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Question Number 187651 by Mastermind last updated on 19/Feb/23

$$\mathrm{Find}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{this}\mathrm{function}$$$${\mathrm{x}}^2-13\mathrm{x}+36=0$$$$$$$$$$$$\mathrm{Help}!$$

Answered by Ar Brandon last updated on 19/Feb/23

$$\mathrm{Let}\mathrm{y}=x^2-13x+36$$$$\Rightarrow x^2-13x+36-\mathrm{y}=0$$$$\Rightarrow x=\frac{13\pm \sqrt{169-4(36-\mathrm{y})}}{2}$$$$=\frac{13\pm \sqrt{25+4\mathrm{y}}}{2}$$$$25+4\mathrm{y}⩾0\Rightarrow \mathrm{y}⩾-\frac{25}{4}$$$$\mathrm{Hence}\mathrm{range}f\left(x\right)⩾-\frac{25}{4}$$

Commented by Mastermind last updated on 20/Feb/23

$$\mathrm{Thank}\mathrm{you}\mathrm{BOSS},{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{what}\mathrm{i}\mathrm{also}$$$$\mathrm{got}\mathrm{but}\mathrm{with}\mathrm{diff}.\mathrm{method}.$$

Answered by mr W last updated on 20/Feb/23

$$x^2-13x+36$$$$={(x-\frac{13}{2})}^2+36-{\left(\frac{13}{2}\right)}^2$$$$⩾36-{\left(\frac{13}{2}\right)}^2=-\frac{25}{4}$$$$\Rightarrow rangeis[-\frac{25}{4},+\mathrm{\infty })$$

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