If x is a real number,what is the maximum val | vedclass

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(D) Let $y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$.$y = \frac{(3x^2 + 9x + 7) + 10}{3x^2 + 9x + 7} = 1 + \frac{10}{3x^2 + 9x + 7}$.For $y$ to be maxim

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$41$

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(D) Let $y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$.$y = \frac{(3x^2 + 9x + 7) + 10}{3x^2 + 9x + 7} = 1 + \frac{10}{3x^2 + 9x + 7}$.For $y$ to be maximum,the denominator $p = 3x^2 + 9x + 7$ must be minimum.The minimum value of a quadratic expression $ax^2 + bx + c$ (where $a > 0$) is given by $\frac{-D}{4a} = \frac{-(b^2 - 4ac)}{4a}$.Here,$a = 3$,$b = 9$,$c = 7$.$D = b^2 - 4ac = 9^2 - 4(3)(7) = 81 - 84 = -3$.$p_{	ext{min}} = \frac{-(-3)}{4(3)} = \frac{3}{12} = \frac{1}{4}$.Thus,$y_{	ext{max}} = 1 + \frac{10}{1/4} = 1 + 10 	imes 4 = 1 + 40 = 41$.

If $x$ is a real number,what is the maximum value of $\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$?

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