Question

Let y
x² + 3x+1
3x+1xe R, then
x²+x+1​

Answer 1

Answer:

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Step-by-step explanation:

Let y=x2+3x+1x2+x+1∀x∈R. Find the range of y

I have reached: x2(y−1)+x(y−3)+(y−1)=0

I also know that the denominator of y=f(x) is greater than 0.

How do I continue from here? I am unable to find a suitable condition to proceed.

Edit: vertex of function in the numerator is at (−32,−54)

and of the function in the denominator is at: (−12,34)

Answer 2

Answer:

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