How do you find the range of $f (x) = x^{2} - x - 2$ $f(x)=x^2-x-2$ ?

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How do you find the range of $f (x) = x^{2} - x - 2$ $f(x)=x^2-x-2$ ?

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Answer 1 · 2015-09-18T17:10:08

The range is $[- \frac{9}{4}, + \infty)$ $[-9/4,+oo)$

Explanation:

The domain is $R$ $R$ and we have that

$y = (x^{2} - x + \frac{1}{4}) - 2 - \frac{1}{4} \Rightarrow y = {(x - \frac{1}{2})}^{2} - \frac{9}{4} \Rightarrow y + \frac{9}{4} = {(x - \frac{1}{2})}^{2} \Rightarrow y + \frac{9}{4} \geq 0 \Rightarrow y \geq - \frac{9}{4}$ $y=(x^2-x+1/4)-2-1/4=>y=(x-1/2)^2-9/4=>y+9/4=(x-1/2)^2=> y+9/4>=0=>y>=-9/4$

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How do you find the range of $f (x) = x^{2} - x - 2$ $f(x)=x^2-x-2$ ?

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