Question

Alguém pode me ajudar nessa por favor?

Answer 1

Vamos descobrir qual a equação.
$f(x)=ax^2+(1-4a)x\ \ \ \ f(6)=1\\ f(6)=a\cdot6^2+(1-4a)\cdot6\\ \\1=a\cdot36+(6-24a)\\ \\1=36a+6-24a\\ \\36a+6-24a=1\\ \\36a-24a=1-6\\ \\12a=-5\\ \\a=\frac{-5}{12}\\ \\a=-\frac{5}{12}$
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Sendo $a=- \frac{5}{12}$, a equação será:
$f(x)=ax^2+(1-4a)x\\ \\f(x)= -\frac{5}{12}x^2+\Big[1-4\cdot\Big(-\frac{5}{12} \Big)\Big]x\\ \\f(x)= -\frac{5}{12}x^2+\Big[1+\frac{20^{:4}}{12_{:4}}\Big]x\\ \\f(x)= -\frac{5}{12}x^2+\Big[1+\frac{5}{3}\Big]x\\ \\f(x)= -\frac{5}{12}x^2+\Big[ \frac{3}{3} +\frac{5}{3}\Big]x\\ \\f(x)= -\frac{5}{12}x^2+\Big[\frac{3+5}{3}\Big]x\\ \\f(x)= -\frac{5}{12}x^2+\frac{8}{3}x$
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Para determinar a altura máxima devemos calcular $y_{v}$
$a=-\frac{5}{12}\ \ \ \ b =\frac{8}{3}\ \ \ \ c =0 \\ \\\Delta=b^2-4ac\\ \\\Delta= \big(\frac{8}{3}\big)^2-4\cdot\big(-\frac{5}{12}\big)\cdot0\\ \\\Delta=\frac{64}{9}-0\\ \\\Delta=\frac{64}{9}$
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$y_v= -\frac{\Delta}{4a} \\ \\y_v=- \frac{64/9}{4 \cdot \big(-\frac{5}{12} \big)} =-\frac{64/9}{\big(-\frac{20^{:4}}{12_{:4}} \big)} =- \frac{64/9}{-5/3} =-\frac{64}{9}\cdot \big(-\frac{3}{5} \big)= \frac{192^{:3}}{45_{:3}}= \frac{64}{15}\\ \\y_v\cong4,27$