Question

Să se determine functia de gr. 2 f:R-->R. f(x)= x pătrat -(2m+1)x+3. m€R. al cărui grafic are abscisa vârfului egală cu 7/4

Answer 1

$f : \mathbb_{R} \rightarrow \mathbb_{R} , \quad f (x) = x^2-(2m+1)x+3, \quad m\in \mathbb{R} \\ \\ Avem: V\Big( -\dfrac{b}{2a}, -\dfrac{\Delta}{4a}\Big) \\ \\ -\dfrac{b}{2a} \rightarrow abscisa varfului \Big| \Rightarrow -\dfrac{b}{2a} = \dfrac{7}{4} \Rightarrow - \dfrac{-(2m+1)}{2\cdot 1} = \dfrac{7}{4} \Rightarrow \\ \\ \Rightarrow \dfrac{2m+1}{2} = \dfrac{7}{4} \Rightarrow 4\cdot(2m+1) = 7\cdot 2 \Rightarrow 8m+4 = 14 \Rightarrow$
$\Rightarrow 8m = 14-4 \Rightarrow 8m = 10 \Rightarrow m = \dfrac{10}{8} \Rightarrow m = \dfrac{5}{4} \\ \\ \\ \Rightarrow f(x) = x^2-\Big(2\cdot \dfrac{5}{4} +1\Big)x+3 \Rightarrow f(x) = x^2-\Big( \dfrac{5}{2} +1\Big)x+3 \Rightarrow \\ \\ \Rightarrow \boxed{f(x) = x^2- \dfrac{7}{2} x+3}$