Bardzo wazne naprawdę pomocy daje naj!!!!!!!!!!.... Question from @oliviamalys97

oliviamalys97 @oliviamalys97

June 2022 1 11 Report

Bardzo wazne naprawdę pomocy daje naj!!!!!!!!!!

kontonawszystko

Odpowiedź:

zad. 1

$y = 2 {x}^{2} + 4x - 21 \\ a = 2 \\ b = 4 \\ c = - 21 \\$

a) postać kanoniczna

$f(x) = a(x - p {)}^{2} + q$

p i q to współrzędne wierzchołka

$p = - \frac{b}{2a} = - \frac{4}{2 \times 2} = - \frac{4}{4} = - 1 \\$

$∆ = {b}^{2} - 4ac = {4}^{2} - 4 \times 2 \times ( - 21) = 16 + 168 = 184 \\ q = - \frac{∆}{4a} = - \frac{184}{4 \times 2} = - \frac{184}{8} = - 23$

$f(x) = 2(x + 1 {)}^{2} - 23$

$W(-1;-23)$

oś symetrii

$x = - \frac{b}{2a} = - \frac{4}{2 \times 2} = - \frac{ 4}{4} = - 1$

monotoniczność funkcji

funkcja maleje (-∞ ; -1>

funkcja rośnie <-1;∞)

zad. 2

$a)3 {x}^{2} - 27 = 0 \\ a = 3 \\ b = 0 \\ c = - 27 \\ ∆ = {b}^{2} - 4ac = 0 - 4 \times 3 \times ( - 27) = 0 + 324 = 324 \\ \sqrt{∆} = \sqrt{324} = 18 \\ x = \frac{ - b - \sqrt{∆} }{2a} = \frac{ 0 - 18}{2 \times 3} = \frac{ - 18}{6} = - 3 \\ x = \frac{ - b + \sqrt{∆} }{2a} = \frac{18}{6} = 3$

$b)2 {x}^{2} - 3x - 1 = 0 \\ a = 2 \\ b = - 3 \\ c = - 1 \\ ∆ = {b}^{2} - 4ac = ( - {3)}^{2} - 4 \times 2 \times ( - 1) = 9 + 8 = 17 \\ \sqrt{∆} = \sqrt{17} \\ x = \frac{ - ( - 3) - \sqrt{17} }{2 \times 2} = \frac{3 - \sqrt{17} }{4} \\ x = \frac{ - ( - 3) + \sqrt{17} }{2 \times 2} = \frac{3 + \sqrt{17} }{4}$

$c)9 {x}^{2} - x = 0 \\ a = 9 \\ b = - 1 \\ c = 0 \\ ∆ = ( - 1 {)}^{2} - 4 \times 9 \times 0 = 1 \\ \sqrt{∆} = \sqrt{1} = 1 \\ x = \frac{ - ( - 1) - 1}{2 \times 9} = \frac{1 - 1}{18} = \frac{0}{18} = 0 \\ x = \frac{ - ( - 1) + 1}{18} = \frac{2}{18} = \frac{1}{9}$

$d)2 {x}^{2} + 5x - 3 > 0 \\ a = 2 \\ b = 5 \\ c = - 3 \\ ∆ = {5}^{2} - 4 \times 2 \times ( - 3) = 25 + 24 = 49 \\ \sqrt{∆} = \sqrt{49} = 7 \\ x = \frac{ - 5 - 7}{4} = \frac{ - 12}{4} = - 3 \\ x = \frac{ - 5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \\ x \: e \: (-∞;-3) \: u \: ( \frac{1}{2} ;∞)$

zad. 3

$y = 2 {x}^{2} - x + 1 \\ p = - \frac{b}{2a} = - \frac{( - 1)}{2 \times 2} = \frac{1}{4} \\ p \: \: e \: \: < - 2 ;2 > \\ f( - 2) = 2 \times ( - 2 {)}^{2} - ( - 2) + 1 = 8 + 2 + 1 = 11 \\ f(2) = 2 \times {2}^{2} - 2 + 1 = 8 - 2 + 1 = 7 \\ f( \frac{1}{4}) = 2 \times ( \frac{1}{4} {)}^{2} - \frac{1}{4} + 1 = 2 \times \frac{1}{8} - \frac{1}{4} + 1 = 1$