*(-1)^2 + b*(-1) + c=a-b+c

a-b+c=1

f(0)=1

f(0)=a*(0)^2 + b*(0) + c=c

c=1

f(1)=3

f(1)=a*(1)^2 + b*(1) + c=a+b+c

a+b+c=3, but c=1

a+b+1=3

a+b=2

We  have also the expression a-b+c=1 and c=1

a-b+1=1

a-b=0

a=b, but a+b=2=>a+a=2

2a=2

a=1

b=1

c=1

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = ax^2+bx+c be the 2nd degree function.

Then

f(-1) = a(-1)^2 +b(-1)+c  . Or  a-b+c = 1 ....(1)

f(0) = ao^2+b*0^2+c = 1. Or c = 1.............(2)

f(1) = a*1^2+b*1++c = 3. Or a+b+c = 3.

From (2) c = 1. So we rewrite (1) and as:

a-b+1 =1 Or a-b = 0...........(4)

a+b+1 = 3. Or a+b = 2.......'(5)

Add (4) and (5) : 2a = 2. So a =1. So from (1) a-b =0, 1-b = 0. or b=1.

Therefore a=b=c=1 .

So f(x) = ax^2+bx+c is the required second degree function.

 

 

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