Question

Fie familia de functii fm:R→R, fm(x)=(m-2)x²-(4m-9)x+3m-7, m≠2.
a) Sa se determine m astfel incat curba G fm intersecteaza axa OX in doua puncte situate la distanta 3.
b) Sa se arate ca parabolele G fm trec prin doua puncte fixe.

Answer 1

$\displaystyle a)~Cu~alte~cuvinte~|x_1-x_2|=3.\\ \\ Avem~\Delta=(2m-5)^2.~Deci~radacinile~ecuatiei~sunt \\ \\ \frac{4m-9 \pm \sqrt{(2m-5)^2}}{2(m-2)} = \frac{4m-9 \pm |2m-5|}{2(m-2)}. \\ \\ Una~dintre~radacini~este~\frac{4m-9+2m-5}{2(m-2)}= \frac{6m-14}{2(m-2)}= \frac{3m-7}{m-2}, \\ \\ iar~cealalta~este~ \frac{4m-9-(2m-5)}{2(m-2)}= \frac{2m-4}{2(m-2)}=1.$
$\displaystyle Deci~ \left| \frac{3m-7}{m-2}-1\right|=3 \Leftrightarrow \left| \frac{2m-5}{m-2} \right|=3. \\ \\ i)~\frac{2m-5}{m-2}=3 \Leftrightarrow 2m-5=3m-6 \Leftrightarrow m=1. \\ \\ ii)~ \frac{2m-5}{m-2}=-3 \Leftrightarrow 2m-5=-3m+6 \Leftrightarrow m= \frac{11}{5}.$

$\displaystyle b)~Pentru~a~determina~cele~doua~puncte,~vom~da~doua~valori \\ \\ particulare~pentru~m~si~vom~cauta~punctele~de~intersectie~ale \\ \\ graficelor~corespunzatoare.~Punctele~cautate~trebuie~sa~se~afle~ \\ \\ printre~ele. \\ \\ Sa~luam~m=3~ai~m=4. \\ \\ f_3(x)=x^2-3x+2 \\ \\ f_4(x)=2x^2-7x+5. \\ \\ Punctele~de~intersesctie~se~gasesc~rezolvand~ecuatia \\ \\ f_3(x)=f_4(x) \Leftrightarrow x^2-3x+2=2x^2-7x+5 \Leftrightarrow x^2-4x+3=0; \\ \\ cu~radacinile~1~si~3.$

$\displaystyle Acum~trebuie~sa~demonstram~a~f_m(1)~si~f_m(3)~sunt~constante. \\ \\Intr-adevar~f_m(1)=0~si~f_m(3)=2. \\ \\ Prin~urmare~G_{f_m}~trece~prin~punctele~(1,0)~si~(3,2)~\forall~m \in \mathbb{R}-\{2\}.$

Answer 2

$f_m(x) = (m-2)x^2-(4m-9)x+3m-7,\quad m\neq 2\\ \\ a) \quad |x_1 - x_2| = 3\\ \\ (x_1-x_2)^2 = x_1^2-2x_1x_2+x_2^2 \\ (x_1-x_2)^2 = x_1^2+x_2^2 - 2x_1x_2 \\ (x_1-x_2)^2 = (x_1+x_2)^2 - 2x_1x_2 - 2x_1x_2$

$(x_1-x_2)^2 = (x_1+x_2)^2 - 4x_1x_2\\ \sqrt{(x_1-x_2)^2} = \sqrt{(x_1+x_2)^2 - 4x_1x_2}\\ \\ \Rightarrow \boxed{|x_1-x_2| = \sqrt{(x_1+x_2)^2 - 4x_1x_2}}\rightarrow \text{formula}$

$\Rightarrow \sqrt{(x_1+x_2)^2 - 4x_1x_2} = 3\Big|^2\\ \\ (x_1+x_2)^2 - 4x_1x_2 = 9 \\ \\ \text{Aplicam relatiile lui Viete:}$

$\Big(\dfrac{4m-9}{m-2}\Big)^2 - 4\cdot\dfrac{3m-7}{m-2}= 9\Big|\cdot (m-2)^2 \\ \\ (4m-9)^2 - 4(3m-7)(m-2) = 9(m-2)^2 \\ \Big(4(m-2)-1\Big)^2-4\Big(3(m-2)-1\Big)(m-2) = 9(m-2)^2 \\ \\ m-2 = t \\ \\ (4t-1)^2-4(3t-1)t = 9t^2 \\ 16t^2-8t + 1 - 12t^2 + 4t - 9t^2 = 0 \\-5t^2 -4t+1 = 0 \\ \\ \Delta = 16 + 20 = 36 = 6^2$

$t_{1,2} = \dfrac{4\pm 6}{-10}\\ \\ t_1 = -1 \Rightarrow m-2 = -1 \Rightarrow m_1 = 1 \\ t_2 = \dfrac{1}{5} \Rightarrow m-2 = \dfrac{1}{5} \Rightarrow m_2 = \dfrac{11}{5} \\ \\ \\ \Rightarrow \boxed{m \in \left\{1,\dfrac{11}{5}\right\}}$


$b)\quad f_m(x) = (m-2)x^2-(4m-9)x+3m-7 \\ \\ \text{Consideram punctul fix M(a,b) ce apartine familiilor}\\ \text{de parabole.}$

$M(a,b) \in Gf \Rightarrow f_m(a) = b\Rightarrow\\ \\ \Rightarrow (m-2)a^2-(4m-9)a + 3m-7 = b \\ \\ ma^2-2a^2-4ma+9a+3m-7-b = 0 \\ \\ m(a^2-4a+3) +(-2a^2+9a-b-7) = 0 \\ \\ a^2-4a+3 = 0 \quad \text{si}\quad -2a^2+9a-b-7 = 0 \\ \\ (a-1)(a-3) = 0 \quad \text{si}\quad -2a^2+9a-b-7 = 0$

$\bullet ~ a = 1 \Rightarrow -2\cdot 1^2 + 9\cdot 1-b-7 = 0 \Rightarrow b = 0 \\ \\ \Rightarrow A(1,0) \\ \\ \bullet ~ a = 3 \Rightarrow -18+27-b-7 = 0 \Rightarrow b = 2 \\ \\ \Rightarrow B(3,2) \\ \\ (1,0) \text{ si } (3,2) \text{ sunt cele 2 puncte fixe.}$

Punctele fixe ale unei familii de parabole sunt acele puncte care apartin graficului pentru orice valoare a parametrului m.