Proste zadania w załączniku na dzisiaj.... Question from @Alavandares

October 2018 1 16 Report

Proste zadania w załączniku na dzisiaj

Roma

$\Delta = b^2 - 4ac \\\ \Delta = 0 \ => \ x_0 = \frac{-b}{2a} \\\ \Delta > 0 \ => \ x_1 = \frac{-b-\sqrt{\Delta}}{2a} \ i \ x_2 = \frac{-b + \sqrt{\Delta}}{2a} \\\ \Delta > 0 \ => \ funkcja \ kwadratowa \ nie \ ma \ miejsc \ zerowych$

Wykresem funkcji kwadratowej jest parabola o wierzchołku P = (p; q); gdzie

$p=\frac{-b}{2a}, \ q=\frac{-\Delta}{4a}$

Funkcję kwadratową można zapisać w postaci ogólnej kanonicznej lub iloczynowej.

$posta\'c \ og\'olna: \ f(x) = ax^2 + bx + c$

$posta\'c \ kanoniczna: \ f(x) = a \cdot(x - p)^2 + q \\\ gdzie \ p=\frac{-b}{2a}, \ q=\frac{-\Delta}{4a}$

$posta\'c \ iloczynowa: \\\\ - \ je\.zeli \ \Delta = 0, \ to \ f(x) = a \cdot(x-x_0)^2 \\\ gdzie \ x_0 \ to \ miejsce \ zerowe$

$- \ je\.zeli \ \Delta >0, \ to \ f(x) = a \cdot (x - x_1)(x - x_2) \\\ gdzie \ x_1, \ x_2 \ to \ miejsca \ zerowe$

$- \ je\.zeli \ \Delta < 0, \ to \ tr\'ojmian \ kwadratowy \ nie \ ma \ postaci \\\ iloczynowej$

------------------------------------------

Zad. 5

$y = 3x^2 - x + 9 \\\ a = - 3; \ b = - 1; \ c = 9 \\\ \Delta = (-1)^2 - 4 \cdot (- 3) \cdot 9 = 1 - 108 = - 107 < 0 \\\ Zatem \ funkcja \ kwadratowa \ nie \ ma \ postaci \ iloczynowej$

$y = - x^2 + 6x = - x \cdot (x - 6)$

Zad. 6

$x_1 = 2; \ x_2 = - 3; \ a = 4 \\\\ y = 4 \cdot (x - 2)(x + 3)$

$x_1 = 0; \ x_2 = 4; \ a = 1 \\\\ y = 1 \cdot (x - 0)(x -4) = x \cdot (x - 4)$

$x_1 = -1; \ x_2 = 0; \ a = \frac{1}{4} \\\\ y = \frac{1}{4} \cdot (x + 1)(x - 0) = \frac{1}{4} \cdot (x + 1) \cdot x = \frac{1}{4}x \cdot (x + 1)$

$x_1 = 2; \ x_2 = -3; \ a = 4\\\\ y = 4 \cdot (x - 2)(x +3)$

$y= x^2 - 4x + 7 \\\ a = 1; \ b = - 4; c = 7 \\\\ \Delta = (-4)^2 - 4 \cdot 1 \cdot 7 = 16 - 28 = - 12 \\\\ p=\frac{-(-4)}{2 \cdot 1} = \frac{4}{2} = 2\\\\ q=\frac{-(-12)}{4 \cdot 1} = \frac{12}{4} = 3\\\\ y = 1 \cdot(x - 2)^2 + 3 = (x - 2)^2 + 3$

$y= 6x^2 + 7 \\\ a = 6; \ b = 0; c = 7 \\\\ \Delta = 0^2 - 4 \cdot 6 \cdot 7 = -168 \\\\ p=\frac{0}{2 \cdot 6} = \frac{0}{12} = 0\\\\ q=\frac{-(-168)}{4 \cdot 6} = \frac{168}{24} = 7\\\\ y = 6 \cdot(x - 0)^2 + 7 = 6x^2 + 7$

Zad. 2

$y= (x- 3)^2 - 4 \\\ a = 1; \ p = 3; q = -4 \\\\ P = (3; \ - 4)$

$y= - 3 \cdot(x- 5)^2 +6 \\\ a = -3; \ p = 5; q = 6 \\\\ P = (5; \ 6)$

$y= (x- 7)^2 \\\ a = 1; \ p = 7; q = 0 \\\\ P = (7; \ 0)$

Zad. 5

$P = (2; \ 4) \\\ a = 4 \\\ Zatem: \ p = 2; q = 4 \\\\ y = 4 \cdot (x - 2)^2 + 4$

$P = (0; \ 1) \\\ a = 2 \\\ Zatem: \ p = 0; q = 1 \\\\ y = 2 \cdot (x - 0)^2 + 1 = 2x^2 + 1$

$P = (-3; \ 0) \\\ a = 1\\\ Zatem: \ p = -3; q = 0 \\\\ y = 1 \cdot(x +3)^2 + 0 = (x+ 3)^2$

Zad. 4

$y = (x-3)(x - 4) \\\ x_1 = 3; \ x_2 = 4$

$y =-3 \cdot (x - 5)(x + 6) \\\ x_1 = 5; \ x_2=-6$

$y = 2 \cdot (x + 6)x \\\ x_1 = -6; \ x_2=0$

Zad. 1

Wykresy - patrz załącznik

$\underline{y = x^2 + 1} \\\ x = - 3 \ => \ y = (-3)^2 + 1 = 9 + 1 = 10 \\\ x = - 2 \ => \ y = (-2)^2 + 1 = 4 + 1 = 5 \\\ x = - 1 \ => \ y = (-1)^2 + 1 = 1 + 1 = 2 \\\ x = 0 \ => \ y = 0^2 + 1 = 0 + 1 = 1 \\\ x = 1 \ => \ y = 1^2 + 1 = 1 + 1 = 2 \\\ x = 2 \ => \ y = 2^2 + 1 = 4 + 1 = 5$

$\underline{y = -x^2 - 1} \\\ x = - 3 \ => \ y = -(-3)^2 - 1 = -9 - 1 = -10 \\\ x = - 2 \ => \ y = -(-2)^2 - 1 = -4 - 1 = -5 \\\ x = - 1 \ => \ y = -(-1)^2 - 1 = -1 - 1 = -2 \\\ x = 0 \ => \ y = -0^2 - 1 = 0 - 1 = -1 \\\ x = 1 \ => \ y = -1^2 - 1 = -1 - 1 = -2 \\\ x = 2 \ => \ y = -2^2 - 1 = -4 - 1 = -5$