The range of f(x) = frac{x^2 + 34x - 71}{x^2 | vedclass

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(B) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.Rearranging the terms,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$.$x^2(y - 1) + x(2y - 34) - 7y + 71 =

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$( - \infty, 5] \cup [9, \infty)$

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(B) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.Rearranging the terms,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$.$x^2(y - 1) + x(2y - 34) - 7y + 71 = 0$.For $x$ to be a real number,the discriminant $D = B^2 - 4AC \ge 0$.$D = (2y - 34)^2 - 4(y - 1)(-7y + 71) \ge 0$.$4(y - 17)^2 - 4(-7y^2 + 71y + 7y - 71) \ge 0$.$(y^2 - 34y + 289) - (-7y^2 + 78y - 71) \ge 0$.$y^2 - 34y + 289 + 7y^2 - 78y + 71 \ge 0$.$8y^2 - 112y + 360 \ge 0$.Dividing by $8$,we get $y^2 - 14y + 45 \ge 0$.$(y - 5)(y - 9) \ge 0$.Thus,the range is $y \in ( - \infty, 5] \cup [9, \infty)$.

The range of $f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ is

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