6. Minimum value determined by the formula function f (x) = 2x ²-8x + p was 20. Value f (2) is.
7. Shape factor of the quadratic equation 4x ²-13x = -3 is.
8. Quadratic function whose graph passes through the point (-12.0) and has a turning point (-15.3) is.
9. Roots of a quadratic equation: 4x ² + px +25 = 0 are x1 and x2, if the roots of the quadratic equation x1 ² + x2 ² = 12.5 then the value of p is.
10. Equation x ²-4x +3 = 0 and x ² +4 x-21 = 0, has a root persekutuan. Akar the alliance is

$$6)\ \ \ f(x)=2x^2-8x+p\\the\ minimum\ value =20\ \ \ \Leftrightarrow\ \ \ y_{\ of\ vertex}=20\ \ \ \Leftrightarrow\ \ \ - \frac{\Delta}{2a} =20\\\\\Delta=(-8)^2-4\cdot2\cdot p=64-8p\ \ \Leftrightarrow\ \ - \frac{64-8p}{2\cdot2} =20\ \ \Leftrightarrow\ \ -16+2p=20\\\\2p=36\ \ \ \Leftrightarrow\ \ \ p=18\ \ \ \Rightarrow\ \ \ \ f(x)=2x^2-8x+18\\\\f(2)=2\cdot2^2-8\cdot2+18=2\cdot4-16+18=8+2=10$$

$$7)\ the\ shape\ factor\ of\ the\ quadratic\ equation\ 4x^2-13x = -3\\ is\ a=4\ \ \ (\ a>0\ \ \ \rightarrow\ \ \ the\ shape\ is\ \cup\ )\\\\8)\ \ \ the\ turning\ point=(-15;3)\ \ \ \Rightarrow\ \ \ f(x)=a(x+15)^2+3\\\\ the\ graph\ passes\ through\ the\ point\ (-12.0) \ \Rightarrow\ \ 0=a(-12+15)^2+3\\\\\Rightarrow\ \ \ a\cdot3^2=-3\ \ \ \Rightarrow\ \ \ a=- \frac{3}{9} =- \frac{1}{3} \ \ \ \Rightarrow\ \ \ f(x)=- \frac{1}{3}(x+15)^2+3$$

$$\Rightarrow\ \ \ f(x)=- \frac{1}{3}(x^2+30x+225)+3=- \frac{1}{3}x^2-10x-72\\\\9)\ \ \ 4x^2+px+25=0\\\\\Delta=p^2-4\cdot4\cdot25=p^2-400\\\\two\ solutions\ \ \Leftrightarrow\ \ \Delta>0\ \ \Leftrightarrow\ \ p^2-40>0\ \ \Leftrightarrow\ \ (p-20)(p+20)>0\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Leftrightarrow\ \ \ p\in(-\infty;\ -20)\ \cap\ (20;\ +\infty)\\-$$

$$the\ Vieta’s\ formulas\ to\ the\ quadratic\ equation\ ax^2+bx+c=0\\\\x_1+x_2=- \frac{b}{a} \ \ \ and\ \ \ x_1\cdot x_2= \frac{c}{a} \\-\\\\x_1+x_2=- \frac{p}{4} \ \ \ and\ \ \ x_1\cdot x_2= \frac{25}{4} \\\\x_1^2+x_2^2=x_1^2+2\cdot x_1\cdot x_2 +x_2^2-2\cdot x_1\cdot x_2 =(x_1+x_2)^2-2\cdot x_1\cdot x_2 \\\\x_1^2+x_2^2=(x_1+x_2)^2-2\cdot x_1\cdot x_2 \ \ \ \Leftrightarrow\ \ \ 12.5=(- \frac{p}{4} )^2-2\cdot \frac{25}{4} \\\\$$

$$12.5= \frac{p^2}{16} +12.5 \ \ \ \Leftrightarrow\ \ \ \frac{p^2}{16}=0 \ \ \ \Leftrightarrow\ \ \ p^2=0 \ \ \ \Leftrightarrow\ \ \ p=0\\\\\\10)\ \ \ x^2-4x+3=0\ \ \ and\ \ \ x^2+4x-21=0\\\\ x^2-4x+3=x^2+4x-21\ \ \Leftrightarrow\ \ -4x-4x=-21-3\\\\\ \ \Leftrightarrow\ \ -8x=-24\ \ \Leftrightarrow\ \ x=3$$

RELATED: