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11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)f(x)=x2−4xf(x)=x^2-4xf(x)=x2−4xmenu_book•(ii)f(x)=x2−5x+6f(x)=x^2-5x+6f(x)=x2−5x+6menu_book•(iii)f(x)=−x2+2x−8f(x)=-x^2+2x-8f(x)=−x2+2x−8menu_book•(iv)f(x)=x2−4x+4f(x)=x^2-4x+4f(x)=x2−4x+4menu_book•(v)f(x)=x2+2x−8.3f(x)=x^2+2x-8.3f(x)=x2+2x−8.3menu_book•(vi)f(x)=6−x−x2f(x)=6-x-x^2f(x)=6−x−x2
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11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)f(x)=x2−4xf(x)=x^2-4xf(x)=x2−4xmenu_book•(ii)f(x)=x2−5x+6f(x)=x^2-5x+6f(x)=x2−5x+6menu_book•(iii)f(x)=−x2+2x−8f(x)=-x^2+2x-8f(x)=−x2+2x−8menu_book•(iv)f(x)=x2−4x+4f(x)=x^2-4x+4f(x)=x2−4x+4menu_book•(v)f(x)=x2+2x−8.3f(x)=x^2+2x-8.3f(x)=x2+2x−8.3menu_book•(vi)f(x)=6−x−x2f(x)=6-x-x^2f(x)=6−x−x2
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Accedmychevron_right11thchevron_rightmathchevron_rightTheory Of Quadratic Functionschevron_rightExercise 3.1

(vi)

Maximum point (−12,254)\boxed{\text{Maximum point }\left(-\frac{1}{2},\frac{25}{4}\right)}Maximum point (−21​,425​)​ Domain R,  Range (−∞,254]\boxed{\text{Domain }\mathbb{R},\;\text{Range }\left(-\infty,\frac{25}{4}\right]}Domain R,Range (−∞,425​]​
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