Question

se considera functia f:R->R, f(x)= $x^{2}$ - 3x + a, a ∈ R. Determinati a stiind ca valoarea minima a functiei este $\frac{3}{4}$.

ajutor urgent :)

Answer 1

$\frac{-(delta)}{4a}= \frac{3}{4} =\ \textgreater \ \frac{-(9-4*1*a)}{4}= \frac{3}{4}=\ \textgreater \ \frac{4a-9}{4}= \frac{3}{4} =\ \textgreater \ 4(4a-9)=4*3=\ \textgreater$ $16a-36=12=\ \textgreater \ 16a=48=\ \textgreater \ a=3$

Tine minte:

$\frac{-b}{2a}$ - punct de minim/maxim
$\frac{-(delta)}{4a}$ - minimul/maximul fct.

Answer 2

$a\ \textgreater \ 0, \quad deci, intradevar avem un minim al parabolei \\ Notam necunoscuta cu m, ca sa nu ne incurcam la calcule. \\ \\ V\Big( -\dfrac{b}{2a}, -\dfrac{\Delta}{4a}\Big) \rightarrow f\Big( -\dfrac{b}{2a}\Big) = -\dfrac{\Delta}{4a}= f_{min} \\ \\ Calculam pe -\dfrac{\Delta}{4a} fiindca este mai usor de calculat.$

[tex] f(x) = x^2-3x+m,\quad m\in \mathbb_{R} $\\ \\ f_{min} = -\dfrac{\Delta}{4a}\Rightarrow \dfrac{3}{4} = -\dfrac{(b)^2-4\cdot a\cdot c}{4\cdot a} \Rightarrow \dfrac{3}{4} = -\dfrac{(-3)^2-4\cdot 1\cdot m}{4\cdot 1} \Rightarrow \\ \\ \Rightarrow \dfrac{3}{4} = -\dfrac{9-4m}{4} \Rightarrow 3 = 4m-9 \Rightarrow 4m = 12 \Rightarrow m = \dfrac{12}{4} \Rightarrow \\ \\ \Rightarrow m = 3 \Rightarrow S = \Big\{ 3\Big\} [/tex]