Question

Varful parabolei.. Valoarea minima ... punctul de minim... f:R->R,f(x)=5x2-10x+16​

Answer 1

Răspuns:

xm=-b/2a ;   ym=-Δ/4a ;   xm=10/10/=-1 ;                                             Δ=b²-4ac=100-4 x 5 x 16=100-320=-220   ;  ym=   220/20=11 ;

Explicație pas cu pas:

Answer 2

$f:\mathbb{R}\to \mathbb{R},\quad f(x) = 5x^2-10x+16\\ \\\\\left.\begin{aligned} f(x) &= 5x^2-10x+16 \\ &= 5x^2-10x+5+11 \\ &= 5(x^2-2x+1)+11 \\ &= \underset{\geq 0}{\underbrace{5(x-1)^2}}+11\end{aligned}\right|\,\Rightarrow \,f_{\text{min}} = 11$