Answer: f(x) = 3x^2-7x+5
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Step-by-step explanation:
Assume that f(x) takes the following form
f(x) = ax^2+bx+c
Then replace each x with x+1
f(x+1) = a(x+1)^2 + b(x+1) + c
Expand out the right hand side and move terms a bit to get this:
f(x+1) = a(x+1)^2 + b(x+1) + c
f(x+1) = a(x^2+2x+1) + b(x+1) + c
f(x+1) = ax^2+2ax+a + bx+b + c
f(x+1) = (ax^2) + (2ax+bx) + (a+b+c)
f(x+1) = ax^2 + (2a+b)x + (a+b+c)
Now equate this with the expression given 3x^2-x+1
ax^2 + (2a+b)x + (a+b+c) = 3x^2 - x + 1
We see that
after equating the corresponding coefficients.
Use the value of 'a' to find b in the second equation.
2a+b = -1
2(3)+b = -1
6+b = -1
b = -1-6
b = -7
Then we can find c.
a+b+c = 1
3+(-7)+c = 1
-4+c = 1
c = 1+4
c = 5
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Summary:
We found that
This means we go from f(x) = ax^2+bx+c to f(x) = 3x^2-7x+5
To check this, simplifying f(x+1) = 3(x+1)^2-7(x+1)+5 should get you to f(x+1) = 3x^2-x+1
I'll let you do this part.