Question

determine se existem os zeros das frações quadráticas usando a forma canônica:
A) f(x )= x²-x-2
B) f(x) =3x²+x-2
C) f(x)² = x²-2x+1
por favor me mandem o calculo!

Answer 1

forma canônica

Lembre que a formula é

y=a[(x-h)²+k      onde    h=-b/2a  (Xv)      e        k=-Δ/4a   (Yv)

A)
$y= x^{2} -x-2 \\ \\ h=- \frac{b}{2a} = \frac{-(-1)}{2} = \frac{1}{2} \\ \\ K=- \frac{(b^2-4ac)}{4a} =- \frac{(1+8)}{4} =- \frac{9}{4} \\ \\ a(x-h)^2+k= \\ \\ (x- \frac{1}{2} )^2- \frac{9}{4}= \\ \\ (x- \frac{1}{2} )^2= \frac{9}{4}$

$x- \frac{1}{2}=\pm \sqrt{ \frac{9}{4} } \\ \\ x- \frac{1}{2} =\pm \frac{3}{2} \\ \\ x'- \frac{1}{2} = \frac{3}{2} \\ \\ x'= \frac{3}{2} + \frac{1}{2} \\ \\ x'= \frac{4}{2} \\ \\ x'=2$

$x"- \frac{1}{2} =- \frac{3}{2} \\ \\ x"=- \frac{3}2} + \frac{1}{2} \\ \\ x"= \frac{-2}{2} \\ \\ x"=-1 \\ \\ S=\{-1,2\}$

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B)
$y=3 x^{2} +x-2 \\ y=a(x-h)^2+k \\ \\ h=- \frac{b}{2a} =- \frac{1}{6} \\ \\ k=- \frac{(b^2-4ac)}{4a} = -\frac{(1+24)}{12} =- \frac{25}{12} \\ \\ 3(x+ \frac{1}{6} )^2- \frac{25}{12} = \\ \\ (x+ \frac{1}{6})^2= \frac{25}{12}$

$(x+ \frac{1}{6} )^2= \frac{25}{12} . \frac{1}{3} \\ \\ (x+ \frac{1}{6} )^2= \frac{25}{36} \\ \\ (x+ \frac{1}{6} )=\pm \sqrt{ \frac{25}{36} } \\ \\ x+ \frac{1}{6} =\pm \frac{5}{6} \\ \\ x'+ \frac{1}{6} = \frac{5}{6} \\ \\ x'= \frac{5}{6} - \frac{1}{6} \\ \\ x'= \frac{4}{6} = \frac{2}{3} \\ \\ x"=- \frac{5}{6} - \frac{1}{6} \\ \\ x"=- \frac{6}{6} =-1 \\ \\ S=\{-1, \frac{2}{3} \}$

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C)
$y= x^{2} -2x+1 \\ \\ h=- \frac{b}{2a} = \frac{2}{2} =1 \\ \\ k=- \frac{(b^2-4ac)}{4a} =- \frac{(4-4)}{4} = \frac{0}{4} =0 \\ \\ y=a(x-h)^2+k \\ \\y= (x-1)^2+0 \\ \\ (x-1)^2=0 \\ \\ (x-1)=\pm \sqrt{0} \\ \\ x-1=0 \\ \\ x'=x"=1 \\ \\ S=\{1\}$